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Chapter, Chapter, , 2, , Atomic, Atomic, Structure, Structure, Remember, Before beginning this chapter, you should be able, to:, • understand basic concept of matter and atom., • identify presence of subatomic particles., , Key Ideas, After completing this chapter, you should be, able to:, • understand the concept of divisibility of atom and, the characteristics of subatomic particles., • study the stability of nucleus and nuclear reactions., • , conceptuliaze the model of atom based on the, discovery of electrons., • , learn the planetary model of atom based on the, concept of nucleus., • , study the concept of stationary orbits based on, quantisation of energy and approach towards the, modern atomic model., F I G U R E 2 . 1 Figure Caption

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2.2, , Chapter 2, , INTRODUCTION, The concept that atoms are the fundamental building blocks of matter dates back to very ancient, times. However, the ideas regarding atoms of those times had no experimental evidence and, remained as mere speculation. Therefore, these ideas had to lay dormant for a long period until, John Dalton proposed his atomic theory on the basis of certain observations and experimental, results. The basic principle of his theory was that it regarded an atom as the ultimate particle of, matter., Dalton’s atomic theory has been successful in giving a convincing explanation for the various laws, of chemical combination, such as, the law of conservation of mass, the law of definite proportions, and the law of multiple proportions. However, Dalton’s idea that the atom is an indivisible particle, of matter has been disproved by later discovery of radioactivity. Several series of experiments on, radioactivity which were carried out later proved the presence of various subatomic particles in, an atom. Atoms are found to be mainly composed of three types of fundamental particles, namely,, positively-charged protons, neutral particles known as neutrons and negatively-charged electrons., The discovery of these fundamental particles paved the way for further research on the internal, structure of an atom which obviously explains the enormous diversity of chemistry involved in a, wide range of chemical reactions., , DISCOVERY OF FUNDAMENTAL PARTICLES, The electron was the first fundamental particle that was discovered. The credit for the discovery of, the electron goes to J.J. Thomson based on his experiments carried out in a discharge tube., Sir William Crookes was the first scientist who designed the discharge tube which was called, Crooke’s discharge tube or cathode ray tube. It is a long glass tube having two metal plates connected, to the oppositely charged poles of a battery. The pressure inside the discharge tube can be adjusted, by means of an exhaust pump., This discharge tube was later slightly modified by J.J. Thomson. When high voltage (HV) was, applied between the cathode and the anode with a small hole at the centre of a partially evacuated, tube at a pressure of 0.01 mm of Hg, a bright spot of light was formed on the zinc sulphide screen, kept at the opposite end of the discharge tube. This was caused by the rays which originated from, the cathode called cathode rays., Anode, , Cathode, , Cathode rays, Exhaust pump, _, , +, , HV source, , FIGURE 2.1 Cathode ray tube, , J. J. Thomson conducted some experiments with a discharge tube for studying the properties of, cathode rays.

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Atomic Structure, , Cathode anode, –, , Gas at low, pressure, , +, , Bright spot, Fluorescent, material, , Cathode, _ + Perforation, HV source, , FIGURE 2.2 J.J. Thomson’s cathode ray tube, , TABLE 2.1 Experiments involved in discovery of fundamental particles, Experiments, Placing a small object in between the cathode and anode, Shadow of, object, , Metal object, , Properties based on observation, Formation of a shadow of the object on the, opposite side of the cathode, Cathode rays travel in straight lines, , •, Anode, , Cathode, •, , +, , –, , HV source, , Placing a light paddle wheel between cathode and anode, Light paddle wheel, , Rotation of light paddle wheel. Small particles, having mass and kinetic energy, , +, , –, •, , •, , •, Anode, , Cathode, , HV source, , Bending of rays towards the positive plate, , Passing through electric field, Cathode anode, , Fluorescent, material, (ZnS), , –, , – +, HV, , Negatively-charged particles, , +, • Bright spot, Schematic diagram, , Passing through magnetic field applied perpendicular to the, path of the cathode rays, , Deflection perpendicular to the applied, magnetic field, , Cathode Anode, S, , Fluorescent, material, (ZnS), , N, –, , +, HV, , Schematic diagram, , •, , Bright spot, , The above experiments were carried out with different, gases in the discharge tube., , No change in properties, The properties do not depend on the nature, of gas taken in the discharge tube., Specific charge (e/m value) remains, the same., , 2.3

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2.4, , Chapter 2, , The discovery of negatively-charged electron was later followed by the experiment conducted, by Robert Millikan in 1909 to determine the quantity of charge on an electron., , Millikan’s Oil Drop Experiment, Some fine oil droplets were allowed to be sprayed into the chamber by an atomiser. The air in the, chamber was subjected to ionisation by X-rays. The electrons produced by the ionisation of air, attached themselves to the oil drops. When sufficient amount of electric field is applied which can, balance the gravitational force acting on an oil drop, the drop remains suspended in the air., From this experiment, Millikan observed that the smallest charge found on them was approximately, 1.59 × 10–19 coulombs and the charge on each drop was always an integral multiple of that value., On the basis of this observation, he concluded that 1.59 × 10–19 coulombs is the smallest possible, charge and considered that value as the charge of the electron., Oil spray, , Atomiser to produce, oil droplets, , (+), , Microscope, , X-rays produce, charges on, the oil drops, , Electrically charged, plates, , (–), , FIGURE 2.3 A schematic representation of the apparatus used by Millikan to determine, the charge of an electron, , DISCOVERY OF PROTONS, The presence of positively-charged particles in an atom has been predicted by Goldstein based on, the electrical neutrality of an atom. The discovery of proton by Goldstein was done on the basis of, the cathode ray experiment conducted by using a perforated cathode., Just like the cathode rays, some rays were found to emanate from an anode. These are called anode, rays or canal rays., Anode rays were found as a stream of positively-charged particles in contrast to the cathode rays. When, hydrogen gas is taken in a discharge tube, these positively-charged particles were found to be protons., Cathode rays, from cathode, , Positive rays, from anode, –, , +, , –, , +, +, Anode, , Positive, rays, +, +, +, , +, –, , +, , Red glow, , –, , Perforated cathode, –, , +, HV source, , –, , FIGURE 2.4 Discovery of Protons, , Properties of Anode Rays, 1. Anode rays travel in straight lines., 2. A,  node rays possess positive charge since they were found to deflect towards negatively-charged, electrodes.

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Atomic Structure, , 3. The properties of anode rays depend upon the nature of the gas taken in the discharge tube., 4. The mass of the particles was same as the atomic mass of the gas inside the discharge tube., The discovery of electrons and protons as subatomic particles inside the atom lead to the conception, of atomic models which depict the arrangement of fundamental particles in an atom., Various atomic models have been proposed by different scientists, like, J.J. Thomson, Rutherford,, Bohr and Sommerfeld., , Thomson’s Atomic Model, J.J. Thomson proposed his atomic model soon after his discovery of electrons as listed hereunder., 1. A,  n atom contains negatively-charged particles called electrons embedded uniformly throughout, a thinly spread positively-charged mass., 2. S ince the atom is electrically neutral, the total negative charge of electrons is balanced by the, total positive charge., Electrons, , Positivelycharged, sphere, , FIGURE 2.5 Thomson’s atomic model, , Thomson’s model of an atom is popularly known as ‘plum pudding model’ or ‘apple pie model’, or ‘watermelon model’., , Validity of Thomson’s Model, Thomson’s model could successfully explain the electrical neutrality of atom. However, it failed, to explain how the positively-charged particles are shielded from the negatively-charged electrons, without getting neutralised., , EXAMPLE, Write different isotopes of oxygen, carbon and chlorine., SOLUTION, The isotopes of oxygen are: 8 O16 , 8 O17 , 8 O18 ., The isotopes of chlorine are:, , 17, , Cl 35 ,, , 17, , Cl 37 ., , The isotopes of carbon are: 6 C12 , 6 C13 , 6 C14 ., , 2.5

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2.6, , Chapter 2, , EXAMPLE, What was the basis for the proposal of Dalton’s atomic theory?, SOLUTION, Laws of chemical combination, such as, the law of conservation of mass, the law of definite proportions, and the law of multiple proportions, were the basis for the proposal of Dalton’s atomic theory., EXAMPLE, (i), (ii), (iii), , What are A, B, C, D, and E in the given figure?, What is the purpose of C?, Explain the role of D in the phenomenon taking place in the, discharge tube., , A, , E, , B, D, , C, , SOLUTION, (i), (ii), (iii), , A is cathode, B is perforated anode, C is suction pump, D is zinc sulphide screen and E, is cathode ray., C is suction pump which can help in reducing the pressure in the discharge tube., D is zinc sulphide screen. Zinc sulphide is a fluorescent material. When cathode rays, strike the zinc sulphide screen, bright spots are formed on the screen., , EXAMPLE,  n atom of an element is represented as ZXA. After the emission of a β-particle, another element, A, Y is formed. Represent Y with atomic number and mass number., SOLUTION, z, , XA → z+1YA + β particle., , EXAMPLE,  alculate the specific charges (e/m) of the following particles and then arrange the particles in the, C, ascending order of their specific charges., (a) Electron, (b) Proton, (c) α-particle, SOLUTION, , , , 1.6 ×10−19, coulomb/kg, 9.1×10−31, = 0.176 × 1012 = 17.6 × 1010 coulomb/kg, , Specific charge of a proton, , =, , Specific charge of an electron =, , 1.6 ×10−19, coulomb/kg = 0.96 × 108 coulomb/kg, 1.67 ×10−27, , 2 ×1.6 ×10−19, 1.6 ×108, =, = 0.472 × 108 coulomb/kg, −27, 2 ×10 (1.67 + 1.72), 3.39, Hence, ascending order of specific charges of electron, proton and α-particles is, α-particle < proton < electron., , Specific charge of α - particle =

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Atomic Structure, , EXAMPLE,  alculate the mass of a charged particle in CGS units if its charge is x coulomb and specific charge, C, is y coulomb/g., SOLUTION, The mass of the particle in CGS units is e g = x g ., e, y, m, EXAMPLE,  he isotopes of an element have mass numbers: A, A + 1, A + 2. The ratio of abundance of these, T, isotopes is 3 : 2 : 4. Calculate the average atomic mass of the element., SOLUTION, Average atomic mass =, , A × 3 + ( A + 1)2 + ( A + 2)4 3A + 2A + 2 + 4 A + 8, 10, =, =A+, 9, 9, 9, , Rutherford’s α-ray Scattering Experiments, In order to test the validity of Thomson’s atomic model, Rutherford conducted α-ray scattering, experiment., In this experiment, α-particles were allowed to pass through a pair of positively-charged parallel, plates and the resultant narrow beam of α-particles was allowed to strike the gold foil which was, surrounded by zinc sulphide screen., The observations or results of this experiment completely disproved Thomson’s model., Thin gold foil, α-particles, Perforation, , Zinc sulphide, screen, , +, +, , Bright spot, , Sources of, Lead, α-particles, container, Spherical screen, , FIGURE 2.6 α-ray Scattering Experiment, , TABLE 2.2 Observations and conclusions of α-ray scattering experiment, Observations, Most of the particles passed straight through the gold, foil without any deflection, Very few α-particles completely rebounded and few, α-particles showed large deflection, , Conclusions, Presence of large empty space in an atom, Presence of a central, positively-charged core known, as nucleus, , Rutherford’s Atomic Model, The atom is mostly composed of empty space. The entire positive charge and mass of the atom is, concentrated in the small, central part known as the nucleus. The size of the nucleus is so small that, its diameter is 105 times less than that of the atom. The diameter of the nucleus has been estimated by, , 2.7

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2.8, , Chapter 2, , Rutherford as 10–13 cm in contrast to that of the atom to be 10–8 cm., The electrons present outside the nucleus revolve round the nucleus, with high velocities., , Nucleus, Electron, , The electrons revolve round the nucleus with high velocities to, counterbalance the electrostatic forces of attraction between protons, and electrons. Rutherford’s atomic model resembles the planetary, motion in solar system. Therefore, Rutherford’s model of an atom is, also called planetary model., FIGURE 2.7 T he solar, system, , Validity of Rutherford’s Atomic Model, , Rutherford’s atomic model could very well explain the presence of positively-charged nucleus and, presence of electrons outside the nucleus in the atom. However, the failure of Rutherford’s theory, stemmed from two major objections., 1. T,  his model is in contradiction to the principle of classical electrodynamics. According to this,, any charged particle in a circular motion radiates energy continuously. The electron being a, charged particle in the circular motion loses energy. This should ultimately result in its spiral, path towards nucleus and the atom should then collapse., 2. The second major objection to Rutherford’s model came from the pattern of atomic spectra., When light passes through a prism, it gets split up into its components of different wave lengths,, like, visible, ultraviolet, infrared light, etc. The arrangement of component light energies according to, their wavelengths is called spectrum and spectroscope is the instrument designed to observe the spectra., Since white light is composed of lights of different wavelengths, a continuous band of different, wave lengths is obtained which is called continuous spectrum. Light from the sun or incandescent, bulb gives such type of spectra., Continuous, spectrum, V I B, GY, , (+), , Electric arc, (white light, house), , (–), , Slit, , Prism, , OR, , Detector (photographic plate), , (a), , High, Voltage, , Detector, (photographic plate), , Hydrogen, Arc Slit, gas, , Prism, (b), , FIGURE 2.8 (a) Spectrum produced by the white light, . (b) Atomic spectra of hydrogen, , But when the spectrum is taken for the atoms of the gas present in the discharge tube, it is found, to consist of discrete lines of different colours. This is called line spectrum, which is a discontinuous, spectrum. Figure 2.8 (b) shows the atomic spectra of hydrogen gas., According to Rutherford’s atomic model, electrons revolving around the nucleus should lose, energy continuously. Hence, the spectra of the atom should be a continuous spectrum, whereas the, observed atomic spectrum was a line spectrum.

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Atomic Structure, , In 1913, Danish scientist, Neils Bohr could overcome the limitation of Rutherford’s atomic, model successfully based on the quantum theory of radiation proposed by Max Planck., , Quantum Theory of Radiation, At the end of the 19th century, physicists had an idea that matter and energy are distinctly different., Matter consists of particles which have mass and have specific positions in space., Energy is the form of electromagnetic radiation which has no mass and does not have any specific, position in space. Matter can absorb or emit any quantity of energy., But in the beginning of the 20th century, these ideas were proved to be incorrect on the basis of, some experimental results., German physicist, Max Planck, in 1901 carried out the first important experiment by studying, the radiation emitted by solid bodies heated to incandescence., He concluded from his experimental observation that energy can be absorbed or radiated by a, body in the form of small packets of energy called quanta which are whole number multiples of the, quantity hv where, h = Planck’s constant = 6.625 × 10–34 J.s, v = frequency of the radiation, EXAMPLE, Following conclusions are drawn by observing α-ray scattering experiment. Write the respective, observations based on which these conclusions are drawn., (i) Non-uniform distribution of positive charge., (ii) Presence of positively-charged core or nucleus., (iii) Presence of large empty space in an atom., SOLUTION, Observations,  on-uniform distribution of positive, (i) N, charge, , Conclusions, The angle of deflections of various α-particles were, different, ,  resence of positively-charged core or Very few α-particles completely rebounded and few, (ii) P, nucleus, α-particles showed large deflection, (iii) Presence of large empty space in an, Most of the α-particles passed through the gold foil, atom, without any deflection, , EXAMPLE, Compare Thomson’s atomic model with Rutherford’s atomic model., SOLUTION, (a), (b), , Thomson’s atomic model, In Thomson’s model positively-charged, particles are thinly spread throughout the atom., Electrons are embedded in the thin positivelycharged mass., , Rutherford’s atomic model, In Rutherford’s model, positively-charged particles, are concentrated in a small central part called nucleus., Electrons revolve around the nucleus with a high, velocity., , 2.9

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2.10, , Chapter 2, , EXAMPLE, The wavelength of particular radiation is 700 nm (1 nm = 10−9 m). Find its frequency (ν)., SOLUTION, λ = 700 nm = 700 × 10–9 m = 7 × 10–7m. Also, c = velocity of light = 3 × 108 m/s, c, 3×108, = 0.42 × 1015/s, So, ν = λ =, 7 ×10−7, , This theory proves the particle nature of energy., On the basis of this theory, Bohr proposed his atom model., , Bohr’s Model of an Atom, 1. Electrons revolve around the nucleus in specified circular paths called orbits or shells., 2. Each orbit or shell is associated with a definite amount of energy. Hence, these are also called, energy levels and are designated K, L, M and N, respectively., 3. The energy associated with a certain energy level increases with the increase of its distance, from the nucleus. Hence, if the energy associated with the K, L, M and N shells are E1, E2,, E3……… respectively, then E1 < E2 < E3 ………., etc., 4. As long as the electron revolves in a particular orbit, the electron does not lose its energy., Therefore, these orbits are called stationary orbits and the electrons are said to be in stationary, energy states.,  5. An electron jumps from a lower-energy level to a higher-energy level, by absorbing energy,, but when it jumps from a higher- to lower-energy level, the energy is emitted in the form of, electromagnetic radiation. The energy emitted or absorbed (∆E) is an integral multiple of ‘hν.’, 6. The electron can revolve only in the orbit in which the angular momentum of the electron, (mvr) is quantised, i.e., mvr is a whole number multiple of h/2π. This is known as the principle, of quantisation of angular momentum., , The angular momentum is written as, , mvr =, , nh, ,, 2π, , , where, n is an integer (n = 1,2,3,4,…….) and is called principal quantum number., , m = mass of the electron, , v = velocity of an electron in its orbit, , r = distance of the electron from the nucleus, n=6, n=5, n=4, n=3, n=2, n=1, + K, , L M NOP, , Stationary, orbits oforbits, an atomof an atom, FIGURE 2.9, Stationary

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Atomic Structure, , By applying the concept of quantisation of energy, Bohr calculated the radius and energy of the, nth orbit of hydrogen atom., rn =, , n 2 h2, −2π 2me 4, ,, E, =, n, 4 π 2me 2, n 2h 2, , With the help of these expressions, Bohr gave a satisfactory explanation for the spectra of hydrogen, and hydrogen-like species (ions having one electron, e.g., He+, Li+2, Be+3)., , Limitations of Bohr’s Atomic Model, The following are the limitations of Bohr’s atomic model:, 1. Bohr could not explain the spectral series for the multi-electron atoms., 2. B,  ohr’s model could not give a satisfactory justification for the assumption that electrons can, revolve in those orbits where their angular momentum (mvr) is a whole number multiple of, nh/2π, i.e., he could not justify quantisation of angular momentum., 3. A,  ccording to Heisenberg’s uncertainty principle, it is impossible to determine simultaneously, with certainty the exact position and the momentum of the particle. Bohr assumed that an, electron of an atom is located at a definite distance from the nucleus and revolves around the, nucleus with a definite velocity, i.e., the momentum of the electron is fixed which is against, Heisenberg’s uncertainty principle., 3. T,  he atomic spectral lines split into a number of closely packed lines in the presence of a, magnetic field and an electric field. These effects are called Zeeman effect and Stark effect,, respectively. Bohr failed to explain these effects., 4. W,  hen the hydrogen spectrum was observed with the spectroscope of high resolving power,, it was found that the individual lines in the spectrum consisted of several fine lines lying close, to each other. This is called fine spectrum and he failed to explain the fine structure of the, spectrum., , DISCOVERY OF NEUTRONS, The electrons being particles having negligible mass and massive protons concentrated inside the, nucleus, it could be predicted that the mass of an atom has to be equal to the mass of the total, number of protons present in the atom. However, this was found to be true only in case of a, hydrogen atom., The difference in the predicted mass and actual mass of the atom has been found to be equal to, the mass of the proton or multiples of the mass of a proton. These particles are supposed to have, neutral charge since the atom is electrically neutral. They were called neutrons and they were, discovered by James Chadwick by an experiment involving the bombardment of beryllium nucleus, with α-particles., The discovery of fundamental particles has ultimately resulted in the establishment of a basic, atomic model. The basic model of an atom comprises small positively-charged nucleus at the centre, of the atom and the electrons revolving round the nucleus in orbits., , 2.11

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2.12, , Chapter 2, , TABLE 2.3 Characteristics of fundamental particles, S. No., 1., , 2., , 3., , Fundamental Particles, Electron (e), , Charges, –1.6 × 10–19C, –4.8 × 10–10 esu, + 1.6 × 10–19 C, + 4.8 × 10–10 esu, , Proton (p), , Neutron (n), , 0, , Masses, 9.1 × 10–31 kg, (or), 0. 00055 amu, 1.67 × 10–27 kg, (or), 1.0078 amu, 1.72 × 10–27 kg, (or), 1.0083 amu, , Relative Charges, –1, , +1, , 0, , TABLE 2.4 Atomic number and mass number, Representations, Atomic number, , Z, , Mass number, , A, , Definitions, The number of protons in an atom, The total number of nucleons, i.e.,, number of protons and neutrons in, an atom, , Examples, Cl atom has 17 protons in its atom,, Z = 17, Cl atom has 17 protons and 18, neutrons in its nucleus., ∴ A = 17 + 18 = 35, , An element when represented along with its atomic number and mass number is represented as, A, ZX . Elements are found to exist in their isotopic forms. Isotopes are the atoms of the same element, having different mass number. Based on their percentage abundance of each isotopic form, average, atomic mass of an element is calculated., Examples: 1H1, 1H2, 1H3, etc., It is also found that atoms of different elements have same mass number. These are called isobars., Examples: 18Ar40, 20Ca40, etc., , Electronic Configuration, The systematic arrangement of electrons in the various shells or orbits in an atom is called electronic, configuration., The electrons are arranged in an atom in the various shells around the nucleus. The, last shell or the outermost shell from the nucleus with electrons is called valence shell., The shell inner to this is called penultimate shell and the one inner to the penultimate, shell is called anti-penultimate shell., The filling of electrons in various shells can be done according to Bohr–Bury scheme., According to this, the maximum number of electrons that can be accommodated in any, shell is given by 2n2, where n represents the number of the shell., , Shell 2n2, K, 2, L, 8, M, 18, M, 32, O, 50, , The maximum number of electrons that can be filled in the valence shell is 8, that in the penultimate, shell is 18 and the anti-penultimate shell has a maximum capacity of 32 electrons. The filling of, electrons till atomic number 30 follows the following pattern:

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Atomic Structure, K, 2, 2, 2, 2, 2, , L, 8, 8, 8, 8, , M, , 8, 8, 18, , N, , 2, 2, , EXAMPLE,  n element has protons whose mass is equal to 23,881 times that of an electron. Identify the, A, element and write its electronic configuration., SOLUTION, Mass of a proton is 1837-times than that of an electron., 23,881, Number of protons =, = 13, 1837, Atomic number of the element is 13., So, the element is aluminium and its electronic configuration is 2, 8, 3., , EXAMPLE,  rite the electronic configuration and the atomic number of the atom which becomes stable by, W, gaining 3 electrons in fifth shell., SOLUTION, The electronic configuration of the stable species is 2, 8, 18, 18, 8, Hence, the electronic configuration of the atom which becomes stable after gaining 3 electrons in, the fifth shell is 2, 8, 18, 18, 5 and the atomic number is 51., , EXAMPLE,  hat is the ratio of the amount of energy required to remove an electron from hydrogen and, W, He+ ion?, SOLUTION, 13.6 × Z2, Energy of electron (single electron species) En = −, eV ;, n2, 2, 1, For hydrogen ⇒ –13.6 × 2 =−13.6 eV ; Energy required to remove this electron = +13.6 eV, 1, 22, +, For He ⇒ –13.6 × 2 =− 54.4 eV ; Energy required to remove this electron = +54.4 eV, 1, 13.6 1, = = 1 : 4., Ratio of energy required =, 54.4 4, , 2.13

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2.14, , Chapter 2, , EXAMPLE, Complete the following table:, Valence, shells in, atom, M, N, , Electronic, Charges on configurations, stable ions, of element, +3, +1, +2, 2, 8, 6, , Number of, electrons in, penultimate shell, , Number of, core electrons, , 8, , 18, 18, , Note: Core electrons are inner electrons which exclude valence electrons., SOLUTION, Valence shells, in atom, M, N, N, M, , Charges on, stable ion, +3, +1, +2, −2, , Electronic, configurations, of element, 2, 8, 3, 2, 8, 8, 1, 2, 8, 8, 2, 2, 8, 6, , Number of, electrons in, penultimate shell, 8, 8, 8, 8, , Number of core, electrons, 10, 18, 18, 10

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Atomic Structure, , 2.15, , TEST YOUR CONCEPTS, Very Short Answer Type Questions, , 2. Like atoms are identical in all respects. This, ­statement of Dalton’s atomic theory is contradicted., Which discovery contradicts this?, 3. The value of the Planck’s constant ‘h’ in erg-s is, ______., 4. Why was a gas at low pressure taken by Thomson, while conducting the experiment?, 5. Why is Rutherford’s model called nuclear model?, 6. Mass of the electron is calculated from ____ and, ____ values of electron., 7. Give the mass and charge of fundamental particles, of an atom., , 15. What is Heisenberg’s uncertainty principle?, 16. According to _______, the charges in an atom are, arranged like the pulp and seeds of a watermelon., 17. Which theory supported the particle nature of an, electron?, 18. Give the value of Planck’s constant in, (1) erg-s, (2) joule-s, 19. Give equations to calculate the following:, (a) the radius of the nth orbit of hydrogen atom, (b) energy of the nth orbit of hydrogen atom, 20. What happens when an electron jumps from a, lower energy level to a higher energy level?, 21. What is Zeeman effect?, , 8. What is an atomic model?, , 22. What is Stark effect?, , 9. The equation for the calculation of energy of nth, orbit of hydrogen atom derived by Bohr is ______., , 23. With the increase in the radius of the orbit, the, energy of an electron _____________., , 10. Who discovered protons? Based on what experiment was he able to discover these protons?, 11. What was the mathematical equation given by Max, Planck?, 12. Who discovered neutrons? How was the discovery, made?, 13. Neutrons were discovered by bombarding b­ eryllium, with _____ particles., 14. What name did Max Planck give to energy packets?, , 24. What is an α-particle?, 25. What is a continuous spectrum?, 26. The circular paths in which electrons revolve are, called ______., 27. Why are light rays known as electromagnetic waves?, 28. The ______ consists of well-defined lines of d­ efinite, frequencies., 29. In the formula E = hv, E is ______ and v is ______., , Short Answer Type Questions, 30. ‘Electrons jump from one orbit to another orbit.’, Justify this statement on the basis of Bohr’s theory., , electron from a hydrogen atom to produce an H+, ion? Explain., , 31. On what basis did Bohr propose his atomic model?, , 35. If Rutherford’s atomic model is correct, then the, atom should collapse. Why?, , 32. What are orbits and why are they called stationary, orbits?, 33. Mention the properties of anode rays., 34. What is the amount of energy needed to remove an, , 36. Describe Millikan’s oil drop experiment in brief., 37. According to Rutherford’s atomic model, where, are the protons and electrons located in an atom?, , PRACTICE QUESTIONS, , 1. Which postulate of Dalton’s atomic theory is, ­considered to be correct even today?

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2.16, , Chapter 2, , 38. Why was the presence of neutrons in an atom, ­predicted? How were neutrons discovered?, , 42. The wavelength of a particular radiation is 700 nm, (1 nm = 10–9 m). Find its frequency ( ν )., , 39. Describe J.J. Thomson’s atomic model., , 43. Distinguish between continuous spectrum and, discontinuous spectrum. Give some examples of, sources for these spectra., , 40. Define angular momentum. In the relation, nh, mvr =, , what do m, v, r and h denote?, 2π, 41. On what basis did Bohr assume the concept of, ­stationary orbits for an electron?, , 44. An electron revolving round in an orbit has angular, h, momentum equal to, . Can it lose energy?, 2π, , Essay Type Questions, 45. Explain Bohr’s atomic model., 46. What are the observations and conclusions drawn, by J.J. Thomson while conducting experiments, with a discharge tube for studying the properties of, cathode rays?, , 47. What are the drawbacks of Rutherford’s atomic, model?, 48. Describe Rutherford’s atomic model., 49. State the limitations of Bohr’s model., , CONCEPT APPLICATION, , PRACTICE QUESTIONS, , Level 1, Direction for questions from 1 to 7:, State whether the following statements are true, or false., , 9. Some of the α-rays deflect in acute and obtuse, angles due to the presence of the ______ in the, centre of the atom., , 1. According to Thomson’s atomic model, electrons, revolve round the nucleus., 2. In a discharge tube, anode rays originate when, electrons collide with gas molecules., 3. 8O16 and 8O18 are isotopes while 20Ca40 and 18Ar40, are isobars., 4. Energy is absorbed when the electron jumps from, K to L energy shells., 5. α-ray scattering experiment proved that the positive particles are present in the extra nuclear part of, an atom., 6. Characteristic spectra of atoms are line spectra., 7. An electron in the excited state of an atom is highly, unstable., , 10. According to classical electrodynamics, if an electrically charged particle revolves in a circular path, it, continuously ______ energy., , Direction for questions from 8 to 14:, Fill in the blanks., 8. Anode rays are deflected towards the negative plate, in the presence of an electric field because they, consist of _____ particles., , 11. The energy of an electron present in the first orbit, of an atom is ________ than the energy of electron, in the other orbits., 12. Splitting of spectral lines in the presence, of ­, magnetic field is known as _____, effect., 13. The kinetic energy of an electron present in the, first orbit of an atom is _______ than that of the, electron in the last orbit., 14. The spectra produced by the deexcitation of an, electron is called _______________.

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Atomic Structure, , 15. Column A, A. e value varies, , Column B, ( ) a. hv, , m, , with the nature of, gas, B. , Plum pudding, ( ) b. Rutherford’s atomic, model, model, Mass of the atom ( ) c. , Sun rays, C. , is concentrated at, the centre of atom, D. , Continuous, ( ) d. Bohr’s stationary orbit, spectrum, ( ) e. , Thomson’s atomic, E. mvr = nh, model, , 2π, , D. Quantum, , ( ) f. Anode rays, , Direction for questions from 16 to 45:, For each of the questions, four choices have been, provided. Select the correct alternative., 16. Which of the following concepts was not considered in Rutherford’s atomic model?, (a) the electrical neutrality of atom, (b) the quantisation of energy, (c) electrons revolve around nucleus at very high, speeds, (d) existence of nuclear forces of attraction on the, electrons, 17. When alpha particles are sent through a thin metal, foil, only one out of ten thousand of them rebounded., This observation led to the conclusion that, (a) positively-charged particles are concentrated at, the centre of the atom, (b) more number of electrons is revolving around, the nucleus of the atom, (c) unit positive charge is only present in an atom, (d) a massive sphere with more negative charge and, unit positive charge is present at the centre of, the atom, 18. Canal ray experiment lead to the discovery of, ______., , (a) protons, (c) electrons, , (b) neutrons, (d) nucleus, , 19. In which of the following pairs of shells, energy difference between two adjacent orbits is minimum?, (a) K, L, (b) L, M, (c) M, N, (d) N, O, 20. Assertion A: An electron in the inner orbit is, more tightly bound to the nucleus., , Reason B: The greater the absolute value of, energy of an electron the more tightly the electron, is bound to the nucleus., (a) Both A and B are true but B is not the appropriate reason for A., (b) Both A and B are individually correct and B is, the correct reason for A., (c) A is correct but B is not correct., (d) Both A and B are not correct., 21. The electron revolves only in the orbits in which, nh, nh, (a) mvr >, (b) mvr ≥, 2π, 2π, nh, nh, (c) mvr =, (d) mvr <, 2π, 2π, 22. Which among the following pairs are having different number of valence electrons?, (a) Na+, Al+3, (b) P–3, Ar, +2, (c) Mg , Ar, (d) O–2, F, 23. If two naturally occurring isotopes of an element, are 7X15, 7X11; what will be the percentage composition of each isotope of X occurring, respectively,, if the average atomic weight accounts to 14?, (a) 95, 5, (b) 80, 20, (c) 75, 25, (d) 65, 35, 24. According to quantum theory of radiation, which, is false?, (a) radiations are associated with energy, (b) , radiation is neither emitted nor absorbed, discontinuously, (c) , the magnitude of energy associated with a, quantum is dependent on frequency, (d) photons are quanta of radiation, 25. Select True/False among the following statements:, , (i) Bohr’s theory successfully explained stability of, the atom., , PRACTICE QUESTIONS, , Direction for question 15:, Match the entries given in Column A with, appropriate ones from Column B., , 2.17

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2.18, , Chapter 2, , , (ii) Atoms give line spectra., , (iii) Velocity of electromagnetic waves depends on, the frequency., , (iv) Bohr introduced the concept of orbital., (a) (i) T, (ii) T, (iii) F, (iv) F, (b) (i) T, (ii) F, (iii) T, (iv) T, (c) (i) F, (ii) T, (iii) F, (iv) F, (d) (i) F, (ii) F, (iii) T, (iv) T, 26. Which of the following particles do not produce, electronic spectra?, (a) Li+2, (b) He+2, +2, (c) Be, (d) Na+, 27. An element has two isotopes with mass numbers, 16 and 18. The average atomic weight is 16.5. The, percentage abundance of these isotopes is ______, and ______, respectively., (a) 75, 25, (b) 25, 75, (c) 50, 50, (d) 33.33, 66.67, , PRACTICE QUESTIONS, , 28. Which among the following are isobars?, (a) bXa and bYa+1, (b) bXa and cYb, (c) bXa and b+1Ya, (d) bXa and b–1Ya–1, 29. Some of the elements have fractional atomic masses., The reason for this could be, (a) the existence of isobars, (b) the existence of isotopes, (c) the nuclear reactions, (d) the presence of neutrons in the nucleus, 30. Which of these pairs has almost similar masses?, (a) proton–electron, (b) neutron–electron, 1, (c) electron–1H, (d) neutron−1H1, 31. The energy of an electron revolving in the 3rd orbit, of Be+3 ion is _________ ev, (a) –10.2, (b) –13.6, (c) –24.2, (d) –18.1, 32. Which of the following concepts, was not considered in Rutherford’s atomic model?, (a) the electrical neutrality of atom, (b) the quantisation of energy, (c) electrons revolve around the nucleus at very, high speeds, (d) existence of nuclear forces of attraction on the, electrons, , 33. 7X15, 7X11 are two naturally occurring isotopes, of an element X. What is the percentage of each, ­isotope of X if the average atomic mass is 14?, (a) 95, 5, (b) 80, 20, (c) 75, 25, (d) 65, 35, 34. A trinegative ion of an element has 8 electrons in its, M shell. The atomic number of the element is, (a) 15, (b) 18, (c) 20, (d) 16, 35. Arrange the following statements given by various, scientists in chronological order:, , (1) calculation of energy and radius of orbit, , (2) atoms of the same elements are identical in all, respects, , (3) calculation of diameter of the nucleus and the, atom, , (4) assumption of thinly spread positively-charged, mass, (a) 4 3 1 2, (b) 4 2 3 1, (c) 2 4 3 1, (d) 3 4 2 1, 36. What is the ratio of radii of the first successive odd, orbits of hydrogen atom?, (a) 9 : 1, (b) 1 : 9, (c) 1 : 3, (d) 3 : 1, 37. An electron revolves round the nucleus in the 3rd, orbit and jumped to a higher orbit X showing a, h, difference in angular momentum equal to, . The, π, value of ‘X’ could be, (a) 4, (b) 6, (c) 5, (d) 7, 38. Rutherford’s α-particle scattering experiment, eventually led to the conclusion that, (a) mass and energy are related, (b) the point of impact with matter can be precisely determined, (c) neutrons are buried deep in the nucleus, (d) electrons are distributed in a large space around, the nucleus, 39. Arrange the following steps which are carried out, in μ-ray experiment in the correct sequence:, , (1) Passage of μ-particles through a slit, , (2) bombardment of μ-particles with a gold foil

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Atomic Structure, , deflection of μ-particles, production of μ-particles, 4123, (b) 4 1 3 2, 1432, (d) 1 4 2 3, , 40. Which among the following pairs are having different number of total electrons?, (a) Na+ and Al+3, (b) P–3 and Ar, (c) Mg+2 and Ar, (d) O–2 and F–, 41. The postulates of Bohr’s atomic model are given, below. Arrange them in the correct sequence:, , (1) As long as the electron revolves in a particular orbit,, the electron does not lose its energy. Therefore,, these orbits are called stationary orbits and the, electrons are said to be in stationary energy states., , (2) Electrons revolve round the nucleus in specified, circular paths called orbits or shells., , (3) The energy associated with a certain energy, level increases with the increase of its distance, from the nucleus., , (4) An electron jumps from a lower energy level to, a higher energy level by absorbing energy. But, when it jumps from a higher to lower energy, level, energy is emitted in the form of electromagnetic radiation., , (5) Each orbit or shell is associated with a definite, amount of energy. Hence, these are also called, energy levels and are designated as K, L, M and, N, respectively., (a) 1 3 4 5 2, (b) 2 3 5 1 4, (c) 2 5 3 1 4, (d) 2 1 4 3 5, , 42. The ratio of atomic numbers of two elements A, and B is 1 : 2. The number of electrons present in, the valence shell (3rd) of A is equal to the ­difference, in the number of electrons present in the other two, shells. Steps involved for the calculation of ratio of, number of electrons present in a penultimate shell, to anti-penultimate shell of B are given below., Arrange them in the correct sequence:, , (1) calculation of atomic number of B, , (2) calculation of valence electrons present in A, , (3) calculation of atomic number of A, , (4) calculation of number of electrons present in, the penultimate and anti-penultimate shells of B, , (5) writing electronic configuration of B, (a) 2 3 4 1 5, (b) 2 3 1 5 4, (c) 4 5 2 3 1, (d) 4 2 1 3 5, 43. The equation given by Bohr to calculate radius of, nth orbit of hydrogen atom is, (a) rn =, , n 2h 2, 4π 2me, , (b) rn =, , n 2h 2, 4π 2me, , nh 2, n 2h 2, (d), r, =, n, 4π 2me, 4π 2 m 2 e, 44. The number of electrons present in the valence, shell of an atom with atomic number 38 is, (a) 2, (b) 10, (c) 1, (d) 8, (c) rn =, , 45. The mass number of an atom whose unipositive ion, has 10 electrons and 12 neutrons is, (a) 22, (b) 23, (c) 21, (d) 20, , Level 2, 1. When the same isotopic gas is taken in two discharge tubes, the angle of deflection is found to be, different though the strength of the external electric field applied is the same. Explain., , 5. If the energy released when an electron jumped, from the 4th orbit to the 3rd orbit of hydrogen is ‘x,’, then what would be the energy difference when it, jumps from the 3rd orbit to the 2nd orbit?, , 2. In a canal ray experiment, different gases were, found to produce canal rays with the same specific, charge. Explain., , 6. Electronic spectra can distinguish isobars but not, isotopes. Justify., , 3. When canal rays experiment is conducted with, hydrogen gas, scientists were found to give particles, e, with different, values. Justify., m, 4. Energy of the electron in the atom is negative. Explain., , [Hint: Energy of a free electron is taken as zero.], , 7. If the energy difference between the orbits when, an electron in H atom gets excited to higher energy, orbit from its ground state is 12.1 eV/atom, calculate the frequency of radiation emitted ( l eV, = 1.602 × 10–19J) when electron comes back to, ­second energy level., , PRACTICE QUESTIONS, , , (3), , (4), (a), (c), , 2.19

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2.20, , Chapter 2, , 8. Is the energy difference between successive orbits, the same for all orbits? Justify your answer., , 17. Why is the source of α-particles kept inside the lead, block?, , 9. Though there is only one electron in a hydrogen, atom, the spectrum of hydrogen contains a number, of lines. How do you explain this?, , 18. If Thomson’s model is considered to be correct,, what would be the observation of Rutherford’s, α-ray scattering experiment?, , 10. What is the ratio of the radius of the 1st orbit to 2nd, orbit, if the velocity of the electron in the 1st orbit, is twice that of the 2nd orbit., , 19. The ratio of the atomic numbers of two elements A, and B is 2 : 3. A is an inert gas with the first 3 orbits, completely filled. Identify A and B and write their, electronic configurations., , 11. A particular atom has the 4th shell as its valence, shell. If the difference between the number of electrons between K and N shells and L and M shells, is zero, find the atomic number of the element and, electronic configuration of its stable ion., 12. A stable unipositive ion of an element contains, three fully filled orbits. What is the atomic number, of the element?, , 21. Predict the possible atomic number(s) of an atom in, which the third shell is incompletely filled and maximum 3 more electrons can be added in that shell., , 13. Explain why a blackened platinum strip when, placed at the radius of curvature turns red hot, only, when the cathode taken has concave shape., , 22. The radius of nth orbit of a single electron species is, 0.132 n2 A°. Identify the element., , 14. The average atomic mass of two isotopes with mass, numbers A and A + 2 is A + 0.25. Calculate the, percentage abundance of the isotopes., 15. Spectral line given by an atom is a kind of signature, of the respective atom. Comment on this statement., , [Hint: The nuclear charges of different atoms are, different.], , PRACTICE QUESTIONS, , 20. A stable dipositive ion and a dinegative ion are isoelectronic with an octet configuration in the second shell, of their atoms. Identify the preceding and succeeding, elements and write their electronic configurations., , Directions for questions from 16 to 25:, Application-Based Questions, , 23. What is the frequency of light emitted when an, electron in a hydrogen atom jumps from the 3rd, orbit to the 2nd orbit?, 24. An electron having an angular momentum of, 1.05 × 10–34 joules jumps to another orbit such that, it has an angular momentum of 4.20 × 10–34 joules., Explain the possible transitions., 25. The mass number of a particular element which has, equal number of protons and neutrons is 32. What, is the electronic configuration of the atom and its, stable ion?, , 16. Why was a spherical sulphide screen used in α-ray, scattering experiment?, , Level 3, 1. In Millikan’s oil drop experiment, the distance, between the metal plates, A and B to which an, electric potential is applied such that A is positive, and B is negative is 5 mm. An oil drop is found to, be suspended at a distance of 2 mm from B. Predict, the change in the position of the oil drop when, there is a sudden drop or rise in potential. Justify., 2. Different gases in the discharge tube produce different colours under suitable conditions of pressure, and voltage. Explain., , [Hint: Each element has its own characteristic, atomic spectrum.], , 3. Is the velocity of an electron in all orbits the same, for an atom of a particular element? How does, it vary for different single electron species? Give, ­reasons in support of your answer., 4. What is the ratio of distance between successive, orbits of 1 and 2 to 2 and 3 of hydrogen atom?, , [Hint: radius of nth orbit in hydrogen is 0.529, × n2 A°], 5. If yx A +1 or yx −−12 B+1 were to be used instead of, α-particles in Rutherford’s experiment, which, would be better and why?

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Atomic Structure, , duced show the same deflection under the external, electric field? Give reasons to support your answer., , Directions for questions from 6 to 10:, Application-Based Questions, 6. Draw a comparison between the potential energy, and kinetic energy of electrons in the 1st orbits of, hydrogen and He+ ion. Also comment on the total, energy of the electrons in the above cases., 7. Though the kinetic energy of electrons decreases, with an increase in the distance from the nucleus,, the potential energy of the electron increases. How, do you account for this?, 8. Why is high voltage and low pressure maintained in, the discharge tube?, , 2.21, , 10., , Cathode, , Anode, , +, , −, , +, H.V., , Fluorescent, material, (ZnS), , −, Schematic diagram, , Bright, spot, , , If the given schematic diagram represents Thomson’s, experiment and the corresponding observation,, what would be his atomic model?, , PRACTICE QUESTIONS, , 9. If canal ray experiments are conducted with different isotopes of hydrogen gas, do the canal rays pro-

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2.22, , Chapter 2, , CONCEPT APPLICATION, Level 1, True or false, 1. False, , 2. True, , 3. True, , 5. False, , 6. True, , 7. True, , 4. True, , Fill in the blanks, 8. positively-charged, , 10. loses, , 12. Zeeman, , 9. positive charge, , 11. less, , 13. more, , 14. emission spectrum, , Match the following, 15. A : f, , B:e, , C:b, , , D:, , E:d, , F:a, , H I N T S A N D E X P L A N AT I O N, , Multiple choice questions, 16. b, , 20. a, , 24. b, , 28. c, , 17. a, , 21. c, , 25. a, , 29. b, , 18. a, , 22. c, , 26. b, , 30. d, , 19. d, , 23. c, , 27. a, , 31. Energy, En =, , −13.6Z2, eV 13 6, n2, , −13.6 × 16, E3 =, eV= − 24.2 eV, 9, , 32. According to Rutherford’s theory, an atom is electrically neutral and electrons revolve around the, nucleus. Nuclear forces of attraction exist between, the nucleus and electrons. The only assumption, that Rutherford did not consider is quantisation of, energy., 33. Let the percentage of 7X15 is x., , ∴ The percentage composition of 7X11 is 100 – x, , Average atomic weight = 14 =, , ⇒1400 = 15x + 1100 – 11x, , x(15) + (100 − x )11x, 100, , , ⇒1400 = 4x + 1100 ⇒ 4x = 300, 300, , x=, =75, 4, , ∴ The percentage of 7X11 = 100 – x, , = 100 – 75 = 25 and that of 7X15 is 75, 34. Since electronic configuration of the trinegative ion, is 2, 8, 8 the electronic configuration of the neutral, atom is 2, 8, 5 and its atomic number is 15., 35. (i) Atoms of the same elements are identical in all, respects., (ii) , Assumption of thinly spread positivelycharged mass, , (iii) Calculation of the diameters of the nucleus, and the atom, , (iv) Calculation of energy and radius of orbit

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Atomic Structure, , , Successive first odd orbits are 1 and 3, r1 0.529 × 12 1, =, = = 1: 9, r3 0.529 × 32 9, , 37. Angular momentum in third orbit is given by, , mvr =, , 3h, ⇒, 2π, , , Angular momentum in X orbit is given by, Xh, → (2), 2π, Xh 3h h, −, =, 2π 2π π, , mvr =, , h, h, = ⇒ X=5, (X − 3), 2π π, , , 38. The α-ray scattering experiment led to the discovery of nucleus which in turn helped him conclude, that electrons revolve around the nucleus to overcome the strong electrostatic force of attraction., 39. (i) production of α-particles, (ii) production of a narrow beam of α-particles, , (iii) bombardment of α-particles with gold foil, , (iv) deflection of α-particles, 40. The electronic configuration of Mg+2 is 2, 8, , 41. (i) E,  lectrons revolve around the nucleus in specified circular paths called orbits or shells., (ii) Each orbit or shell is associated with a definite, amount of energy. Hence, these are also called, energy levels and are designated as K, L, M, and N, respectively., , (iii) The energy associated with a certain energy, level increases with the increase of its distance, from the nucleus., , (iv) As long as the electron revolves in a particular orbit,, the electron does not lose its energy. Therefore,, these orbits are called stationary orbits and the, electrons are said to be in stationary energy states., (v) An electron jumps from a lower energy level, to a higher energy level by absorbing energy., But when it jumps from a higher to lower, energy level, energy is emitted in the form of, electromagnetic radiation., 42. (i) calculation of valence electrons present in A, (ii) calculation of atomic number of A, , (iii) calculation of atomic number of B, , (iv) writing electronic configuration of B, (v) calculation of number of electrons present in, the penultimate and anti-penultimate shells of, B, 43. The equation given by Bohr to calculate radii of nth, orbit of hydrogen atom is, , , ∴ Total no. of electrons = 10, , rn =, , , The electronic configuration of Ar is 2, 8, 8, , n 2h 2, 4π 2me 2, , , ∴ Total no. of electrons = 18., , 44. Z = 38, electronic configuration = 2, 8, 18, 8, 2, , , Hence, Mg+2 and Ar are having different number, of total electrons., , , ∴ Two valence electrons, 45. Mass number = 11 + 12 = 23, , Level 2, 1. (i) factors which affect angle of deflection in an, electric field, , (ii) conditions for changing the factors which affect, the angle of deflection, 2. (i) factors affecting specific charge, , (ii) conditions where different gases can have the, same specific charge, , 3. (i) e/m depends on number of protons and neutrons, (ii) existence of isotopes, , (iii) variation in e/m for isotopes, (i) comparing energy of free electron and energy, of electron in an atom, (ii) change in energy when an electron is brought, closer to the atom, 4., , H I N T S A N D E X P L A N AT I O N, , 36. rn = 0.529 × n2 A°, , 2.23

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2.24, , Chapter 2, , , (iii) , reason for the change in the energy of an, electron, 6. (i) fundamental particle responsible for the spectra, (ii) , relation between fundamental particle and, structure of spectra, , (iii) difference in number of the above particles, between isobars and isotopes, , (iv) effect of this on structure of spectra, 7. (i) relation between energy and n, (ii) calculation of n2 value from the difference in, energy, , (iii) calculation of frequency from energy, , (iv) calculation of energy difference based on n, values, (v) calculation of frequency, , (vi) ν = 0.456 × 1015 s–1, , H I N T S A N D E X P L A N AT I O N, , 8. (i) factors affecting deflection, (ii) the effect of atomic number on deflection, , (iii) the effect of kinetic energy on deflection, 9. (i) Bohr’s model of atom, (ii) relation between energy absorbed and excitation, , (iii) relation between the path of electron during, deexcitation and energy emitted, , (iv) relation between energy emitted and spectrum, 10. (i) comparing the angular momentum of electron in the orbits, (ii) comparison of radius, , (iii) r1 : r2 = 1 : 4, 11. (i) number of electrons in K and L shells when N, shell is the valence shell, , (ii) calculation of number of electrons in K, L, M, and N shells, , (iii) calculation of atomic number, , (iv) number of electrons to be lost to form stable ion, (v) atomic number = 20, 12. (i) maximum number of electrons present in an, orbit (Bohr–Bury scheme)., (ii) electronic configuration of neutral atom, , (iii) 37, 13. (i) factors responsible for the strip to turn red hot, (ii) The path in which the electrons travel from, concave cathode., , 14. (i) (n1 × A) + (n2 × (A + 2)) = (n1 + n2), (A + 0.25) form, (ii) 87.5% and 12.5%, 15. (i) calculation of energy of electrons in He+ and, Li+2 ions, (ii) 4 : 9, 16. to observe scintillations even if the α-rays get, deflected at large angles, 17. α-particles cannot penetrate lead, but β- and γ-rays, can. In order to screen α-particles from β- and, γ-rays, lead block was used., 18. If Thomson’s model is correct, all the α-particles, would have penetrated the gold foil and their angle, of deflection would be insignificant., 19. A → electronic configuration – 2, 8, 18, 8. Thus,, the atomic number = 36., , A:B=2:3, , ∴B=, , 36 × 3, = 54, 2, , , The electronic configuration of, , B = 2, 8, 18, 18, 8., 20. Let the dipositive ion be X+2 and dinegative ion is, Y–2., , The octet in the second shell → 2, 8, , The number of electrons in X = 10 + 2 = 12., , The number of electrons in Y = 10 – 2 = 8., , X is magnesium with an electronic configuration 2,, 8, 2. Y is oxygen with a configuration 2, 6., , The element preceding magnesium is sodium and, the succeeding one is aluminium. They have electronic configurations 2, 8, 1 and 2, 8, 3, respectively. The elements preceding and succeeding for, oxygen are nitrogen and fluorine which have electronic configurations 2, 5 and 2, 7, respectively., 21. Electronic configuration of the atom which can, accommodate three more electrons in the 3rd shell, could be 2, 8, 5 and 2, 8, 15, 2. Hence, the probable atomic numbers are 15 and 27.

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Atomic Structure, , , Initial angular momentum = 1.05 × 10–34 joules, , Kn2, 22. rn =, Z, ∴0.132n 2 =, , , 2.25, , n × 6.625 × 10 −34, ,n = 1, 1.05 × 10 −34 =, 2 × 3.14, , , The electron is present in the 1st orbit originally., , 0.529 × n 2, Z, , 0.529, ∴Z =, =4, 0.132, , , Since the atomic number is 4, the element is, beryllium., 1 1, 23. E 3 − E 2 = −21.72 × 10 −19  − ,  9 4, 5, 36, , ∴ Difference in energy, E3 – E2 = 3.01 × 10–19 J, = 21.72 × 10 −19 ×, , , According to Planck’s equation, ΔE = hv, , 3.01 × 10–19 = 6.625 × 10–34 ν, −19, , , Frequency, (v) =, , 3.01 × 13, = 4.5 × 1014 s −1, 6.625 × 10 −34, , , When the electron gets excited, the angular momentum, , = 4.20 × 10–34 joules., n × 6.625 × 10 −34, ,n = 4, 4.20 × 10 −34 =, 2 × 3.14, , , An electron can lose energy when it is present in, the 4th orbit and not from the 1st orbit., , The possible transitions are 4 → 3, 4 → 2, 4 → 1,, 3 → 2, 3 → 1 and 2 → 1., 25. Mass number = 32, , No. of protons = No. of neutrons = 16, , ∴ Electronic configuration = 2, 8, 6, , Element is sulphur and stable ion is S−2, , Electronic configuration of S−2 is 2, 8, 8, , , mvr =, , nh, 2π, , Level 3, 1. (i) charge acquired by oil drop, (ii) different forces acting on the charged oil drop, when it is at a distance of 2 mm from B., , (iii) relation between position of oil drop and different forces, , (iv) effect of a particular force on the position of, oil drop, (v) change in position of oil drop with change in, potential, 2., , (i) energy of electron, , Excitation and deexcitation, (ii) factors affecting the energy of electron, , (iii) comparison of the energy emitted during the, deexcitation of electron in different atoms, 3. (i) forces acting on moving electron, (ii) position of the electron in the orbit, , (iii) , relationship between position, forces and, velocity, , , (iv) comparison of nuclear charge in different single electron species, (i) effects of high voltage and low pressure on the, gas molecules in the discharge tube, (ii) the effect of velocity of electrons on ionisation, , (iii) relation between the voltage and the velocity, of electrons, , (iv) relation between the pressure and the number, of gas molecules, (v) the effect of the number of gas molecules on, the impact of collision, 4., , 5. (i) the factors affecting angle of deflection, (ii) the characteristics in which the two particles, differ, 6. The electrons revolve round the nucleus with high, velocities to counter balance the nuclear force of, attraction. As nuclear force of attraction in the, 1st orbit of He+ is more than that in H atom., , H I N T S A N D E X P L A N AT I O N, , 24. Angular momentum of an orbit

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2.26, , Chapter 2, , Kinetic energy of electron in He+ is more (due to, greater velocity) than in H atom. When an electron approaches towards an atom, it loses potential energy because it works towards the force of, ­attraction. The greater the force of attraction, the, more is the loss of potential energy. Hence, the, electron in He+ has lesser potential energy than the, electron in H atom. But the loss of PE is more significant than the change in KE. Hence, total energy, of helium is less than that of hydrogen., 7. The kinetic energy of an electron is proportional, to its velocity. With the increase in distance from, the nucleus, the velocity of the electron decreases, as the electron has to overcome a lesser nuclear, force of attraction. An electron loses its potential, energy when it approaches towards an atom that, is a nucleus because it works towards the force of, attraction during this process. Hence, the potential energy of the electron decreases with decrease, in distance between the nucleus and the electron, and increases with increase in distance between the, nucleus and electron., , H I N T S A N D E X P L A N AT I O N, , 8. Low pressure means that less number of gas molecules is present in the discharge tube. If the num-, , ber of molecules is very less, the collisions between, the electrons which move towards the anode with a, high velocity and the gas molecules become effective. These collisions lead to the dislodgement of, electrons from gaseous molecules and formation of, cathode rays takes place. Moreover, high voltage, increases the kinetic energy of the electrons which, in turn increases the probability of removal of electrons from the gaseous molecules., 9. Protium (1H1), deuterium (1H2) and tritium (1H3), are naturally occurring isotopes of hydrogen., Unipositive ions of protium, deuterium and tritium, differ in their mass., , Thus, their e/m ratio is different. Therefore, they, deflect at different angles in an external electric, field., 10. Since in the given experiments, it is observed that, positively-charged particles are detached from the, molecules under low pressure and high voltage,, Thomson’s model would be the other way round,, i.e., positively-charged particles would be embedded in a thinly spread negatively-charged mass.