The existence of atoms has been proposed since the time of early Indian and Greek philosophers (400 B.C.) who were of the view that atoms are the fundamental building blocks of matter.

→The word 'atom' has been derived from the Greek word 'a-tomio' which means 'uncut-able' or 'non-divisible'. ↔

The atomic theory of matter was first proposed on a firm scientific basis by John Dalton, a British school teacher in 1808. His theory, called Dalton's atomic theory, regarded the atom as the ultimate particle of matter (Unit 1).
Dalton's atomic theory was able to explain the law of conservation of mass, law of constant composition and law of multiple proportion very successfully.
⇉ Postulates Dalton's atomic theory -
(i) All matter is made of very tiny particles called atoms.
(ii) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.
(iii) Atoms of a given element are identical in mass and chemical properties.
(iv) Atoms of different elements have different masses and chemical properties.
(v) Atoms combine in the ratio of small whole numbers to form compounds.
(vi) The relative number and kinds of atoms are constant in a given compound.
Limitations of Dalton's atomic theory :
Atoms of the same or different types have a strong tendency to combine together to form a new group of atoms.
For example, hydrogen, nitrogen, oxygen gases exist in nature as group of two atoms. This indicates that the smallest unit capable of independent existence is not an atom, but a group of atoms.
With the discovery of sub-atomic particles, e.g., electrons, neutrons and protons, the atom can no longer be considered indivisible. However, it failed to explain the results of many experiments, for example, it was known that substances like glass or ebonite when rubbed with silk or fur get electrically charged.
⇔

Discovery of sub-atomic Particles

"Like charges repel each other and unlike charges attract each other".
In mid 1850s many scientists mainly Faraday began to study electrical discharge in partially evacuated tubes, known as cathode ray discharge tubes.
Cathode Ray Tube
⇉The cathode ray tube (CRT), invented in 1897 by the German physicist Karl Ferdinand Braun, is an evacuated glass envelope containing an electron gun a source of electrons and a fluorescent light, usually with internal or external means to accelerate and redirect the electrons.
Light is produced when electrons hit a fluorescent tube.
The electron beam is deflected and modulated in a manner that allows an image to appear on the projector. The picture may reflect electrical wave forms (oscilloscope), photographs (television, computer monitor), echoes of radar-detected aircraft, and so on.
The single electron beam can be processed to show movable images in natural colours.

The experiment Cathode Ray Tube (CRT) conducted by J. J. Thomson, is one of the most well-known physical experiments that led to electron discovery.
In addition, the experiment could describe characteristic properties, in essence, its affinity to positive charge, and its charge to mass ratio. This paper describes how J is simulated. J. Thomson experimented with Cathode Ray Tube.
The major contribution of this work is the new approach to modelling this experiment, using the equations of physical laws to describe the electrons’ motion with a great deal of accuracy and precision. The user can manipulate and record the movement of the electrons by assigning various values to the experimental parameters.
The Cathode ray experiment was a result of English physicists named J. J. Thomson experimenting with cathode ray tubes.
During his experiment he discovered electrons and it is one of the most important discoveries in the history of physics. He was even awarded a Nobel Prize in physics for this discovery and his work on the conduction of electricity in gases. ⇔
→ J. J. Thomson took a tube made of glass containing two pieces of metal as an electrode. The air inside the chamber was subjected to high voltage and electricity flowing through the air from the negative electrode to the positive electrode. Voltage 5000-10000V .
Cathode Ray Tube
J. J. Thomson designed a glass tube that was partly evacuated, i.e. all the air had been drained out of the building. He then applied a high electric voltage at either end of the tube between two electrodes. He observed a particle stream (ray) coming out of the negatively charged electrode (cathode) to the positively charged electrode (anode). This ray is called a cathode ray and is called a cathode ray tube for the entire construction.
(i) When the discharge tube contains any gas at normal pressure, nothing is observed even by applying high voltage (5,000 -10,000 V) between the electrodes. The gas remains non-conducting.
(ii) The pressure of the gas inside the tube is decreased by pumping out the gas with the help of a vacuum pump. When the pressure of the gas is decreased to about l0^-2 atm (about 1 mm to 10 mm of Hg), the gas becomes conducting and light is emitted by the residual gas in the tube. The colour of the light depends upon the nature of the gas taken.
(iii) When the pressure of the gas in the discharge tube is further reduced, the glow becomes weak. At about 10^-4 atm pressure (about 0.01 mm of Hg), the glow in the tube stops but the gas continues to conduct electricity. Moreover, the glass tube at the anode end begins to glow (fluoresces) with a faint greenish light.
↔ J. J. Thomson Experiment – The Discovery of Electron
The Cathode ray experiment was a result of English physicists named J. J. Thomson experimenting with cathode ray tubes. During his experiment he discovered electrons and it is one of the most important discoveries in the history of physics. He was even awarded a Nobel Prize in physics for this discovery and his work on the conduction of electricity in gases.
→Procedure of the Experiment 1. Apparatus is set up by providing a high voltage source and evacuating the air to maintain the low pressure inside the tube.
2. High voltage is passed to the two metal pieces to ionize the air and make it a conductor of electricity.
3. The electricity starts flowing as the circuit was complete.
4. To identify the constituents of the ray produced by applying a high voltage to the tube, the dipole was set up as an add-on in the experiment.
5. The positive pole and negative pole were kept on either side of the discharge ray.
6. When the dipoles were applied, the ray was repelled by the negative pole and it was deflected towards the positive pole.
This was further confirmed by placing the phosphorescent substance at the end of the discharge ray. It glows when hit by a discharge ray.
By carefully observing the places where fluorescence was observed, it was noted that the deflections were on the positive side. So the constituents of the discharge tube were negatively charged.
Conclusion
After completing the experiment J.J. Thomson concluded that rays were and are basically negatively charged particles present or moving around in a set of a positive charge. This theory further helped physicists in understanding the structure of an atom. And the significant observation that he made was that the characteristics of cathode rays or electrons did not depend on the material of electrodes or the nature of the gas present in the cathode ray tube. All in all, from all this we learn that the electrons are in fact the basic constituent of all the atoms.
Most of the mass of the atom and all of its positive charge are contained in a small nucleus, called a nucleus. The particle which is positively charged is called a proton. The greater part of an atom’s volume is empty space.
The number of electrons that are dispersed outside the nucleus is the same as the number of positively charged protons in the nucleus. This explains the electrical neutrality of an atom as a whole. →
⇉ Uses of Cathode Ray Tube 1. Used as a most popular television (TV) display.
2. X-rays are produced when fast-moving cathode rays are stopped suddenly.
3. The screen of a cathode ray oscilloscope, and the monitor of a computer, are coated with fluorescent substances. When the cathode rays fall off the screen pictures are visible on the screen.
Properties / Characteristics of cathode rays-
a. Cathode rays consist of negatively charged electrons.
b. Cathode rays themselves are not visible but their behavior can be observed with help of fluorescent or phosphorescent materials.
c. In absence of electrical or magnetic field cathode rays travel in straight lines.
d. In presence of electrical or magnetic field, behaviour of cathode rays is similar to that shown by electrons.
e. The characteristics of the cathode rays do not depend upon the material of the electrodes and the nature of the gas present in the cathode ray tube. Charge to mass (e/m) ratio of an electron was determined by Thomson. ⇔

The charge to mass ratio of an electron as 1.758820 x 10¹¹ C kg^-1 .
Charge on an electron was determined by R A Millikan by using an oil drop experiment. The value of the charge on an electron is -1.6 x 10^-19C.
The mass on an electron was determined by combining the results of Thomson’s experiment and Millikan’s oil drop experiment.
The mass of an electron was determined to be 9.1094 x10^-31kg.

⇉ While carrying out the discharge tube experiment, Thomson observed that the particles of the cathode deviate from their path. He noticed the amount of deviation in the presence of an electrical or magnetic field depends on various related parameters. They are a. Particles with a greater magnitude of the charge experienced greater interaction with the electric or magnetic field. Thus, they exhibited greater deflection. b. Lighter particles experienced greater deflection. Thus, deflection is inversely proportional to the mass of the particle. c. Deflection of particle from their path is directly proportional to the strength of the electrical and the magnetic field present.
⇔ Millikan’s Oil Drop Method In this method, oil droplets in the form of mist, produced by the atomiser, were allowed to enter through a tiny hole in the upper plate of electrical condenser. The downward motion of these droplets was viewed through the telescope, equipped with a micrometer eye piece. ⇉By measuring the rate of fall of these droplets, Millikan was able to measure the mass of oil droplets. ⇔The air inside the chamber was ionized by passing a beam of X-rays through it. The electrical charge on these oil droplets was acquired by collisions with gaseous ions. The fall of these charged oil droplets can be retarded, accelerated or made stationary depending upon the charge on the droplets and the polarity and strength of the voltage applied to the plate. By carefully measuring the effects of electrical field strength on the motion of oil droplets, Millikan concluded that the magnitude of electrical charge, q, on the droplets is always an integral multiple of the electrical charge, e, that is, q = n e, where n = 1, 2, 3... . The mass on an electron was determined by combining the results of Thomson’s experiment and Millikan’s oil drop experiment. The mass of an electron was determined to be 9.1094 x10^-31kg.

Discovery of Proton

The presence of negatively charged electrons in an atom suggests that there must be some positively charged particles because the atom on the whole is electrically neutral.
→ In l836, E. Goldstein discovered that in addition to cathode rays, a new kind of rays are also found streaming behind the cathode in discharge tube experiments (Fig.). These rays travelled in opposite direction to the cathode rays.
These rays are also deflected by the magnetic and electric fields like cathode rays" But the deflection of anode rays is in the opposite direction to that of cathode rays.
↔For example, these rays were attracted towards the negative plate in the electric field as shown in Fig. This means that these rays consist of positively charged particles and were also named positive rays or canal rays or anode rays.
Discovery of protons and canal rays: Modified cathode ray tube experiment was carried out which led to the discovery of protons.

Discovery of Electron
In 1830, Michael Faraday showed that if electricity is passed through a solution of an electrolyte, chemical reactions occurred at the electrodes, which resulted in the liberation and deposition of matter at the electrodes.

A cathode ray tube is made of glass containing two thin pieces of metal, called electrodes, sealed in it. The electrical discharge through the gases could be observed only at very low pressures and at very high voltages.
The pressure of different gases could be adjusted by evacuation of the glass tubes.
When sufficiently high voltage is applied across the electrodes, current starts flowing through a stream of particles moving in the tube from the negative electrode (cathode) to the positive electrode (anode).
These were called cathode rays or cathode ray particles.
The flow of current from cathode to anode was further checked by making a hole in the anode and coating the tube behind anode with phosphorescent material zinc sulphide.
When these rays, after passing through anode, strike the zinc sulphide coating, a bright spot is developed on the coating
Ref fig
→ The experiment in its simplest form consists of a cylindrical hard glass tube (about 50 cm long) closed it both ends [Fig.]. It is known as discharge tube or Crookes tube.It is frtted with two metallic electrodes. The tube is connected to a side tube, through which it can be evacuated to any desired pressure with the help of a vacuum pump. The discharge tube is filled with the gas under study and the two electrodes are connected to a source of high voltage. ↔

The results of these experiments are summarised below.
⇉
(i) The cathode rays start from cathode and move towards the anode.
(ii) These rays themselves are not visible but their behaviour can be observed with the help of certain kind of materials (fluorescent or phosphorescent) which glow when hit by them.
Television picture tubes are cathode ray tubes and television pictures result due to fluorescence on the television screen coated with certain fluorescent or phosphorescent materials.
(iii) In the absence of electrical or magnetic field, these rays travel in straight lines.
(iv) In the presence of electrical or magnetic field, the behaviour of cathode rays are similar to that expected from negatively charged particles, suggesting that the cathode rays consist of negatively charged particles, called electrons.
(v) The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube. Thus, we can conclude that electrons are basic constituent of all the atoms. ⇔

Charge to Mass Ratio of Electron

In 1897, British physicist J.J. Thomson measured the ratio of electrical charge (e) to the mass of electron (me ) by using cathode ray tube and applying electrical and magnetic field perpendicular to each other as well as to the path of electrons (Fig.). When only electric field is applied, the electrons deviate from their path and hit the cathode ray tube at point A (Fig.). Similarly when only magnetic field is applied, electron strikes the cathode ray tube at point C. By carefully balancing the electrical and magnetic field strength, it is possible to bring back the electron to the path which is followed in the absence of electric or magnetic field and they hit the screen at point B.

While carrying out the discharge tube experiment, Thomson observed that the particles of the cathode deviate from their path. He noticed the amount of deviation in the presence of an electrical or magnetic field depends on various related parameters. They are -

a. Particles with a greater magnitude of the charge experienced greater interaction with the electric or magnetic field. Thus, they exhibited greater deflection.
b. Lighter particles experienced greater deflection. Thus, deflection is inversely proportional to the mass of the particle.
c. Deflection of particle from their path is directly proportional to the strength of the electrical and the magnetic field present.

→ Charge to Mass Ratio of Electron The charge to mass ratio of the electron is given by:
e/m = 1.758820× 10¹¹ C kg^-1. Where,
m = mass of an electron in kg = 9.10938356 × 10 ^-31 kilograms.
e = magnitude of the charge of an electron in coulombs = 1.602 x 10^-19coulombs. →
Where me is the mass of the electron in kg and e is the magnitude of the charge on the electron in coulomb (C). Since electrons are negatively charged, the charge on electron is -e.

Charge on the Electron R.A. Millikan (1868-1953) devised a method known as oil drop experiment (1906-14), to determine the charge on the electrons. He found the charge on the electron to be – 1.6 × 10^–19C. The present accepted value of electrical charge is – 1.602176 × 10^–19 C. The mass of the electron (me) was determined by combining these results with Thomson’s value of e/me ratio. _{M = 9.1094×10^-31 kg}

Discovery of Protons and Neutrons

Electrical discharge carried out in the modified cathode ray tube led to the discovery of canal rays carrying positively charged particles. → Properties / Characteristics of cathode rays-
a. Cathode rays consist of negatively charged electrons.
b. Cathode rays themselves are not visible but their behavior can be observed with help of fluorescent or phosphorescent materials.
c. In absence of electrical or magnetic field cathode rays travel in straight lines.
d. In presence of electrical or magnetic field, behaviour of cathode rays is similar to that shown by electrons.
e. The characteristics of the cathode rays do not depend upon the material of the electrodes and the nature of the gas present in the cathode ray tube. Charge to mass (e/m) ratio of an electron was determined by Thomson.
↔ Discovery of Proton

→ Characteristics of positively charged particles (canal rays) -
(i) Unlike cathode rays, mass of positively charged particles depends upon the nature of gas present in the cathode ray tube. These are simply the positively charged gaseous ions.
(ii) The charge to mass ratio of the particles depends on the gas from which these originate.
(iii) Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.
(iv) The behaviour of these particles in the magnetic or electrical field is opposite to that observed for electron or cathode rays.
↔ Chadwick (1932) by bombarding a thin sheet of beryllium by αparticles. When electrically neutral particles having a mass slightly greater than that of protons were emitted. He named these particles as neutrons. (031102IIb)

⇒ What are the sub atomic particles of an atom?
There are three subatomic particles: protons, neutrons and electrons. Two of the subatomic particles have electrical charges: protons have a positive charge while electrons have a negative charge.
Q: What is a quark? A quark is a subatomic particle found inside the protons and neutrons.
Write important characteristics of protons
a. Protons are positively charged subatomic particles. b. to Ernest Rutherford discovered protons c. The mass of a proton is 1.676 x 10-27 grams. d. The charge of a proton is +1.602 x 10-19 Coulombs.
Write important characteristics of Electrons
a. Electrons are negatively charged subatomic particles. b. An equal number of electrons and protons are found in the atoms of all elements. c. J. J. Thomson is credited with the discovery of electrons. He also accurately calculated the mass and the charge of an electron. d. The mass of an electron is negligible as compared to the mass of a proton, equal to (1/1837) times the mass of a proton. e. The charge of an electron is equal to -1.602 x 10-19 Coulombs. Click here to learn more about electrons.
Write important characteristics of Neutrons
a. Neutrons are named for their neutral nature as they do not carry any charge. Neutrons are neutrally charged subatomic particles.
b. The neutron was discovered by James Chadwick in 1932.
c. The masses of two different isotopes of an element vary due to the difference in the number of neutrons in their respective nuclei.
d. The mass of a neutron is 1.676 x10-27 grams.
What are nucleons?
Protons and neutrons make up the nucleus. Hence, these particles together are also called nucleons.
What were the major problems before the scientists after the discovery of sub-atomic particles ?
The major problems were-
a. to account for the stability of atom
b. to compare the behaviour of elements in terms of both physical and chemical properties
c. to explain the formation of different kinds of molecules by the combination of different atoms and
d. to understand the origin and nature of the characteristics of electromagnetic radiation absorbed or emitted by atoms.

⇉Models of atom
Dalton's atomic model: Dalton's Billiard Ball (Solid Sphere) Model.
J.J. Thomson's model: Plum Pudding model.
Ernest Rutherford's model: Nuclear model.
Niels Bohr's model: Planetary model.
Erwin Schrodinger's model: Electron Cloud Model/Quantum Model.
Wave mechanical model.
1. John Dalton’s atomic model: Dalton’s Billiard Ball (Solid Sphere) Model
Dalton’s atomic theory was able to explain the law of conservation of mass, law of constant composition, and law of multiple proportions very successfully."
2. J.J. Thomson’s model: Plum Pudding model
“J. J. Thomson, in 1898, proposed that an atom possesses a spherical shape in which the positive charge is uniformly distributed.”
3. Ernest Rutherford’s model: Nuclear model
“Rutherford and his students (Hans Geiger and Ernest Marsden) bombarded very thin gold foil with α–particles. Rutherford’s famous α–particle scattering experiment.”
4. Niels Bohr’s model: Planetary model
"He worked on the dual character of electromagnetic radiation."
5. Erwin Schrodinger’s model: Electron Cloud Model/Quantum Model
“Quantum mechanics was developed independently in 1926 by Werner Heisenberg and Erwin Schrödinger.”
6. Wave mechanical model
The model, which is the basis of the modern understanding of the atom, is known as the quantum mechanical or wave mechanical model. The fact that there are only certain allowable states or energies that an electron in an atom can have is similar to a standing wave.
“The wave mechanical model proposed that the electrons act like particles as well as waves of energy.”
Wave mechanical model was developed by Schrödinger. This model is based on the particle and wave nature of electron. The motion of electron around nucleus is circular motion and may be considered to be analogous to the standing waves, the waves which are generated by plucking the stretched string.
⇔

Thomson’s Model of atom
J. J. Thomson, in 1898, proposed that an atom possesses a spherical shape (radius approximately 10^{-10 m} ) in which the positive charge is uniformly distributed.
In 1904 he proposed that an atom consists of a uniform sphere in which positive charge is uniformly distributed. The electrons are embedded into it in such a way as to give the most stable electrostatic arrangement (Fig.) The radius of the sphere is of the order of 10^-10 m, which is equal to the size of the atom.
This model was much like pudding or cake (of positive charge) with raisins (electrons) embedded into it. Therefore, this model is also known as raisin pudding model. An important feature of this model is that the mass of the atom is considered to be evenly spread over the atom.
This model explains some of the known properties and electrical neutrality of atom.
Limitations of Thomson's model
If the positive charge of the atom was uniformly distributed, then positively charged α -particies with a considerable mass (4 a.m.u.) would pass through weak electric field largely undeflected or slightly deflected. The expected deflections on the basis of Thomson's model are shown. (031101Ib). However, he noticed that some of the α -particles experienced strong deflections. Even some particles returned back from the foil. Thus, Thomson's model could not provide answers for these observations and therefore, was discarded.
Thomson was awarded Nobel Prize for physics in 1906, for his theoretical and experimental investigations on the conduction of electricity by gases.
radioactivity and the radioactive elements. (031102IIa)
Wilhalm Röentgen (1845-1923) in 1895 showed that when electrons strike a material in the cathode ray tubes, produce rays which can cause fluorescence in the fluorescent materials placed outside the cathode ray tubes. Since Röentgen did not know the nature of the radiation, he named them X-rays and the name is still carried on.
It was noticed that X-rays are produced effectively when electrons strike the dense metal anode, called targets.
These are not deflected by the electric and magnetic fields and have a very high penetrating power through the matter and that is the reason that these rays are used to study the interior of the objects.
These rays are of very short wavelengths (∼0.1 nm) and possess electro-magnetic character.

Henri Becqueral (1852-1908) observed that there are certain elements which emit radiation on their own and named this phenomenon as radioactivity and the elements known as radioactive elements. This field was developed by Marie Curie, Piere Curie, Rutherford and Fredrick Soddy.

It was observed that three kinds of rays i.e., α -, β- and γ -rays are emitted. Rutherford found that α -rays consists of high energy particles carrying two units of positive charge and four unit of atomic mass. He concluded that α - particles are helium nuclei as when α - particles combined with two electrons yielded helium gas. β-rays are negatively charged particles similar to electrons. The γ -rays are high energy radiations like X-rays, are neutral in nature and do not consist of particles. Alpha particles are large and less penetrating power compared to beta particles Beta particles are smaller than alpha particles but have high penetrating power than beta particles. Gama particles has highest penetrating power.

⇒ Half life periods of few elements- there are 37 known elements with no stable isotopes and these are known as the "radioactive elements.
Carbon

Write note on alpha, beta and gamma perticles There exist three major types of radiations emitted by the radioactive particles - Alpha, Beta and Gama
These radiations are released from the nucleus of an atom. (031102Ic)

Rutherford's scattering experiments.
In order to understand the arrangement of electrons and protons in an atom, Rutherford and his students (Hans Geiger and Ernest Masden) in 1909 performed a series of experiments known as Rutherford's scattering experiments. They bombarded a target of atoms by subatomic projectiles.
These projectiles called alpha (α) particles, were obtained from a radioactive substance.
α particles are high energy positively charged helium ions having charge +2 and mass 4 u. They bombarded alpha (α) particles emitted from a radioactive substance on a piece of thin foil of gold or some other heavy metals. In this experiment, a piece of radioactive substance (radium) is placed in a lead block. The block is constructed in such a way with slits that only a narrow beam of α -particles could escape. The beam of high energy α particles was directed at a thin gold foil (thickness about 100 nm). In order to detect the α -particles after scattering, a movable circular screen coated with zinc sulphide is placed around the gold foil. When α -particles strike thin zinc sulphide screen, these produce flashes of light or scintillations which can be detected.
By examining different portions of the screen, it was possible to determine the proportions of the α particles which got deflected through various angles. The following observations were made from these experiments:
(i) Most of the α particles (nearly 99.7%) passed through the gold foil undeflected. (ii) A small fraction of α particles got deflected through small angles.
(iii) Very few (about one in 20,000) did not pass through the foil at all but suffered large deflections (more than 90⁰) or even came back suffering a deflection of 180⁰.
The radius of the sphere is of the order of 10 ^-10 m, which is equal to the size of the atom.

Rutherford's Nuclear Model of Atom
(i) In an atom, the entire mass and the positive charge is concentrated in a very small region at the centre known as nucleus.
(ii) The positive charge of the nucleus is due to protons.
(iii) The mass of the nucleus is due to protons and neutrons. The neutrons are neutral particles This neutral particle was discovered later on by Chadwick in 1932.
(iv) The nucleus is surrounded by negatively charged electrons which balance the positive charge on the nucleus. Thus, the atom is electrically neutral.
(v) The electrons are not stationary but are revolving around the nucleus at very high speeds like planets revolving around the sun.
(vi) The electrons and the nucleus are held together by electrostatic forces of attraction.
(vii) Most of the space in an atom between the nucleus and the revolving electrons is empty.
Protons and neutrons present in the nucleus are collectively called nucleons.

Drawbacks of Rutherford’s model of atom :
a. According to Rutherford’s model of atom, electrons which are negatively charged particles revolve around the nucleus in fixed orbits. Thus, the electrons undergo acceleration. According to electromagnetic theory of Maxwell, a charged particle undergoing acceleration should emit electromagnetic radiation. Thus, an electron in an orbit should emit radiation. Thus, the orbit should shrink. But this does not happen.
b. The model does not give any information about how electrons are distributed around nucleus and what are energies of these electrons.

Atomic Number, Atomic Mass, Isotope, Isobar and Isotone

Atoms (except hydrogen) contain mainly three fundamental particles proton, electron and neutron. Since all elements and compounds have mass; atom also has mass. The total mass of an atom is concentrated in a nucleus which has very small volume (compared to the total volume of the atom).
Electrons are arranged in the outer part of nucleus. The positive charge of nucleus is due to protons (which are positively charged) because neutrons are electrically neutral. The charge on the proton is equal but opposite to that of electron.
The number of protons present in the nucleus of an atom is called atomic number (Z) e.g. Number of protons in the nucleus of the very first element of periodic table hydrogen atom is one, and therefore atomic number of hydrogen element is one. As atom is electrically neutral, number of protons and number of electrons in an atom are always equal.
Atomic number (Z) = number of protons in the nucleus of atom OR number of electrons in a neutral atom.
Mass of nucleus is due to proton and neutron. (as mass of the electron is negligible). Protons and neutrons which are present in the nucleus are collectively known as 'nucleons.
atomic mass number (A) of an element is equal to the number of nucleons. Atomic mass = protons(Z) + neutrons (n)
e.g. Atomic mass of sodium is 23. Its atomic number is 11, therefore number of neutrons = 23 - 11 = 12
Atomic mass and atomic number of any clement (X) are indicated by accepted symbols.
Accordingly element is represented by X with superscript on left hand side as the atomic mass number (A) and subscript (Z) on left hand side as the atomic number.

Define the following - Atomic Number, Atomic Mass, Isotope, Isobar and Isotone The atomic number of an atom is equal to the number of protons in the nucleus of an atom or the number of electrons in an electrically neutral atom.
Atomic Mass can be defined as 1/12 of the mass of a carbon-12 atom in its ground state. The mass of an atom can be accounted for by the sum of the mass of protons and neutrons which is almost equal to the atomic mass. This small change is due to the binding energy mass loss.
1 amu = 1.66 ×10−24 g
Why is carbon 12 used as atomic mass reference? Since the chemical atomic weights of carbon 12 are almost equal to those of the natural mix of oxygen, it was selected as the standard. Since no other nuclide has an identical whole-number mass on this scale except carbon-12. Six protons, six neutrons, and six electrons make up carbon-12.
Why was oxygen 16 replaced by carbon 12 as the reference for measuring atomic masses? Carbon can form very large bonds whereas oxygen cannot since oxygen has 2 valence electrons so it can only react with elements which can give 2 electrons to oxygen. On the other hand, carbon has 4 valence electrons which means Carbon can combine with many more elements by forming covalent bonds.
Isotopes are the atoms of an element which have the same number of protons but different numbers of neutrons, e.g. Isotopes of Hydrogen are Protium (Z=1, A=1), Deuterium (Z=1, A=2), Tritium (Z=1, A=3)
carbon-14, carbon-13, and carbon-12 are all isotopes of carbon. Carbon-14 contains a total of 8 neutrons, carbon-13 contains a total of 7 neutrons, and carbon-12 contains a total of 6 neutrons.
uranium-235 and uranium-239 are two different isotopes of the element uranium. Who was the first person to find multiple stable isotopes of an element? The British physicist J.J. Thomson was the first to discover evidence for multiple isotopes of the element neon (Ne) in the year 1912. Later, the English chemist and physicist F.W. Aston discovered many other stable isotopes of elements with the help of a mass spectrograph.
What are the two types of isotopes? There are two different types of isotopes, stable and radioactive. Stable isotopes are ones that can exist in their free state without breaking down spontaneously. Radioactive isotopes are ones that are too unstable to sustain themselves.
Isobars have same mass number but different atomic numbers.
Examples of Isobars Potassium, Argon, and calcium contain atoms of the same mass number 40.

⇒ Potassium-40 (⁴⁰K) is a radioactive isotope of potassium which has a long half-life of 1.25 billion years. It makes up about 0.012% (120 ppm) of the total amount of potassium found in nature.
⁴⁰ K,

⇒ Argon-38 isotope is used for as a tracer in geological dating.

Electromagnetic radiations:
The radiations which are associated with electrical and magnetic fields are called electromagnetic radiations. When an electrically charged particle moves under acceleration, alternating electrical and magnetic fields are produced and transmitted. These fields are transmitted in the form of waves. These waves are called electromagnetic waves or electromagnetic radiations.

Properties of electromagnetic radiations:

a. Oscillating electric and magnetic field are produced by oscillating charged particles. These fields are perpendicular to each other and both are perpendicular to the direction of propagation of the wave.
b. They do not need a medium to travel. That means they can even travel in vacuum.

Characteristics of electromagnetic radiations:
a. Wavelength: It may be defined as the distance between two neighbouring crests or troughs of wave as shown. It is denoted by λ.
b. Frequency (ν): It may be defined as the number of waves which passthrough a particular point in one second.
c. Velocity (v): It is defined as the distance travelled by a wave in one second. In vacuum all types of electromagnetic radiations travel with the same velocity. Its value is 3X10⁸ m sec ^-1.
It is denoted by v
d. Wave number: Wave number is defined as the number of wavelengths per unit length.
Velocity = frequency x wavelength c = ν λ.
What are Electromagnetic radiations? Write four characteristics.
Planck's Quantum Theory
The radiant energy is emitted or absorbed not continuously but discontinuously in the form of small discrete packets of energy called ‘quantum’. In case of light , the quantum of energy is called a ‘photon’.
The energy of each quantum is directly proportional to the frequency of the radiation, i.e. E∝ ν or
E= hν where h= Planck’s constant = 6.626 x 10^-34 Js
Energy is always emitted or absorbed as integral multiple of this

→

↔

Neutral particles ( Neutrons)
were discovered by Chadwick (1932).

In 1932, the physicist James Chadwick conducted an experiment in which he bombarded Beryllium with alpha particles from the natural radioactive decay of Polonium.

Spectrum (031102IIIa)

When a white light is passed through a prism, it splits into a series of coloured bands known as spectrum.
Spectrum is of two types: continuous and line spectrum
a. The spectrum which consists of all the wavelengths is called continuous spectrum.
b. A spectrum in which only specific wavelengths are present is known as a line spectrum. It has bright lines with dark spaces between them.
Electromagnetic spectrum is a continuous spectrum. It consists of a range of electromagnetic radiations arranged in the order of increasing wavelengths or decreasing frequencies. It extends from radio waves to gamma rays. Spectrum is also classified as emission and line spectrum.

The study of emission or absorption spectra is called spectroscopy.
Emission spectrum:
The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum.
Absorption spectrum
is the spectrum obtained when radiation is passed through a sample of material. The sample absorbs radiation of certain wave lengths. The wavelengths which are absorbed are missing and come as dark lines. The study of emission or absorption spectra is referred as spectroscopy.
Differences between Emission and Absorption Spectra

Spectroscopy

is the study of the absorption and emission of light and other radiation by matter. It involves the splitting of light (or more precisely electromagnetic radiation) into its constituent wavelengths (a spectrum), which is done in much the same way as a prism splits light into a rainbow of colours.

⇒ Infrared (IR) Spectroscopy. ...
Ultraviolet-Visible (UV/Vis) Spectroscopy. ...
Nuclear Magnetic Resonance (NMR) Spectroscopy. ...
Raman Spectroscopy. ...
X-Ray Spectroscopy.

What are 5 uses of spectroscopy?
Monitoring diffused oxygen content in freshwater and aquatic ecosystems.
Determining the atomic structure of a sample.
Determining the metabolic structure of a muscle.
Studying spectral emission lines of distant galaxies.
Altering the structure of drugs to improve the effectiveness. ⇔

Hydrogen Spectrum
It show quantized electronic structure of an atom. The hydrogen atoms of the molecule dissociate as soon as an electric discharge is passed through a gaseous hydrogen molecule.
It results in the emission of electromagnetic radiation initiated by the energetically excited hydrogen atoms. The hydrogen emission spectrum comprises radiation of discrete frequencies. These series of radiation are named after the scientists who discovered them.
Wavelength absorbed ,
When a hydrogen atom absorbs a photon, it causes the electron to experience a transition to a higher energy level, for example, n = 1, n = 2. When a photon is emitted through a hydrogen atom, the electron undergoes a transition from a higher energy level to a lower, for example, n = 3, n = 2. During this transition from a higher level to a lower level, there is the transmission of light occurs. The quantized energy levels of the atoms, cause the spectrum to comprise wavelengths that reflect the differences in these energy levels. For example, the line at 656 nm corresponds to the transition n = 3 n = 2.

Hydrogen emission spectrum: (031102IIIc) In the year 1885, on the basis of experimental observations, Balmer proposed the formula for correlating the wave number of the spectral lines emitted and the energy shells involved. This formula is given as:

This series of the hydrogen emission spectrum is known as the Balmer series. This is the only series of lines in the electromagnetic spectrum that lies in the visible region. The value, 109,677 cm^-1, is called the Rydberg constant for hydrogen. Ref Fig. The Balmer series is basically the part of the hydrogen emission spectrum responsible for the excitation of an electron from the second shell to any other shell. Similarly, other transitions also have their own series names. Some of them are listed below,
The transition from the first shell to any other shell – Lyman series
The transition from the second shell to any other shell – Balmer series
The Transition from the third shell to any other shell – Paschen series
The transition from the fourth shell to any other shell – Brackett series
The transition from the fifth shell to any other shell – Pfund series
What is the formula for the radius of a stationary orbit? radius of stationery orbit is r = n²h²/4π²mZe²
Spectral Lines for atomic hydrogen:
Ref fig.

How is the hydrogen emission spectrum formed?

Firstly a hydrogen molecule is broken into hydrogen atoms. The electron in a hydrogen atom absorbs energy and gets excited. The electron jumps from a lower energy level to a higher energy level and when it comes back to its original state, it gives out energy which forms a hydrogen spectrum.
What are the first three series of hydrogen emission spectrum?
Lyman, Balmer and Paschen series
Name the series which falls in the visible region of the hydrogen emission spectrum? Balmer series. It is a hydrogen spectral series found in the hydrogen spectrum.

Bohr’s Model of an Atom
In 1913, Niels Bohr proposed a theory for the hydrogen atom, based on quantum theory that some physical quantities only take discrete values. Electrons move around a nucleus, but only in prescribed orbits, and If electrons jump to a lower-energy orbit, the difference is sent out as radiation.
Salient features of Niels Bohr atomic model are: Electrons revolve around the nucleus in stable orbits without emission of radiant energy. Each orbit has a definite energy and is called an energy shell or energy level. An orbit or energy level is designated as K, L, M, N shells.

Postulates of Bohr’s Model of an Atom
1. In an atom, electrons (negatively charged) revolve around the positively charged nucleus in a definite circular path called orbits or shells.
2. Each orbit or shell has a fixed energy and these circular orbits are known as orbital shells.
3. The energy levels are represented by an integer (n=1, 2, 3…) known as the quantum number. This range of quantum number starts from nucleus side with n=1 having the lowest energy level. The orbits n=1, 2, 3, 4… are assigned as K, L, M, N…. shells and when an electron attains the lowest energy level, it is said to be in the ground state.
4. The electrons in an atom move from a lower energy level to a higher energy level by gaining the required energy and an electron moves from a higher energy level to lower energy level by losing energy.

Bohr’s model for hydrogen atom:
(a) The stationary states for electron are numbered n = 1,2,3.......... These integral numbers (Section 2.6.2) are known as Principal quantum numbers.
(b) The radii of the stationary states are expressed as: r_n = n² aₒ (2.12) where aₒ = 52.9 pm. Thus the radius of the first stationary state, called the Bohr orbit, is 52.9 pm.

⇒ Normally the electron in the hydrogen atom is found in this orbit (that is n=1). As n increases the value of r will increase. In other words the electron will be present away from the nucleus.

c) The most important property associated with the electron, is the energy of its stationary state. It is given by the expression.

⇒ Explanation of Line Spectrum of Hydrogen
Line spectrum observed in case of hydrogen atom, as mentioned in section 2.3.3, can be explained quantitatively using Bohr’s model. According to assumption 2, radiation (energy) is absorbed if the electron moves from the orbit of smaller Principal quantum number to the orbit of higher Principal quantum number, whereas the radiation (energy) is emitted if the electron moves from higher orbit to lower orbit. The energy gap between the two orbits is given by equation
ΔE = E_f – E_i
) Combining equations
(031102.0ex2b)
In case of absorption spectrum, n_f > n_i and the term in the parenthesis is positive and energy is absorbed. On the other hand in case of emission spectrum n_i > n_f , Δ E is negative and energy is released. The expression is similar to that used by Rydberg derived empirically using the experimental data available at that time.
Further, each spectral line, whether in absorption or emission spectrum, can be associated to the particular transition in hydrogen atom. In case of large number of hydrogen atoms, different possible transitions can be observed and thus leading to large number of spectral lines. The brightness or intensity of spectral lines depends upon the number of photons of same wavelength or frequency absorbed or emitted.

Where E1 and E2 are the energies of the lower and higher allowed energy states respectively. This expression is commonly known as Bohr’s frequency rule.

Limitations of Bohr’s model of atom:
Bohr’s theory was quite successful to explain the atomic spectrum of Hydrogen atom and many other one electron species like He¹⁺ and Li²⁺ ions. It also enabled to calculate the energy of hydrogen electron in the different energy levels. However the theory suffered from certain limitations which are described as follows:
→1. The theory could not explain the atomic spectra of the atoms containing more than one electron or multi-electron atoms.
2. Bohr’s theory predicts only one spectral line for the electronic transition between two energy levels. However if examined carefully, certain single lines are found to contain a number of finer lines. Bohr’s theory failed to explain the fine structure of the spectral lines.
3. This theory could not explain Zeeman effect and Stark effect.
4. Bohr’s theory failed to explain the shapes of molecules formed by the combination of atoms.
5. Bohr’s Theory could not explain de-Broglie’s relationship and Heisenberg’s Uncertainty Principle.

The Zeeman effect is the effect of splitting of a spectral line into several components in the presence of a static magnetic field.

Stark effect –

the splitting of spectral lines observed when the radiating atoms, ions, or molecules are subjected to a strong electric field.
→
Difference between Bohr and Quantum Model Bohr model was proposed by Niels Bohr in 1915. Quantum model is the modern model of an atom. The key difference between Bohr and quantum model is that Bohr model states that electrons behave as particles whereas quantum model explains that the electron has both particle and wave behavior.

Particle Nature of Electromagnetic Radiation: Planck’s Quantum Theory Some of the experimental phenomenon such as diffraction and interference can be explained by the wave nature of the electromagnetic radiation.
Diffraction is the spreading out of waves as they pass through an aperture or around objects. It occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave.
interference The phenomenon in which two or more waves superpose to form a resultant wave of greater, lower or the same amplitude. The interference of waves results in the medium taking on a shape resulting from the net effect of the two individual waves.

⇒
Dual behavior of electromagnetic radiation- The light possesses both particle and wave like properties, i.e., light has dual behavior. whenever radiation interacts with matter, it displays particle like properties. (Black body radiation and photoelectric effect) Wave like properties are exhibited when it propagates (interference and diffraction)
However, following are some of the observations which could not be explained with the help of even the electromagentic theory of 19th century physics (known as classical physics):

(i) the nature of emission of radiation from hot bodies (black -body radiation)
(ii) ejection of electrons from metal surface when radiation strikes it (photoelectric effect)
(iii) variation of heat capacity of solids as a function of temperature
(iv) Line spectra of atoms with special reference to hydrogen.

⇒ These phenomena indicate that the system can take energy only in discrete amounts. All possible energies cannot be taken up or radiated.
It is noteworthy that the first concrete explanation for the phenomenon of the black body radiation mentioned above was given by Max Planck in 1900.

⇒ Let us first try to understand this phenomenon, which is given below:
a. Hot objects emit electromagnetic radiations over a wide range of wavelengths. At high temperatures, an appreciable proportion of radiation is in the visible region of the spectrum. b. As the temperature is raised, a higher proportion of short wavelength (blue light) is generated. For example, when an iron rod is heated in a furnace, it first turns to dull red and then progressively becomes more and more red as the temperature increases. As this is heated further, the radiation emitted becomes white and then becomes blue as the temperature becomes very high.
c. This means that red radiation is most intense at a particular temperature and the blue radiation is more intense at another temperature. It means intensities of radiations of different wavelengths emitted by hot body depend upon its temperature.
d. when the surface of an object is irradiated with light (electromagnetic radiation), a part of radiant energy is generally reflected as such, a part is absorbed and a part of it is transmitted. The reason for incomplete absorption is that ordinary objects are as a rule imperfect absorbers of radiation.

⇒ An ideal body, which emits and absorbs radiations of all frequencies uniformly, is called a black body and the radiation emitted by such a body is called black body radiation.
In practice, no such body exists.
Carbon black approximates fairly closely to black body.
A black body is also a perfect radiator of radiant energy. Furthermore, a black body is in thermal equilibrium with its surroundings. It radiates same amount of energy per unit area as it absorbs from its surrounding in any given time. The amount of light emitted (intensity of radiation) from a black body and its spectral distribution depends only on its temperature.
At a given temperature, intensity of radiation emitted increases with the increase of wavelength, reaches a maximum value at a given wavelength and then starts decreasing with further increase of wavelength, as shown in Fig. 2.8.
Also, as the temperature increases, maxima of the curve shifts to short wavelength. Several attempts were made to predict the intensity of radiation as a function of wavelength.
But the results of the above experiment could not be explained satisfactorily on the basis of the wave theory of light.

⇒

Black body:
An ideal body, which emits and absorbs all frequencies, is called a black body. The radiation emitted by such a body is called black body radiation.
It is called black body because the device used for this experiment was black in color and experiment was impossible without it.
To stay in thermal equilibrium, a black body must emit radiation at the same rate as it absorbs, so it must also be a good emitter of radiation, emitting electromagnetic waves of as many frequencies as it can absorb, i.e. all the frequencies. The radiation emitted by the blackbody is known as blackbody radiation.
To explain characteristics of the blackbody radiation let us understand following laws.
a. Wien’s displacement law b. Planck’s law c. Stefan-Boltzmann law

Wien’s Displacement Law
Wien’s displacement law states that the blackbody radiation curve for different temperature peaks at a wavelength is inversely proportional to the temperature.
Graph (a) The higher the temperature of a body, the higher the area under the curve, i.e., more amount of energy is emitted by the body at a higher temperature.
(b) The energy emitted by the body at different temperatures is not uniform. For both long and short wavelengths, the energy emitted is very small.
(c) For a given temperature, there is a particular wavelength (l_m) for which the energy emitted (El) is maximum.
(d) With an increase in the temperature of the black body, the maxima of the curves shift towards shorter wavelengths.

Planck’s Law
It states that electromagnetic radiation from heated bodies is not emitted as a continuous flow but is emitted in discrete units or quanta of energy, the size of which involves a fundamental physical constant (Planck’s constant).

Stefan-Boltzmann Law The Stefan-Boltzmann law explains the relationship between total energy emitted and the absolute temperature. the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature.
E α T⁴

⇒ What is black body radiation?
The radiation emitted by the blackbody is known as blackbody radiation. To stay in thermal equilibrium, a black body must emit radiation at the same rate as it absorbs, so it must also be a good emitter.
What is a black body? It is an ideal body that absorbs all incident electromagnetic waves or radiation, regardless of the angle of incidence or frequency. As it absorbs all colours of light, it is named as black “body”.
What are the laws that explain the characteristics of a black body? Wien’s displacement law, Planck’s Law and Stefan-Boltzmann laws are the laws that explain the character of a black body.
What is Wien’s Displacement Law? According to Wien’s Displacement Law, the blackbody radiation curve for different temperature peaks at a wavelength is inversely proportional to the temperature.
What is Planck’s Law?
Planck’s Law states that electromagnetic radiation from heated objects is not released as a continuous form but is made up of discrete quanta or units of energy. This size possesses a fundamental physical constant called Planck’s constant.
What is an oscillator
An oscillator is a mechanical or electronic device that works on the principles of oscillation: a periodic fluctuation between two things based on changes in energy. Oscillation is going back and forth repeatedly between two positions or states. An oscillation can be a periodic motion that repeats itself in a regular cycle, such as the side-to-side swing of a pendulum, or the up-and-down motion of a spring with a weight.

Max Planck arrived at a satisfactory relationship by making an assumption that absorption and emmission of radiation arises from oscillator i.e., atoms in the wall of black body. Their frequency of oscillation is changed by interaction with oscilators of electromagnetic radiation. Planck assumed that radiation could be sub-divided into discrete chunks of energy.
He suggested that atoms and molecules could emit or absorb energy only in discrete quantities and not in a continuous manner. He gave the name quantum to the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E ) of a quantum of radiation is proportional to its frequency (nue) and is expressed by equation E = h ν
The proportionality constant, ‘h’ is known as Planck’s constant and has the value 6.626×10^-34 J s.
With this theory, Planck was able to explain the distribution of intensity in the radiation from black body as a function of frequency or wavelength at different temperatures.

Photoelectric effect:
The phenomenon of ejection of electrons from the surface of metal when light of suitable frequency strikes it is called photoelectric effect.
The ejected electrons are called photoelectrons.
Light of a particular frequency strikes a clean metal surface inside a vacuum chamber. Electrons are ejected from the metal and are counted by a detector that measures their kinetic energy.
Experimental results observed for the experiment of Photoelectric effect a. When beam of light falls on a metal surface electrons are ejected immediately.
b. Number of electrons ejected is proportional to intensity or brightness of light c. Threshold frequency (ν⁰): For each metal there is a characteristic minimum frequency below which photoelectric effect is not observed. This is called threshold frequency.
d. If frequency of light is less than the threshold frequency there is no ejection of electrons, no matter how long it falls on surface or how high is its intensity.
e. Photoelectric work function (W^o) : The minimum energy required to eject electrons is called photoelectric work function.
W^o= hv^o
f. Energy of the ejected electrons :

Quantisation -
The energy can take any one of the values from the following set, but cannot take on any values between them. E = 0, hν, 2hν, 3hν....nhν.....

Photoelectric Effect
In 1887, H. Hertz performed a very interesting experiment in which electrons (or electric current) were ejected when certain metals (for example potassium, rubidium, caesium etc.) were exposed to a beam of light as shown in Fig. The phenomenon is called Photoelectric effect.
The results observed in this experiment were:
(i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface, i.e., there is no time lag between the striking of light beam and the ejection of electrons from the metal surface.
(ii) The number of electrons ejected is proportional to the intensity or brightness of light.
(iii) For each metal, there is a characteristic minimum frequency, ν ₒ(also known as threshold frequency) below which photoelectric effect is not observed.
At a frequency ν >ν ₒ(, the ejected electrons come out with certain kinetic energy.
The kinetic energies of these electrons increase with the increase of frequency of the light used.
All the above results could not be explained on the basis of laws of classical physics.
According to latter, the energy content of the beam of light depends upon the brightness of the light. In other words, number of electrons ejected and kinetic energy associated with them should depend on the brightness of light. It has been observed that though the number of electrons ejected does depend upon the brightness of light, the kinetic energy of the ejected electrons does not.

Heisenberg’s uncertainty principle:
It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. The product of their uncertainties is always equal to or greater than h/4Π.
Heisenberg’s uncertainty principle rules out the existence of definite paths or trajectories of electrons and other similar particles.

⇒ Failure of Bohr’s model:
a. It ignores the dual behavior of matter. b. It contradicts Heisenberg’s uncertainty principle.

Classical mechanics

is based on Newton’s laws of motion.
It successfully describes the motion of macroscopic particles but fails in the case Of microscopic particles.
Reason:
Classical mechanics ignores the concept of dual behaviour of matter especially for sub-atomic particles and the Heisenberg’s uncertainty principle.

What is Classical mechanics?

Classical mechanics describes the behavior of macroscopic bodies, which have relatively small velocities compared to the speed of light.
Quantum mechanics describes the behavior of microscopic bodies such as subatomic particles, atoms, and other small bodies.
Classical mechanics is the study of macroscopic bodies. The movements and statics of macroscopic bodies are discussed under classical mechanics. Classical mechanics has three different branches.
They are, namely, Newtonian mechanics, Lagrangian mechanics, and Hamiltonian mechanics.
These three branches are based on the mathematical methods and quantities used to study the motion.
For an example, Newtonian mechanics uses vectors such as displacement, velocity, and acceleration to study the motion of the object, whereas Lagrangian mechanics uses energy equations and rate of energy change to study.
The proper method is selected depending on the problem to be solved. Classical mechanics is applied in places such as planetary motion, projectiles, and most of the events in daily lives.
In classical mechanics, energy is treated as a continuous quantity.
A system can take any amount of energy in classical mechanics.

⇒ What is Lagrangian mechanics? There is an alternative approach known as lagrangian mechanics which enables us to find the equations of motion when the newtonian method is proving difficult. In lagrangian mechanics we start, as usual, by drawing a large, clear diagram of the system, using a ruler and a compass.

The Newtonian force-momentum formulation is vectorial in nature, it has cause and effect embedded in it. The Lagrangian approach is cast in terms of kinetic and potential energies which involve only scalar functions and the equations of motion come from a single scalar function, i.e. Lagrangian.

⇒ What is Quantum Mechanics?
a. Quantum mechanics is the study of microscopic bodies. The term “quantum” comes from the fact that energy of a microscopic system is quantized.
b. The photon theory is one of the cornerstones of quantum mechanics.
c. It states that the energy of light is in the form of wave packets.
d. Heisenberg, Max Plank, Albert Einstein are some of the prominent scientists involved in developing the quantum mechanics.
e. Quantum mechanics falls into two categories.
(i) The first one is quantum mechanics of non-relativistic bodies. This field studies the quantum mechanics of particles with relatively small speeds compared to the speed of light.
(ii) The other form is relativistic quantum mechanics, which studies particles moving with speeds compatible with the speed of light. f. Heisenberg’s uncertainty Principal is also a very important theory behind quantum mechanics. It states that the linear momentum of a particle and the position of that particle in the same direction cannot be measured simultaneously with 100% accuracy.

⇒ Significance of Uncertainty Principle a. it rules out existence of definite paths or trajectories of electrons and other similar particles. The trajectory of an object is determined by its location and velocity at various moments. The position of an object and its velocity fix its trajectory. Since for a sub-atomic object such as an electron, it is not possible simultaneously to determine the position and velocity at any given instant to an arbitrary degree of precision, it is not possible to talk of the trajectory of an electron.
b. The effect of Heisenberg Uncertainty Principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects. This can be seen from the following examples. If uncertainty principle is applied to an object of mass, say about a milligram (10–6 kg), then (031102.0b)

Quantum mechanics

a. Quantum mechanics is a theoretical science that deals with the study of the motions of the microscopic objects that have both observable wave like And particle like properties.
b. Quantum mechanics is based on a fundamental equation which is called Schrodinger equation.

Schrodinger’s equation:

Schrodinger wave equation is a mathematical expression describing the energy and position of the electron in space and time, taking into account the matter wave nature of the electron inside an atom.
It is based on three considerations. They are
a. Classical plane wave equation
b. Broglie’s hypothesis of matter-wave
c. Conservation of energy

The Schrodinger equation gives a. a detailed account of the form of the wave functions or probability waves that control the motion of some smaller particles.
b. The equation also describes how these waves are influenced by external factors. c. The equation makes use of the energy conservation concept that offers details about the behaviour of an electron that is attached to the nucleus.
d. By calculating the Schrödinger equation, we obtain Ψ and Ψ2, which helps us determine the quantum numbers, as well as the orientations and the shape of orbitals, where electrons are found in a molecule or an atom.
e. There are two equations, which are a time-dependent Schrödinger equation and a time-independent Schrödinger equation.
The time-dependent Schrödinger equation is represented as - Fig.

⇒ Quantum Mechanical Model-
The branch of science that takes into account dual behavior i.e. particle and wave nature of matter is called quantum mechanics.

Features of Quantum Mechanical Model:

1. The energy of an electron is quantized i.e. an electron can only have certain specific values of energy.
2. The quantized energy of an electron is the allowed solution of the Schrödinger wave equation and it is the result of wave like properties of electron.
Ψ -psi. 3. As per Heisenberg’s Uncertainty principle, the exact position and momentum of an electron cannot be determined. So the only probability of finding an electron at a position can be determined and it is |Ψ| |² at that point where Ψ represents the wave-function of that electron.
(Ψ -psi) 4. An atomic orbital is the wave-function (Ψ) of an electron in an atom. Whenever an electron is described by a wave-function, it occupies atomic orbital. As an electron can have many wave-functions, there are many atomic orbitals for the electron. Every wave-function or atomic orbital have some shape and energy associated with it. All the information about the electron in an atom is stored in its orbital wave function Ψ and quantum mechanics makes it possible to extract this information out of Ψ.
5. The probability of finding an electron at a point within an atom is proportional to the square of the orbital wave function i.e., | Ψ|² at that point. | Ψ |² is known as probability density and is always positive.

Revision of some facts:

First active study of thermal radiation laws occurred in the 1850’s and the theory of electromagnetic waves and the emission of such waves by accelerating charged particles was developed in the early 1870’s by James Clerk Maxwell, which was experimentally confirmed later by Heinrich Hertz.
Here, we will learn some facts about electromagnetic radiations. James Maxwell (1870) suggested that when electrically charged particle moves under accelaration, alternating electrical and magnetic fields are produced and transmitted. These fields are transmitted in the forms of waves called electromagnetic waves or electromagnetic radiation.
In earlier days (Newton) light was supposed to be made of particles (corpuscules). It was only in the 19th century when wave nature of light was established. Maxwell was again the first to reveal that light waves are associated with oscillating electric and magnetic character (Fig. 2.6). Although electromagnetic wave motion is complex in nature, we will consider here only a few simple properties.

(i) The oscillating electric and magnetic fields produced by oscillating charged particles are perpendicular to each other and both are perpendicular to the direction of propagation of the wave. Simplified picture of electromagnetic wave is shown in Fig.
(ii) Unlike sound waves or waves produced in water, electromagnetic waves do not require medium and can move in vacuum.
(iii) It is now well established that there are many types of electromagnetic radiations, which differ from one another in wavelength (or frequency). These constitute what is called electromagnetic spectrum (Fig.).

Regions of electromagnetic spectrum

Radio frequency region around 10⁶ Hz, used for broadcasting;
microwave region around 10¹⁰ Hz used for radar;
infrared region around 10¹³ Hz used for heating;
ultraviolet region around 10¹⁶ Hz a component of sun’s radiation. The small portion around 10¹⁵ Hz, is what is ordinarily called visible light. It is only this part which our eyes can see (or detect).
Special instruments are required to detect non-visible radiation. Fig.
(iv) Different kinds of units are used to represent electromagnetic radiation. These radiations are characterised by the properties, namely, frequency (ν nue) and wavelength (λ lambda).
The SI unit for frequency is hertz (Hz, s^–1), after Heinrich Hertz.
It is defined as the number of waves that pass a given point in one second.
Wavelength should have the units of length and as you know that the SI units of length is meter (m).
Since electromagnetic radiation consists of different kinds of waves of much smaller wavelengths, smaller units are used.
Fig. shows various types of electromagnetic radiations which differ from one another in wavelengths and frequencies. In vaccum all types of electromagnetic radiations, regardless of wavelength, travel at the same speed, i.e., 3.0 × 10⁸ m s^–1 (2.997925 × 10⁸ m s^–1, to be precise). This is called speed of light and is given the symbol ‘c‘.

2.6.1 Orbitals and Quantum Numbers
A large number of orbitals are possible in an atom.
An orbital of smaller size means there is more chance of finding the electron near the nucleus. Similarly shape and orientation mean that there is more probability of finding the electron along certain directions than along others.
Atomic orbitals are precisely distinguished by what are known as quantum numbers. Each orbital is designated by three quantum numbers labelled as n, l and m_l.

Types of Quantum Numbers Principal Quantum Number
Azimuthal Quantum Number
Magnetic Quantum Number
Spin electron Quantum Number.
Principal Quantum number - a. It is denoted by n. It refers to the electron shell with the most electrons also giving the electron's likely distance from the nucleus.
b. Bigger value of 'n' signifies a larger orbital size and thus a larger atomic radius.
c. Any positive integer can be used as the value of 'n.' An atom can't have either zero or negative energy, it can't be zero or a negative integer. d. when n = 1, the ground state or lowest energy means the shell is designated as the innermost shell or first primary shell. An electron can gain energy and hop to higher shells by absorbing energy or photons, thus increasing the value of 'n'. If it loses energy, it returns to lower shells, and the value of 'n' drops, and photons are emitted.
Azimuthal Quantum Number -
a. The azimuthal quantum number showing the angular momentum of an orbital is defined for the determination of the shape of an orbital.
b. The value of Azimuthal Quantum Number is equal to the total number of angular nodes in the orbital and is indicated by the symbol 'l.'

⇒ What are angular nodes? The planes or planar areas around the nucleus where the probability of finding an electron is zero are called angular nodes. The value of the angular nodes does not depend upon the value of the principal quantum number. It only depends on the value of the azimuthal quantum number.

c. The azimuthal quantum number can suggest an s, p, d, or f subshell. This value is determined by the primary quantum number. The azimuthal quantum number varies between 0 and l (symbol). (n-1).
e.g. if n = 3, the azimuthal quantum number can be 0, 1, or 2.
The resulting subshells will be
‘s’ subshell when l=0.
'p' subshell When l=1 and
'd' subshell when l=2. Thus three subshells would be 3s, 3p, and 3d. Let us say value of n is 4.
Thus the possible values of l are 0, 1, 2 and 3
The resulting subshells will be
‘s’ subshell when l=0.
'p' subshell When l=1
'd' subshell when l=2 and
'f' subshell when l=3
Magnetic Quantum Number a. The magnetic quantum number determines the subshell's overall number of orbitals and their orientation. The symbol ‘ml’is used to represent it.
b. This number shows the orbital angular momentum projected along a given axis. The azimuthal (or orbital angular momentum) quantum number determines the value of the magnetic quantum number.
The value of ‘m’ is dependent on the value of ‘l’. Magnetic quantum numbers can have a total of (2l + 1) values. The value of ml for a particular value of l is in the range of -l to +l. As a result, it is indirectly affected by the value of n. So various values would be
0= s one s orbital 1= P (-1,0,1) three values and orbitals would be P_x , P_y and p_z
2= d (-2,-1,0,1,2) five values and orbitals would be d_xy, d_yz, d_xz, d_(x2-y2) and d_z2
3 = f (-3,-2,-1,0,1,2,3) seven values and orbitals would be (031102.0c) <1-- fx(x2-y2), fy(x2-y2), fxyz, fz3, fyz, fxz2,fz(x2-y2). -->
Electron Spin Quantum Number
In 1925, George Uhlenbeck and Samuel Goudsmit proposed the presence of the fourth quantum number known as the electron spin quantum number (m). An electron spins around its own axis, much in a similar way as earth spins around its own axis while revolving around the sun.
Spin angular momentum of the electron — a vector quantity, can have two orientations relative to the chosen axis. These two orientations are distinguished by the spin quantum numbers ms which can take the values of +½ or –½. These are called the two spin states of the electron and are normally represented by two arrows, ↑ (spin up) and ↓ (spin down).
Two electrons that have different ms values (one +½ and the other –½) are said to have opposite spins. An orbital cannot hold more than two electrons and these two electrons should have opposite spins.
a. The values of n, l, and ml have no effect on the electron spin quantum number. The symbol ‘m’ represents the value of this number, which indicates the direction in which the electron is spinning.
b. The value of ms gives direction of electron's spin. The electron spin quantum number has two values: 1/2 and -1/2. The positive number of ms denotes an upward spin on the electron, often known as pin up,' and is represented by the symbol ↑. If m is negative, the electron in question has a downward spin, or spin down,' which is represented by the symbol.

⇒ Nodal point - The nodal point is a point at which there is a zero probability of finding the electron.
There are two types of nodes: Radial nodes and angular nodes.
The radial nodes calculate the distance from the nucleus, while the angular node determines the direction.
Angular nodes = total nodes - radial nodes.
How to calculate radial nodes
Total Nodes=n-1. From knowing the total nodes we can find the number of radial nodes by using. Radial Nodes= n-1-l.
No. of angular nodes = l
Total number of nodes = n – 1
Nodal planes are defined as planes of zero probability region to find the electron.
Shapes of orbitals
There are four different kinds of orbitals, denoted s, p, d and f each with a different shape.
An s-orbital is spherical with the nucleus at its centre,
a p-orbitals is dumbbell-shaped and
four of the five d orbitals are cloverleaf shaped. The fifth d orbital is shaped like an elongated dumbbell with a doughnut around its middle.
The orbitals in an atom are organized into different layers or electron shells.

Nodes
The region where this probability density function reduces to zero is called nodal surfaces or simply nodes. In general, it has been found that ns-orbital has (n – 1) nodes, that is, number of nodes increases with increase of principal quantum number n. In other words, number of nodes for 2s orbital is one, two for 3s and so on.

⇒ Energies of Orbitals
The energy of the orbitals in hydrogen atom increases as follows : 1s < 2s = 2p < 3s = 3p = 3d <4s = 4p = 4d = 4f < ….
Although the shapes of 2s and 2p orbitals are different, an electron has the same energy when it is in the 2s orbital as when it is present in 2p orbital.

What are degenerate orbitals?
The orbitals having the same energy are called degenerate.
What is ground state?
The ground state of a quantum-mechanical system is its stationary state of lowest energy; the energy of the ground state is known as the zero-point energy.

⇒ The 1s orbital in a hydrogen atom, as said earlier, corresponds to the most stable condition and is called the ground state and an electron residing in this orbital is most strongly held by the nucleus.

An electron in the 2s, 2p or higher orbitals in a hydrogen atom is in excited state.

⇒ The energy of an electron in a multielectron atom, unlike that of the hydrogen atom, depends not only on its principal quantum number (shell), but also on its azimuthal quantum number (subshell). That is, for a given principal quantum number, s, p, d, f ... all have different energies.
Within a given principal quantum number, the energy of orbitals increases in the order s The main reason for having different energies of the subshells is the mutual repulsion among the electrons in multi-electron atoms.
The only electrical interaction present in hydrogen atom is the attraction between the negatively charged electron and the positively charged nucleus.

In multi-electron atoms, besides the presence of attraction between the electron and nucleus, there are repulsion terms between every electron and other electrons present in the atom.

Thus the stability of an electron in a multi-electron atom is because total attractive interactions are more than the repulsive interactions.

⇒ In general, the repulsive interaction of the electrons in the outer shell with the electrons in the inner shell are more important. On the other hand, the attractive interactions of an electron increases with increase of positive charge (Ze) on the nucleus.
Due to the presence of electrons in the inner shells, the electron in the outer shell will not experience the full positive charge of the nucleus (Ze). The effect will be lowered due to the partial screening of positive charge on the nucleus by the inner shell electrons.
This is known as the shielding of the outer shell electrons from the nucleus by the inner shell electrons, and the net positive charge experienced by the outer electrons is known as effective nuclear charge (Z_effe).
Despite the shielding of the outer electrons from the nucleus by the inner shell electrons, the attractive force experienced by the outer shell electrons increases with increase of nuclear charge.
In other words, the energy of interaction between, the nucleus and electron (that is orbital energy) decreases (that is more negative) with the increase of atomic number (Z).
Both the attractive and repulsive interactions depend upon the shell and shape of the orbital in which the electron is present.
For example electrons present in spherical shaped, s orbital shields the outer electrons from the nucleus more effectively as compared to electrons present in p orbital. Similarly electrons present in p orbitals shield the outer electrons from the nucleus more than the electrons present in d orbitals, even though all these orbitals are present in the same shell.
Further within a shell, due to spherical shape of s orbital, the s orbital electron spends more time close to the nucleus in comparison to p orbital electron which spends more time in the vicinity of nucleus in comparison to d orbital electron.
In other words, for a given shell (principal quantum number), the Zeff experienced by the electron decreases with increase of azimuthal quantum number (l), that is, the s orbital electron will be more tightly bound to the nucleus than p orbital electron which in turn will be better tightly bound than the d orbital electron.
The energy of electrons in s orbital will be lower (more negative) than that of p orbital electron which will have less energy than that of d orbital electron and so on. Since the extent of shielding from the nucleus is different for electrons in different orbitals, it leads to the splitting of energy levels within the same shell (or same principal quantum number), that is, energy of electron in an orbital, as mentioned earlier, depends upon the values of n and l.
Mathematically, the dependence of energies of the orbitals on n and l are quite complicated but one simple rule is that, the lower the value of (n + l) for an orbital, the lower is its energy. If two orbitals have the same value of (n + l), the orbital with lower value of n will have the lower energy. The Table shows arrangement of Orbitals with Increasing Energy on the Basis of (n+l)
Rule illustrates the (n + l ) rule and Fig. depicts the energy levels of multielectrons atoms.
It may be noted that different subshells of a particular shell have different energies in case of multi-electrons atoms.
However, in hydrogen atom, these have the same energy. Lastly it may be mentioned here that energies of the orbitals in the same subshell decrease with increase in the atomic number (Zeff).
For example, energy of 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium and that of lithium is greater than that of sodium and so on, that is, E_2s(H) > E_2s (Li) > E_2s (Na) > E_2s (K).

What is effective nuclear charge?
Effective nuclear charge – The attractive positive charge of nuclear protons acting on valence electrons.
The effective nuclear charge is always less than the total number of protons present in a nucleus due to the shielding effect. Zeff = Z - S Where Z is the atomic number and S is the number of shielding electrons.
In periodic table there is a. an increase across a period (due to increasing nuclear charge with no accompanying increase in shielding effect).
b. decrease down a group (although nuclear charge increases down a group, shielding effect more than counters its effect).

⇒ Filling of Orbitals in Atom
Aufbau Principle
The word ‘aufbau’ in German means ‘building up’. The building up of orbitals means the filling up of orbitals with electrons. The principle states : In the ground state of the atoms, the orbitals are filled in order of their increasing energies. In other words, electrons first occupy the lowest energy orbital available to them and enter into higher energy orbitals only after the lower energy orbitals are filled. As you have learnt above, energy of a given orbital depends upon effective nuclear charge and different type of orbitals are affected to different extent. Thus, there is no single ordering of energies of orbitals which will be universally correct for all atoms. However, following order of energies of the orbitals is extremely useful: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 6p, 7s...

Aufbau Principle It states that electrons are filled into atomic orbitals in the increasing order of orbital energy level. According to the Aufbau principle, the available atomic orbitals with the lowest energy levels are occupied before those with higher energy levels.
The word ‘Aufbau’ has German roots and can be roughly translated as ‘construct’ or ‘build up’.
Salient Features of the Aufbau Principle
a. According to the Aufbau principle, electrons first occupy those orbitals whose energy is the lowest. Thus electrons enter the orbitals having higher energies only when orbitals with lower energies have been completely filled.
b. The order in which the energy of orbitals increases can be determined with the help of the (n+l) rule, where the sum of the principal and azimuthal quantum numbers determines the energy level of the orbital.
c. The order in which the orbitals are filled with electrons is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, and so on.

Write atomic number, symbol atomic mass and electronic configuration of the following elements. Sodium, potassium, phosphorus, sulphur, calcium, helium, argon and neon.

( SVT031101.014 (1) )

⇒ Pauli Exclusion Principle The number of electrons to be filled in various orbitals is restricted by the exclusion principle,
given by the Austrian scientist Wolfgang Pauli (1926).
According to this principle : No two electrons in an atom can have the same set of four quantum numbers. Pauli exclusion principle can also be stated as : “Only two electrons may exist in the same orbital and these electrons must have opposite spin.”

Explain Pauli Exclusion Principle
The Pauli exclusion principle states that in a single atom, no two electrons will have an identical set or the same quantum numbers (n, l, ml, and ms).
To put it in simple terms, every electron should have or be in its own unique state (singlet state). There are two salient rules that the Pauli exclusion principle follows:
Only two electrons can occupy the same orbital.
The two electrons that are present in the same orbital must have opposite spins, or they should be antiparallel.
Importance of the Pauli exclusion principle
a. It explains electron shell structure of atoms and the way atoms share electrons.
b. It helps in describing the various chemical elements and how they participate in forming chemical bonds.
c. The principle helps in describing the stability of large systems with many electrons and many nucleons.

⇒ Hund’s Rule of Maximum Multiplicity
This rule deals with the filling of electrons into the orbitals belonging to the same subshell (that is, orbitals of equal energy, called degenerate orbitals). Explain Hund’s Rule of Maximum Multiplicity
Pairing of electrons in the orbitals belonging to the same subshell (p, d or f) does not take place until each orbital belonging to that subshell has got one electron each i.e., it is singly occupied.
So we can say -
a. Before the double occupation of any orbital, every orbital in the sub level is singly occupied.
b. For the maximization of total spin, all electrons in a single occupancy orbital have the same spin.
If two electrons come in contact they would behave like two magnets do, first they get as far away from each other as possible before they have to pair up.

⇒ What are half filled and fully filled orbitals
Half-filled orbitals or also called partially filled orbitals are the orbitals that have half the number of electrons (i.e. one electron). On the other hand, filled or filled orbitals have a full number of electrons (i.e. two electrons).
The exactly half-filled and fully filled orbitals have greater stability than other configurations.

The reason for their stability are symmetry and exchange energy. Reasons of Stability of exactly half-filled and fully filled orbitals
(a) Symmetry:
The half-filled and fully-filled orbitals are more symmetrical than any other configuration and symmetry leads to greater stability.
(b) Exchange Energy:
The electrons present in the different orbitals of the same sub-shell can exchange their positions. Each such exchange leads to the decrease in energy known as Exchange Energy. Greater the number of exchanges, greater the exchange energy and hence greater the stability. As the number of exchanges that take place in the half-filled and fully-filled orbitals is maximum, thus exchange energy is maximum and hence maximum stability.

Electronic Configuration of Atoms
The distribution of electrons into orbitals of an atom is called its electronic configuration
The electronic configuration of different atoms can be represented in two ways. For example :
(i) s p d ...... notation and
(ii) Orbital diagram
In the first notation, the subshell is represented by the respective letter symbol and the number of electrons present in the subshell is depicted, as the super script, e.g.

Hydrogen (H) ¹₁ H
1 → 1s ¹
Helium (He) ⁴₂ He
2 → 1s ²
Lithium (Li) ⁷₃Li
3 → 1s²2s¹
Beryllium (Be) ⁹₄ Be
→ 1s ²2s²
Boron (B) ¹¹₅B
5 → 1s ² 2s ² 2p ¹
Carbon (C) ¹²₆C
6 → 1s ² 2s ² 2p ²
Nitrogen (N) ¹⁴₇ N
7 → 1s ² 2s ² 2p ³
Oxygen (O) ¹⁶₈O
8 → 1s ² 2s ² 2p ⁴
Fluorine (F) ¹⁹₉F
9 → 1s ² 2s ² 2p ⁵
Neon (Ne) ²⁰₁₀Ne
10 → 1s ² 2s ² 2p ⁶
Sodium (Na) ²³₁₁Na
11 → 1s ² 2s ² 2p ⁶ 3s ¹
Magnesium (Mg) ²⁴₁₂Mg
12 → 1s ² 2s ² 2p ⁶ 3s ²
Aluminum (Al) ²⁷₁₃Al
13 → 1s ² 2s ² 2p ⁶ 3s ² 3p ¹
Silicon (Si) ²⁸₁₄Si
14 → 1s ² 2s ² 2p ⁶ 3s ² 3p ²
Phosphorous (P) ³¹₁₅P
15 → 1s ² 2s ² 2p ⁶ 3s ² 3p ³
Sulfur (S) ³²₁₆S
16 → 1s ² 2s ² 2p ⁶ 3s ² 3p ⁴
Chlorine (Cl) ^35.5₁₇Cl
17 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁵
Argon (Ar) ⁴⁰₁₈Ar
18 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁶
Potassium (K) ³⁹₁₉K
19 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁶ 4s¹
Calcium (Ca) ⁴⁰₂₀Ca
20 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁶ 4s²

Orbital diagram
(031102.0c)
Amplitude: The amplitude of a wave is the maximum distance moved by it on either side of the mean position.
Intensity: The intensity of light can be defined as the energy that is transmitted per unit area in one unit of time.
According to the electromagnetic theory, the intensity of a wave is directly proportional to its amplitude, i.e., I  A2
. Since light is also an electromagnetic wave, the intensity of light will depend upon the amplitude of the wave.


SVT031102.1 Go back - Top
SVOENB Sec Formulae-
E=hν
E= mc²
E=hc/λ
λ = c/ν
K.E. = ½ mv²

Mass of one mole of electrons is 9.11 × 10⁻³¹× 6.023×10²³ =5.486×10⁻⁷kg.

Charge of 1 mole of electrons is
1.602×10⁻¹⁹ × 6.023×10²³=9.647×10⁴C.

B2. (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C. (Assume that mass of a neutron = 1.675 × 10^–27 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP. Will the answer change if the temperature and pressure are changed ?
Number of electrons present in 1 molecule of methane (CH₄)
{1(6) + 4(1)} = 10
Number of electrons present in 1 mole i.e., 6.023×10²³ molecules of methane
= 6.023×10²³×10 = 6.023×10²⁴
(a) Number of atoms of ¹⁴C in 1 mole= 6.023×10²³
Since 1 atom of 14C contains (14 – 6) i.e., 8 neutrons, the number of neutrons in 14 g of ¹⁴C is (6.023 × 1023 ) ×8. Or,
14 g of ¹⁴C contains (6.022 × 1023 × 8) neutrons.
Number of neutrons in 7 mg =

Z=6, P=6, A=12; n=A-Z (12-6) = 6

Z=8, P=8, A=16; n=A-Z (16-8) = 8

Z=12, P=12, A=24; n=A-Z (24-12) = 12

Z=26, P=26, A=56; n=A-Z (56-26) = 30

Z=38, P=6, A=88; n=A-Z (88-38) = 50

B4. Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A) (i) Z = 17, A = 35. (ii) Z = 92, A = 233. (iii) Z = 4, A = 9.

B5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber (⊽) of the yellow light. λ=c/v
v=c/λ Where, ν = frequency of yellow light c = velocity of light in vacuum = 3×10⁸ ms^–1
λ = wavelength of yellow light = 580 nm = 580 × 10^–9m
Substituting the values in expression v=c/λ

=1.72×10⁶m

B6. Find energy of each of the photons which (i) correspond to light of frequency 3×10¹⁵ Hz. (ii) have wavelength of 0.50 Å.
(hertz, unit of frequency. The number of hertz (abbreviated Hz) equals the number of cycles per second. )
( i ) Energy (E) of a photon is given by the expression,
E = hv Where, h = Planck’s constant = 6.626 ×10^–34 Js ν = frequency of light = 3×10¹⁵ Hz
Substituting the values in the above expression
E = 6.626 ×10^–34 Js × 3×10¹⁵ Hz
E = 1.988 ×10^–18 J.
B7. Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0×10^–10s.
Period—time taken for one wave cycle to complete.
A wave cycle is a wave's peak to peak or trough to trough, while the wavelength is this measured distance in meters. It is important to note that the distance between peak to peak is the same as the distance between trough to trough. The period of a wave is the amount of time it takes for a wave to complete one cycle.
ν = 1/ period ; =

= 5.0×10^–9s^–1.
(ii) λ = c/ν

3.0×108m–1.

5.0×10–9s–1

=6.0×10^–2m.
(iii) ⊽ = 1/ λ =

1

6.0×10–2m

=16.66 m ^–1.
B8. What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?
λ = 4000pm
= 4000× 10^–12m.
=4.0× 10^–9m.
Energy of one photon = hc/λ
E=

6.626 ×10–34 Js ×3×108m

4.0× 10–9m

= 4.97×10¹⁷ J
Number of photons providing 1 J of energy =1/E

1

4.97×1017J

=2.01×10¹⁶.

B9 A photon of wavelength 4 × 10^–7 m strikes on metal surface, the work function of the metal being 2.13 eV.
Calculate (i) the energy of the photon (eV),
(ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV= 1.6020 × 10^–19J).
(i) Energy of the photon (eV)
E=hν = hc/λ
λ= 4 × 10^–7 m
C=3 × 10⁸m
h= 6.626 ×10^–34 Js

∴   E=

(6.626 ×10–34 Js ) × 3×108m

4 × 10–7m

(ii) Kinetic energy of emission
K.E.= hυ - hυₒ
Hυ = 3.10 eV and hυₒ (work function) =2.13 eV
K.E. = 3.10-2.13 = 0.97eV
(iii) K.E. = ½ mv²
K.E. =.97eV = .97× 1.602×10¹⁹J
K.E. = ½ × 9.11 × 10^-31 kg ×v²
=.97eV = .97× 1.602×10¹⁹J

v² =  

2×K.E.

m

v² =  

2× 0.97× 1.602×1019 kg m2 s-2

9.11×10-31kg

v²=34.12×10¹⁰m²s^-2
v=5.84 ×10⁵ ms^-1
Oz. Is a joule equal to kg m² s^-2?
If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared. Kinetic energy is usually measured in units of Joules (J); one Joule is equal to 1 kg m² s^-2.

B10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol^–1.
I.E. of one sodium atom = hυ= h C/λ
λ = 242nm = 242 × 10⁻⁹m
h= 6.626 ×10^–34 Js
c= 3×10⁸m

∴   E=

(6.626 ×10–34 Js ) × 3×108m

242 × 10−9m

= 8.21 ×10¹⁹J.
I.E per mol= 8.21 ×10¹⁹J. × 6.023×10²³
= 494×10³J
= 494 kJ mol^-1.
B11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second.
λ = 0.57 µm = .57 × 10⁻⁶m
h= 6.626 ×10^–34 Js
c= 3×10⁸m

∴   E=

(6.626 ×10–34 Js ) × 3×108m

.57 × 10−6m

= 3.487 ×10⁻¹⁹J
25 Watt= 25J/s

∴   Number of photons emitted per second=

25

3.487 ×10−19J

= 7.17×10¹⁹s^-1
B12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (νₒ ) and work function (Wₒ ) of the metal.
K.E.= hυ - hυₒ
If the velocity=0, K1.602×10⁻¹⁹Kg =0
0 = hυ - hυₒ
hυ = hυₒ
υ = υₒ
υ=c/λ

∴   =

3×108m

6800 ×10−10m

= 4.41×10¹⁴s^-1
Threshold frequency = 4.41×10¹⁴s^-1
Work fuction = hυₒ
= 6.626 ×10^–34Js × 4.41×10¹⁴s^-1 = 2.92 ×10^-19J .
Oz. What is called work function? Work function is a property of a material, which is defined as the minimum quantity of energy which is required to remove an electron to infinity from the surface of a given solid
B13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?
(031102.11220)
B14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit). B15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state? B16. (i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom. B17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen. B18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10–11 ergs. B19. The electron energy in hydrogen atom is given by En = (–2.18 × 10–18 )/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition? B20. Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s–1. B21. The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength. B22. Which of the following are isoelectronic species i.e., those having the same number of electrons? Na+, K+, Mg2+, Ca2+, S2–, Ar. B23. (i) Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2– (d) F– (ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5 ? (iii) Which atoms are indicated by the following configurations ? (a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1. B24. What is the lowest value of n that allows g orbitals to exist? B25. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron. B26. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element. B27. Give the number of electrons in the species B28. (i) An atomic orbital has n = 3. What are the possible values of l and ml ? (ii) List the quantum numbers (ml and l) of electrons for 3d orbital. (iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f B29. Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n=1, l=0; (b) n = 3; l=1 (c) n = 4; l =2; (d) n=4; l=3. B30. Explain, giving reasons, which of the following sets of quantum numbers are not possible. (a) n = 0, l = 0, ml = 0, ms = + ½ (b) n = 1, l = 0, ml = 0, ms = – ½ (c) n = 1, l = 1, ml = 0, ms = + ½ (d) n = 2, l = 1, ml = 0, ms = – ½ (e) n = 3, l = 3, ml = –3, ms = + ½ (f) n = 3, l = 1, ml = 0, ms = + ½ B31. How many electrons in an atom may have the following quantum numbers? (a) n = 4, ms = – ½ (b) n = 3, l = 0 B32. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit. Go back - Top
B33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum ? B34. Calculate the energy required for the process He+ (g)  He2+ (g) + e– The ionization energy for the H atom in the ground state is 2.18 × 10–18 J atom–1 B35. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long. B36. 2 ×108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm. B37. The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise. B38. A certain particle carries 2.5 × 10–16C of static electric charge. Calculate the number of electrons present in it. B39. In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10–18C, calculate the number of electrons present on it. B40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?
B41. Symbols 35 79Br and 79Br can be written, whereas symbols 79 35Br and 35Br are not acceptable. Answer briefly.
B42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
B43. An ion with mass number 37 possesses one unit of negative charge. If the ion conatins 11.1% more neutrons than the electrons, find the symbol of the ion.
B44. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.
B45. Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.
B46. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser.
B47. Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.
B48. In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10–18 J from the radiations of 600 nm, calculate the number of photons received by the detector.
B49. Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.
B50. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calcualte the frequency of each transition and energy difference between two excited states.
B51. The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
B52. Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant. λ (nm) 500 450 400 v × 10–5 (cm s–1) 2.55 4.35 5.35

B53. The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
Energy of incident radiation= Work function+ K.E. of photoelectrons
Energy of incident radiation, E = hυ = hc/λ
TABLE STYLE="display:inline-table; font-size:1; border-collapse:collapse; vertical-align:middle">

E=

(6.626 ×10–34 Js ) × 3×108m

256.7 × 10–9 m

E= 7.74 ×10^-19 J
TABLE STYLE="display:inline-table; font-size:1; border-collapse:collapse; vertical-align:middle"> Or E=

7.74 ×10-19 J

1.602 × 10–19 m

= 4.83 eV
The Potential energy gives the kinetic energy to the electrons.
Hence K.E. of the electron= 0.35 eV
Work function = 4.83 eV – 0.35eV = 4.48 eV

B54. If the photon of the wavelength 150 pm strikes an atom and one of tis inner bound electrons is ejected out with a velocity of 1.5 × 10⁷ m s^–1, calculate the energy with which it is bound to the nucleus.
Energy of incident photon = hc/λ

Energy of incident photon =

6.626 ×10–34 Js ×3×108ms-1

150×10×-12

m
= 1.325×10^-15J
Energy of ejected photon =

1

2

mv².
Energy of ejected photon =

1 × 9.11 × 10-31kg ×(1.5× 107ms-1)2

2

Energy with which electron was bound to the nucleus= 13.25×10^-16J−1.025×10^-16J
=12.225×10 ^-16J
Or =

12.225×10 -16J

1.602×10−19Kg

= 7.63× 10³eV.
B55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 10¹⁵ (Hz) [1/3²–1/n²] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
B55. Wavelength of transition = 1285 nm λ= 1285 × 10^–9 m (Given)
ν =3.29×10¹⁵(

1

3 2

  − 

1

n 2

)
Since ν =

c

λ

=

3×108m

1285×10–9m

Now υ = 2.33× 10¹⁴ s^-1
By substituting the value of ν in the given expression

B56. Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
r_n B57. Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 106 ms–1, calculate de Broglie wavelength associated with this electron.
B58. Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
B59. If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1, calculate the de Broglie wavelength associated with it.
B60. The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calcualte the wavelength associated with this velocity.
B61. If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.
B62. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists: 1. n = 4, l = 2, ml = –2 , ms = –1/2 2. n = 3, l = 2, ml = 1 , ms = +1/2 Rationalised 2023-24

Oz. What is 1 eV equal to in erg?
electron volt, unit of energy commonly used in atomic and nuclear physics, equal to the energy gained by an electron (a charged particle carrying unit electronic charge) when the electrical potential at the electron increases by one volt. The electron volt equals 1.602 × 10−12 erg, or 1.602 × 10−19 joule.
Electrons have an electric charge of −1, and their mass is approximately about 1/2000 the mass of a neutron or proton. Electron charge is usually denoted by the symbol e. It is a fundamental physical constant that is used to express the naturally occurring unit of electric charge, which is = 1.602 × 10-19 coulomb.
What is the relation between energy and electric charge?
Voltage is measured in volts (V) where 1 volt is 1 joule of energy per coulomb. Therefore, if something has a voltage of 10 volts it must have 10 joules of energy transferred per unit of charge passed. (This is the same as saying 10 joules per coulomb, since 1 coulomb is 1 unit of charge passed.)
1. Arrange s, p and d sub-shells of a shell in the increasing order of effective nuclear charge (Zeff ) experienced by the electron present in them.

1. The increasing order of effective nuclear charge (Zeff) experienced by the electron present in them is s > p > d because s is close to the nucleus, and it will shield the nuclear charge more effectively.v

2. Show the distribution of electrons in the atoms of Oxygen, Nitrogen, Phosphorus and Argon using an orbital diagram.

2. The distribution of electrons using an orbital diagram is given below-

3. The electronic configuration of each element is decided by the Aufbau principle which states that the electrons fill orbitals in order of increasing energy levels. Complete the given equation. The equation is
1s<2s<2p<…<3p<4s<…<4p<5s<…<
5p<…<4f<…<6p<7s<…<6d<…
1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<
5p<6s<4f<5d<6p<7s<5f<6d<7p

4. Write names of p orbitals - Names of p orbitals are
p_x, p_y, p_z

5. Write Names of d orbitals -
Names of d orbitals -
d _xy, d _yz, d _zx, d_(x²-y²), d_z²

6. Write atomic number, atomic mass and electronic configuration of first 20 elements of modern periodic table (Long fom).
(An example of N is given below)
Nitrogen (N):
N →At No 7; At Wt. 14;
Electronic Configuration: N
7 → 1s²2s² 2p³

Hydrogen (H) H
1 → 1s ¹

Helium (He) He
2 → 1s ²

Lithium (Li) Li
3 → 1s²2s¹

Beryllium (Be) Be
4 → 1s ²2s²

Boron (B) B
5 → 1s ² 2s ² 2p ¹

Carbon (C) C
6 → 1s ² 2s ² 2p ²

Nitrogen (N) N
7 → 1s ² 2s ² 2p ³

Oxygen (O) O
8 → 1s ² 2s ² 2p ⁴

Fluorine (F) F
9 → 1s ² 2s ² 2p ⁵

Neon (Ne) Ne
10 → 1s ² 2s ² 2p ⁶

Sodium (Na) Na
11 → 1s ² 2s ² 2p ⁶ 3s ¹

Magnesium (Mg) Mg
12 → 1s ² 2s ² 2p ⁶ 3s ²

Aluminum (Al) Al
13 → 1s ² 2s ² 2p ⁶ 3s ² 3p ¹

Silicon (Si) Si
14 → 1s ² 2s ² 2p ⁶ 3s ² 3p ²

Phosphorous (P) P
15 → 1s ² 2s ² 2p ⁶ 3s ² 3p ³

Sulfur (S) S
16 → 1s ² 2s ² 2p ⁶ 3s ² 3p ⁴

Chlorine (Cl) Cl
17 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁵

Argon (Ar) Ar
18 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁶

Potassium (K) K
19 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁶ 4s¹

Calcium (Ca) Ca
20 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁶ 4s²

7. Write atomic number and electronic configuration of elements with atomic number 21 to 30 20 of modern periodic table (long from).
7. Scandium (Sc) 21 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁶ 4s² 3d ¹

Titanium (Ti) Ti
22 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁶ 4s² 3d²

Vanadium (V) 23 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁶ 4s² 3d ³

Chromium (Cr) 24 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁶ 4s ¹ 3d ⁵

Manganese (Mn) 25 → 1s ² 2s ² 2p ⁶ 3s ² 3p⁶ 4s² 3d⁵

Iron (Fe) 26 → 1s ² 2s ² 2p ⁶ 3s ² 3p ⁶ 4s² 3d ⁶

Cobalt (Co) 27 → 1s ² 2s ² 2p ⁶ 3s ² 3p ⁶ 4s² 3d⁷

Nickel (Ni) 28 → 1s ² 2s ² 2p ⁶ 3s ² 3p ⁶ 4s² 3d⁸

Copper (Cu) 29 → 1s ² 2s ² 2p ⁶ 3s ² 3p ⁶ 4s1 3d¹⁰

Zinc (Zn) 30 → 1s ² 2s ² 2p ⁶ 3s ² 3p ⁶ 4s ² 3d¹⁰

031102.2

8. The arrangement of orbitals on the basis of energy is based upon their (n+l) value. Lower the value of (n+1), the lower is the energy. For orbitals having the same values of (n+l), the orbital with a lower value of n will have lower energy.
I. Based upon the above information, arrange the following orbitals in the increasing order of energy.
(a) 1s, 2s, 3s, 3p
(b) 4s, 3s, 3p, 4d
(c) 5p, 4d, 5d, 4f, 6s
(d) 5f, 6d, 7s, 7p
9. The arrangement of orbitals on the basis of energy is based upon their (n+l) value. Lower the value of (n+1), the lower is the energy. For orbitals having the same values of (n+l), the orbital with a lower value of n will have lower energy.
Based upon the above information, solve the questions given below :
(a) Which of the following orbitals has the lowest energy?
4d, 4f, 5s, 5p
(b) Which of the following orbitals has the highest energy?
5p, 5d, 5f, 6s, 6p

8.
1s < 2s < 3s < 3p
3s < 3p < 4s < 4d
4d < 5p < 6s < 4f < 5d
7s < 5f < 6d < 7p
9.
5s has the lowest energy.
5f has the highest energy.

10. What is the difference between the terms orbit and orbital?

10.

Orbit	Orbitals
An orbit is the simple planar representation of an electron.	An orbital refers to the dimensional motion of an electron around the nucleus in a three-dimensional motion.
It can be simply defined as the path that gets established in a circular motion by revolving the electron around the nucleus	An orbital can simply be defined as the space or the region where the electron is likely to be found the most.
The shape of molecules cannot be explained by an orbit as they are non-directional by nature.	The shapes of the molecules can be found as they are directional by nature.
An orbit that is well-defined goes against the Heisenberg principle	An ideal orbital agrees with the theory of Heisenberg’s Principles.

11. Wavelengths of different radiations are given below:
λ (A) = 300 nm
λ (B) = 300 μm
λ (C) = 3 nm
λ (D) =30 Å
Arrange these radiations in the increasing order of their energies.

11.
Since, E ∝ 1/ λ
Therefore, λ (B) < λ (A) <λ (C) = λ (D)
300μm < 300 nm < 3 nm = 30Å
12. What do you understand by degenerate orbitals? Explain degree of degeneracy of various orbitals.

. Electron orbitals having the same energy levels are called degenerate orbitals. p orbital has 3 degenerate orbitals. All three have the same energy levels. Each orbital is first assigned with only one electron. The second electron will be of opposite spin. Each orbital is filled and the total is six electrons.
p sub level - Orbitals in the 2p sublevel are degenerate orbitals – Which means that the 2px, 2py, and 2pz orbitals have the exact same energy, as illustrated in the diagram provided below.
d sub level - Similarly, the 3p_x, 3p_y, and 3p_z are degenerate orbitals. And at the 3d energy level, 5 orbitals are are degenerate orbitals with the same energy.
Degeneracy – The total number of different states of the same energy is called degeneracy. It is also known as the degree of degeneracy.
The degree degeneracy of p orbitals is 3
The degree degeneracy of d orbitals is 5
The degree degeneracy of f orbitals is 7

13. What is the electron configuration of Ni and Ni⁺⁺?

13 . Ni atom’s electronic configuration is
[Ar]¹⁸ 3d⁸ 4s². It will lose two electrons from the 4s orbital to form Ni ²⁺ with the electronic configuration
[Ar]¹⁸ 3d⁸ 4s⁰.

031102.3
14. MCQs 1. An orbital is having n = 3 and l = 1, the number of electrons it can occupy are (a) 6 (b) 14 (c) 10 (d) 2 D 2. The force that allows electrons to move around the nucleus is (a) nuclear force (b) weak nuclear force (c) electrostatic force (d) gravitational force C *3. According to Bohr’s atomic model, the radius of the orbit is directly proportional to (a) n2 (b) 1/n2 (c) 1/n (d) n A 4. How many electrons can exist with the principal quantum number’s value of 4? 16 4 32 12 Answer: c) 32 Explanation: The number of orbitals within an orbit is n2. But as each orbital can accommodate 2 electrons, the number of electrons that can exist with the “n” as the principal quantum number is 2n2. So here 2n2=2(4)2=2(16)=32. *5. Write the values for l, n, and m for Ψ 3,1,0. 1, 3, 0 3, 1, 0 0, 3, 1 1, 0, 3 Answer: a) 1, 3, 0 Explanation: The representation of the Schrodinger wave function is given by Ψn,l,m. Therefore by comparing Ψn,l,m and Ψ3,1,0 we get that n=3, l=1, and m=0. Here n, l, and m are principal, azimuthal, and magnetic quantum numbers respectively. 6. Who discovered the neutron? a. J.J Thomson b. Rutherford c. Chadwick d. Priestley 7. A sub-shell with n = 6, l = 2 can accommodate how many maximum electrons? a. 12 electrons b. 10 electrons c. 36 electrons d. 72 electrons Answer: b) 10 electrons Explanation: n = 6, l = 2, therefore 6d will have 5 orbitals. Hence maximum of 10 electrons can be accommodated as each orbital can have a maximum of 2 electrons. *9. In Bohr’s hydrogen atom model, the ratio of kinetic energy to the total energy of the electron in a quantum state n is 1 -1 2 -2 *10. In a Hydrogen atom, the energy of the first excited state is – 3.4 eV. What is the KE of the same orbit of the Hydrogen atom? -13.6 eV 13.6 eV 3.4 eV 6.8 eV Answer: c) 3.4 eV Explanation: For a hydrogen atom, the kinetic energy is the negative of the total energy. The potential energy is twice the total energy. The first excited state energy of orbital = -3.4 eV and kinetic energy of same orbital = -(-3.4 eV) = 3.4 eV Therefore, the kinetic energy of the same orbit of a hydrogen atom is 3.4 eV. *11. The ionization enthalpy of a hydrogen atom is 1.312 × 106 J mol-1. The energy to excite the electron from n = 1 to n = 2 is – 6.56 × 105 J mol-1 8.51 × 105 J mol-1 7.56 × 105 J mol-1 9.84 × 105 J mol-1 Answer: d) 9.84 × 105 J mol-1 Explanation: The energy required when an electron shifts from n = 1 to n = 2 E2 = −(1.312 × 106 × (1)²)/(2²) = −3.28 × 105 J mol-1 E1 = −1.312 × 106 J mol-1 ΔE = E2 − E1 = −3.28 × 105−(−13.2 × 106) ΔE = 9.84 × 105 J mol-1 12. According to Bohr’s atomic model, the angular momentum of orbits is multiple of (a) h/2𝜋 (b) 2𝜋/h (c) 2h/𝜋 (d) 𝜋/2h A 13. When electrons move from a lower energy level to a higher energy level, energy is (a) absorbed (b) emitted (c) both (a) and (b) (d) none of the above A *14. Radius of the hydrogen atom on going to the first excited state is _____ of Bohr’s radius. (a) double (b) half (c) 4 times (d) same C 15. The energy of each orbit is (a) same (b) fixed (c) changes with time (d) none of the above B 16. How many atomic orbitals are present in the fourth energy level of an atom? (a) 32 (b) 8 (c) 16 (d) 4 C *17. A subshell of an atom contains a maximum of _____ electrons. (a) 4l – 2 (b) 4l + 2 (c) 2l + 1 (d) 2n2 B 18. When electrons move from the higher energy level to a lower energy level, energy is (a) absorbed (b) emitted (c) both (a) and (b) (d) none of the above B 19. Who discovered the electron? a. Rutherford b. Chadwick c. Thomson d. Goldstein Answer. (c) Thomson Explanation: J.J Thomson discovered electrons in 1897 by performing the cathode ray experiment. Electrons are the smallest of the particles that make up an atom, and they carry a negative charge. The number of protons and electrons is equal in a neutral atom. 20. The principal quantum number describes ____ energy and size of the orbit the shape of the orbital spatial orientation of the orbital the spin of the electron Answer: a) energy and size of the orbit Explanation: Among the four quantum numbers, the principal quantum number describes the size and energy of the orbit. It is represented by the symbol “n”. For shells, K, L, M, N, and O, n are given by 1, 2, 3, 4, and 5. 13. What was the source of alpha particles in the Rutherford scattering experiment? Hydrogen nucleus Argon nucleus Helium nucleus None of these Answer. (c) Helium nucleus Explanation: Alpha particle consists of two neutrons and two protons and is thus identical to the nucleus of a helium atom. It has no charge. The mass of the particle is approximately the sum of the mass of the proton and neutron. Its charge is equal to that of 2 protons. 14. According to Aufbau's principle, which of the following orbital should be filled first? 5d 4p 3p 2s Answer: d) 2s Explanation: As per Aufbau's principle, the orbital or subshell with the lowest energy should be filled first. The ascending order of orbitals energy is given by 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 6p, 7s. So 2s orbital should be filled first. 15. What property of an element determines its chemical behaviour? Size of an element Valency of an element The molar mass of the element None of these Answer. (b) Valency of an element Explanation: An element's valence is the number of hydrogen atoms that can combine with or replace (directly or indirectly) one of the element's atoms. Oxygen, for instance, has six valence electrons but its valence is 2. Some elements may have more than one power combination (or valence), while others may have only one. 16. Which of the following does not match the characteristics of an Isotope? Isotopes of some elements are radioactive Isotopes are the atoms of different elements Isotopes differ in the number of neutrons Isotopes have similar chemical properties Answer. (b) Isotopes are the atoms of different elements Explanation: Isotopes are members of a family of an element that all have the same number of protons but different numbers of neutrons. The number of protons in a nucleus determines the element's atomic number on the Periodic Table. 11. What is the order of the radius of an electron orbit in a hydrogen atom? a) 10-8 m b) 10-9 m c) 10-11 m d) 10-13 m Answer: c Explanation: The radius of an electron orbit in a hydrogen atom is of the order of 10-11 m. It is equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state. 12. What will be the longest wavelength in the Balmer series of hydrogen spectrum? a) 6557 × 10-10 m b) 5557 × 10-10 m c) 9557 × 10-10 m d) 1557 × 10-10 m Answer: a 13. A hydrogen atom in its ground state absorbs 10.2 eV of energy. What is the orbital angular momentum is increased by? a) 4.22 × 10-3 Js b) 2.11 × 10-34 Js c) 3.16 × 10-34 Js d) 1.05 × 10-34 Js Answer: d Explanation: Increase in angular momentum = h2π. h/2π = 6.6 × 10−34× 3.14 x 1 / 2 h/2π = 1.05 × 10-34 Js. 1eV = 1.602×10-19 J 1 J = 6.2415×1018 eV 14. Which of the following is true regarding the Bohr model of atoms? a) Assumes that the angular momentum of electrons is quantized b) Uses Faraday’s laws c) Predicts continuous emission spectra for atoms d) Predicts the same emission spectra for all types of atoms View Answer Answer: a Explanation: Bohr model of atoms assumes that the angular momentum of electrons is quantized. The atom is held between the nucleus and surroundings by electrostatic forces. The other options are not valid. 15. Hydrogen atoms are excited from ground state to the state of principal quantum number 4. Then, what will be the number of spectral lines observed? a) 3 b) 6 c) 5 d) 2 Answer: d Explanation: n = 4. The number of spectral lines emitted = n(n−1)2. n(n−1)2=(4×3)6 = 2 16. When a hydrogen atom is in its first excited level, what is the relation of radius and Bohr radius? a) Twice b) 4 times c) Same d) Half Answer: b Explanation: For the first excited level, n = 2. r2 = (2)2r0 = 4r0. So, when a hydrogen atom is in its first excited level, its radius is 4 times of the Bohr radius. Ques 1. Ques 2. Ques 3. Ques 4. Ques 5. Ques 6. Ques 7. No two electrons have the same set of all four quantum numbers. True False Answer: a) True Explanation: Yes, no two electrons have the same set of all four quantum numbers. This is explained by Pauli’s exclusion principle. At most electrons can have all 3 quantum numbers the same as they are in the same orbital. But the spin quantum number’s values are different. Ques 8. Which of the two will be chemically more reactive, Sulphur(S) with atomic number 16 or Chlorine (Cl) with atomic number 17? Chlorine Sulfur Both are equally reactive Can’t say Answer. (a) Chlorine Ques 9. Which of the following elements does not exhibit electrovalency? Sodium Calcium Carbon Chlorine Answer. (c) Carbon Explanation: Elements that lose electrons show positive electrovalency and those which gain electrons show negative electrovalency. For example, in the formation of sodium chloride (Na+Cl–), the electrovalency of sodium (Na) is +1, while that of chlorine (Cl) is –1. Ques 10. What is the shape of the orbital, whose “l” is 1? Spherical Dumbbell Double dumbbell Complex Answer: b) Dumbbell Explanation: The azimuthal quantum number is given by “l”. When l =0, 1, 2, and 3, they are s-orbital, p-orbital, d-orbital, and f-orbital respectively. The shapes of s-orbital, p-orbital, d-orbital, and f-orbital are Spherical, Dumbbell, Double dumbbell, and Complex respectively. Ques 11. Which scientist gave the concept of fixed energy levels around the nucleus? Ernest Rutherford Neils Bohr J.J.Thomson None of these Answer. (b) Neils Bohr Ques 12. Which of the following pairs are isobars? 17Cl35 & 17Cl37 18Ar40 & 20Ca40 6C12 & 6C14 None of these Answer. (b) 18Ar40 & 20Ca40 Explanation: Isobar is any member of a group of atomic or nuclear species all of which have the same mass number—that is, the same total number of protons and neutrons. Thus, they are isobars. Ques 13. Which radioactive element is used in the treatment of cancer? Iodine-131 Uranium-234 Plutonium-239 Cobalt-60 Answer. (d) Cobalt-60 Explanation: Radioactive elements are made up of atoms whose nuclei are unstable and give off atomic radiation as part of a process of attaining stability. The emission of radiation transforms radioactive atoms into another chemical element, which may be stable or maybe radioactive such that it undergoes further decay. Ques 14. What is the magnetic quantum number of the orbital 2pz? 1 ±1 -1 0 Answer: d) 0 Explanation: The magnetic quantum of orbital ranges from -( l–1) to l–1. Its symbol is given by m. For 2p orbital, there are 3 magnetic quantum numbers namely -1, 0, and 1. For 2pz orbital it's 0, taking that z is the internuclear axis. Ques 15. Total number of nodes for 3d orbital is ________ 3 2 1 0 Answer: b) 2 Explanation: Total number of nodes includes angular and radial nodes. Angular nodes and radial nodes are given by the formula n–l -1 and l respectively. So the total number of nodes is n–l -1+l=n–1. For 3d orbit, “n” is 3, so t total number of nodes is 3–1=2. Ques 16. Which of the following sets of quantum numbers is not valid? n=5, l=2, m=0, s=½ n=1, l=2, m=0, s=½ n=5, l=3, m=2, s=½ n=5, l=2, m=0, s=-½ Answer: b) n=1, l=2, m=0, s=½ Explanation: The set of quantum numbers n=1, l=2, m=0, s=1/2, is not valid because the value of the azimuthal quantum number should lie only in between 0 and n-1, where n is a principal quantum number. So the above set of quantum numbers is not valid. Ques 17. Ques 18. Ques 19. Ques 20. Ques 21. Ques 22 Ques 23. Ques 24. The electrons of the same orbitals are identified by – Spin quantum number Azimuthal quantum number Magnetic quantum number Principal quantum number Answer: a) Spin quantum number Explanation: Electrons having the same orbital are identified by Spin quantum number. Spin Quantum number has two values +1/2 or -½. Spin quantum number explains the direction through which an electron spins in an orbital. An electron in the same orbitals must have opposite spins. Ques 25. An element in the ground state that has 13 electrons in its M-shell is – Nickel Copper Iron Chromium Answer: d) Chromium Explanation: As it is in M shell, n = 3 Number of electrons in M shell = 13 → 3s23p63d5 The electronic configuration is (1s2) (2s2 2p6) (3s2 3p6 3d5) (4s1) The element is therefore chromium. Ques 26. Out of the following element, which one has the least number of electrons in its M-shell? Sc Mn K Ni Answer: c) K Explanation: K = 19 = 1s²2s22p63s23p6s1 3s23p6 = m-shell Hence, k has only 8 electrons in the M shell. Ques 27. What does the magnetic quantum number signify? Shape of orbitals Nuclear Stability Orientation of orbitals Size of orbitals Answer: c) Orientation of orbitals Explanation: The magnetic quantum number signifies the orientation of orbitals Ques 28. A gas absorbs a 355 nm photon and emits two wavelengths. If the first emission is at 680 nm, the other is at: 518 nm 325 nm 1035 nm 743 nm Answer: d) 743 nm Explanation: From the Law of Conservation of energy, the energy of the absorbed photon is equal to the combined energy of two emitted photons. ET = E1 + E2 – (1) Here, E1 is the Energy of the first emitted photon emitted and E2 is the Energy of the second emitted photon. Now, E = (hc)/ (λ) – (2) Substituting values from (2) in (1) we get (hc/λT) = (hc)/ (λ1) + (hc)/ (λ2) Therefore, on solving the equation, we get λ2 = 742.77nm Ques 29. The increasing order for the e/m values for the following are – n, p, e, α n, p, α, e e, p, n, α n, α, p, e Answer: d) n, α, p, e Explanation: The charge to mass ratio (e/m) for the following is as follows – neutron, n = (0/1) = 0 α− particle = (2/4) = 0.5 Proton, p = (1/1) = 1 Electron, p = (1/1837) = 1837. Ques 30. In the ground state, the electronic configuration of the silver atom is – [Kr]4d95s2 [Kr]4d105s1 [Xe]4f145d106s1 [Kr]3d104s1 Answer: b) [Kr]4d105s1 Explanation: The electronic configuration of Ag in the ground state is [Kr]4d105s1. SVT031102.1
SVT031102.2
SVT031102.3
SVT031102.002 (Test) Topics 1. Properties of cathode ray and anode rays. 2. Rutherford model of atom. 3. Bohr's model of atom with limitations. 4. Quantum numbers 5. Electronic configuration of 1-20 elements of periodic table. 6. Aufbau principle, Bohr bury's rule, Pauli's exclusion principle, Hund's rule. 7. Atomic and mass numbers of elements 1-30. 8. Isotopes, isobars, isotones. 9. Hydrogen spectrum 10. Mass and charge of sub atomic particles electrons protons and neutrons and values of Rydberg constant planks constant Class 11 Chemistry Test Structure of atom 1. Write the symbol, atomic number and mass number of 1-30 elements. https://rksvirtuals.com/vdos/rksvirtuals5419.mp4 2. Give electronic configuration of following- He, Na, K, Ar, S, P, N, C, O 3. Explain Aufbau principle, Bohr bury's rule, Pauli's exclusion principle and Hund's rule with examples. 4. Define the following with one example each- Isotopes, isobars and isotones. 5. Write the values of the following- Mass of electron, proton and neutron Charge on electron in coulombs 6. Fill in the blanks a. Rydberg constant = ……………. b. Plancks constant = …………….   1. Write formulae of the following ammonium chloride, ammonium sulphate, aluminium hydroxide, aluminium oxide, aluminium sulphate, aluminium chloride, barium chloride, barium sulphate, calcium hydroxide, calcium chloride, calcium carbonate, copper nitrate, ferric sulphate, magnesium oxide, magnesium hydroxide, potassium sulphate, potassium bromide, Potassium oxide, sodium hydroxide, sodium carbonate, sodium sulphate, sodium oxide, silver sulphide, sodium hydroxide, zinc nitrate, 2. Write electronic configuration of the following - Nitrogen Oxygen Fluorine Neon Sodium Magnesium Aluminum Silicon Phosphorous (Sulfur Chlorine Argon Potassium Calcium Scandium Titanium (Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc. Video section
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MCQs SV301102.0012018
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SV031102001. 4. Given below are two statements:
Statement I : According to Bohr's model of hydrogen atom, the angular momentum of an electron in a given stationary state is quantised.
Statement II : The concept of electron in Bohr's orbit, Violates the Heisenberg uncertainty principle. In the light of the above statements, choose the most appropriate answer from the options given below:
(a) Statement I is correct but Statement II in incorrect. (b) Both Statement I and Statement IIare incorrect. (c) Both Statement I and Statement IIare correct. (d) Statement I is incorrect but Statement II is correct.
C
SV031102002.12. Given below are two statements:
Statement -I:According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus.
Statement-II: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number.
In the light of the above statements, choose the most appropriate answer from the options given below:
(a) Statement-I is false but statement-Il is true. (b) Statement -I is true but statement-Il is false. (c) Both statement I and statement-Il are false. (d) Both statement I and statement-Il are true.
A
SV031102003.13.
Given below are two statements.
Statement-I : Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom.
Statement-II : Bohr's model of hydrogen atom contradicts Heisenberg's uncertainty principle.
In the light of the above statements, choose the most appropriate answer from the options given below.
(a) Both statement-I and statement-Il are true. (b) Statement -I is false but statement-II is true. (c) Both statement-I and statement-Il are false. (d) Statement-I is true but statement-Il is false
A
SV031102004.14. If the Thomson model of the atom was correct, then the result of Rutherford's gold foil experiment would have been,
(a) alpha-particles pass through the gold foil deflected by small angles and with reduced speed. (b) alpha -particles are deflected over a wide range of angles. (c) All alpha -particles get bounced back by 180°. (d) All of the alpha -particles pass through the gold foil without decrease in speed.
A
SV031102005. 15. The number of neutrons and electrons, respectively, present in the radioactive isotope of hydrogen is
(a) 2 and l (b) 2 and 2 (c) 3 and l (d) l and l
A
SV031102006.19. The region in the electromagnetic spectrum where the Balmer series lines appear is
(a) microwave (b) ultraviolet (c) visible (d) infrared.
C
SV031102007.41. The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with
(a) Paschen series (b) Brackett series (c) Lyman series (d) Balmer series.
D
SV031102008. 46. A gas absorbs a photon of 355 nm and emits at two wavelengths. Ifone of the emission is at 680 nm, the other is at
(a) 1035 nm (b) 325 nm (c) 743 nm (d) 518 nm
C
SV031102009. 69. When the excited electron of a H atom from n = 5 drops to the ground state, the maximum number of emission lines observed are ............
10
SV031102010.65. If wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series is .......... nm. (Nearest integer). 492
SV031102011.89. Maximum number of electrons that can be accommodated in shell with n =4 are
(a) 72 (b) 16 (c) 32 (d) 50
C
SV031102012.90. Arrange the following orbitals in decreasing order of energy.
A. n = 3, I = 0, m = 0 B. n = 4, I = 0, m = 0 C. n = 3, 1= 1, m = 0 D. n = 3, 1= 2, m = 1 (a) D > B > C > A (b) B> D > C > A (c) A> C > B > D (d) D> B > A > C
A
SV031102013.92. Given below are two statements. One is labelled as
Assertion A and the other is labelled as Reason R.
Assertion A : Energy of 25 orbital of hydrogen atom is greater than that of 25 orbital of lithium.
Reason R : Energies of the orbitals in the same subshell decrease with increase in the atomic number.
In the light of the above statements, choose the correct answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
A
SV031102014.97. The number of radial and angular nodes in 4d orbital are, respectively
(a) 1 and 2 (b) 3 and 2 (c) 1 and 0 (d) 2 and 1
A
SV031102015.103. The orbital having two radial as well as two angular nodes is
(a) 4/ (b) 4d (c) 5d (d) 3p
C
SV031102016.104. The number of subshells associated with n = 4 and m = -2 quantum numbers is
(a) 8 (b) 4 (c) 16 (d) 2
D
SV031102017.105. The correct statement about probability density (except at infinite distance from nucleus) is
(a) it can be zero for Is-orbital (b) it can be negative for 2p-orbital (c) it can be zero for 3p-orbital (d) it can never be zero for 2s-orbital.
C
SV031102018.122. The number of correct statements from the following is __ .
i. For Is orbital, the probability density is maximum at the nucleus. ii. For 2s orbital, the probability density first increases to maximum and then decreases sharply to zero. iii. Boundary surface diagrams of the orbitals encloses a region of 100%probability of finding the electron. iv. p and d-orbitals have 1 and 2 angular nodes respectively. v. Probability density of p-orbital is zero at the Nucleus
(a) All are correct (b) Two are correct (c) All are wrong (d) 3 are correct
D