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genius PHYSICS, Atomic structure 1, , Important Atomic Models., (1) Thomson's model, J.J. Thomson gave the first idea regarding structure of atom. According to this model., (i) An atom is a solid sphere in which entire and positive charge and it's mass is uniformly distributed and, in which negative charge (i.e. electron) are embedded like seeds in watermelon., Positively charged, sphere, , –, , –, , –, –, , –, , Electron, , –, , Success and failure, Explained successfully the phenomenon of thermionic emission, photoelectric emission and ionization., The model fail to explain the scattering of - particles and it cannot explain the origin of spectral lines, observed in the spectrum of hydrogen and other atoms., (2) Rutherford's model, Rutherford's -particle scattering experiment, Rutherford performed experiments on the scattering of alpha particles by extremely thin gold foils and, made the following observations, , , Number of scattered particles :, b, , r0, , N, , N, , 1, sin 4 ( / 2), , Nucleus, , N(180°), , , , -particle, (energy E ), , (i) Most of the -particles pass through the foil straight away undeflected., (ii) Some of them are deflected through small angles., (iii) A few -particles (1 in 1000) are deflected through the angle more than 90o., (iv) A few  -particles (very few) returned back i.e. deflected by 180o., (v) Distance of closest approach (Nuclear dimension), The minimum distance from the nucleus up to which the -particle approach, is called the distance of, 1, Ze 2, 1, ., ; E  mv 2  K.E. of -particle, closest approach (r0). From figure r0 , 4  0 E, 2, (vi) Impact parameter (b) : The perpendicular distance of the velocity vector ( v ) of the -particle from, the centre of the nucleus when it is far away from the nucleus is known as impact parameter. It is given as, b, , Ze 2 cot( / 2),  b  cot( / 2), 1, , 4  0  mv 2 , 2, 

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genius PHYSICS, , 2 Atomic Structure, , Note : , , If t is the thickness of the foil and N is the number of -particles scattered in a, , particular direction ( = constant), it was observed that, , N, t, N,  constant  1  1 ., N 2 t2, t, , After Rutherford's scattering of -particles experiment, following conclusions were made as regard as, atomic structure :, Atom, , (a) Most of the mass and all of the charge of an atom concentrated in a, very small region is called atomic nucleus., (b) Nucleus is positively charged and it's size is of the order of, Fermi., , 10–15, , +, , m1, , Nucleus, , 10–15 m, , (c) In an atom there is maximum empty space and the electrons revolve, around the nucleus in the same way as the planets revolve around the sun., , 10–10 m, Size of the nucleus = 1 Fermi = 10–15 m, Size of the atom 1 Å = 10–10 m, , Draw backs, (i) Stability of atom : It could not explain stability of atom because according to classical electrodynamic, theory an accelerated charged particle should continuously radiate energy. Thus an electron moving in an, circular path around the nucleus should also radiate energy and thus move into, smaller and smaller orbits of gradually decreasing radius and it should, ultimately fall into nucleus., e–, (ii) According to this model the spectrum of atom must be continuous, where as practically it is a line spectrum., (iii) It did not explain the distribution of electrons outside the nucleus., (3) Bohr's model, , Instability of atom, , Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a, single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom), Bohr's model is based on the following postulates., (i) The electron can revolve only in certain discrete non-radiating orbits, called stationary orbits, for, h, which total angular momentum of the revolving electrons is an integral multiple of, ( ), 2,  h, i.e. L  n ,  2, , ,   mvr ; where n = 1, 2, 3, ……..= Principal quantum number, , , (ii) The radiation of energy occurs only when an electron jumps from one permitted orbit to another., When electron jumps from higher energy orbit (E1) to lower energy orbit (E2) then difference of energies, of these orbits i.e. E1 – E2 emits in the form of photon. But if electron goes from E2 to E1 it absorbs the same, amount of energy., E, E1, , 1, , E1 – E2 = h, , E2, Emission, , Note : , will be, , E1 – E2 = h, , E2, Absorption, , According to Bohr theory the momentum of an e  revolving in second orbit of H 2 atom, , h, , ,  For an electron in the nth orbit of hydrogen atom in Bohr model, circumference of orbit  n ;, where  = de-Broglie wavelength.

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genius PHYSICS, Atomic structure 3, Bohr's Orbits (For Hydrogen and H2-Like Atoms)., (1) Radius of orbit, For an electron around a stationary nucleus the electrostatics force of attraction provides the necessary, centripetal force, i.e., , (Ze )e mv 2, , 4  0 r 2, r, 1, , also mvr , , ……. (i), , nh, 2, , …….(ii), , From equation (i) and (ii) radius of nth orbit, r, , , 1 ,  where k , , 4  0 , , , n 2 h 2 0, n 2h2, n2, rn , , , 0, ., 53, Å, Z, 4 2 kZme 2 mZe 2, , n, , n2,  rn , Z, , Note : , , The radius of the innermost orbit (n = 1) hydrogen atom (z = 1) is called Bohr's radius a0, , i.e. a 0  0 .53 Å ., (2) Speed of electron, From the above relations, speed of electron in nth orbit can be calculated as, vn , , 2kZe 2, Ze 2,  c, , , nh, 2 0 nh  137, , v, ,  Z, 6 Z, m / sec, .  2 . 2  10, n,  n, , where (c = speed of light 3  108 m/s), , Note : , , n, , The ratio of speed of an electron in ground state in Bohr's first orbit of hydrogen atom to, , velocity of light in air is equal to, , e2, 1, , (where c = speed of light in air), 2 0 ch 137, , (3) Some other quantities, For the revolution of electron in nth orbit, some other quantities are given in the following table, Quantity, (1) Angular speed, (2) Frequency, (3) Time period, (4) Angular momentum, (5) Corresponding current, (6) Magnetic moment, , Formula, , Dependency on n and Z, , n , , vn, mz 2 e 4, , rn, 2 02 n 3 h 3, , n , , n , , n, mz 2 e 4, , 2, 4  02 n 3 h 3, , n , , Tn , , 1, , n, , , , 4  02 n 3 h 3, mz 2 e 4, ,  h , L n  mv n rn  n , ,  2 , in  e  n , , mz 2 e 5, 4  02 n 3 h 3, ,  , , M n  in A  in  rn2, , Tn , , Z2, n3, Z2, n3, n3, Z2, , Ln  n, , in , , Z2, n3, , Mn  n

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genius PHYSICS, , 4 Atomic Structure, , (where  0 , (7) Magnetic field, , B, ,  0 in, 2rn, , , , eh,  Bohr magneton), 4m, , m 2 z 3 e 7  0, , B, , 8  03 n 5 h 5, , Z3, n5, , (4) Energy, (i) Potential energy : An electron possesses some potential energy because it is found in the field of, (Ze )(e ), kZe 2, , nucleus potential energy of electron in nth orbit of radius rn is given by U  k ., rn, rn, (ii) Kinetic energy : Electron posses kinetic energy because of it's motion. Closer orbits have greater, kinetic energy than outer ones., As we know, , mv 2 k . (Ze )(e ), kZe 2 | U |, , K, , , , Kinetic, energy, rn, 2rn, 2, rn2, , (iii) Total energy : Total energy (E) is the sum of potential energy and kinetic energy i.e. E = K + U,  E, ,  me 4, n 2 h 2 0, kZe 2, , also rn , ., Hence, E, , ,  8 2 h 2, 2rn, mze 2,  0, , where R , , Note : , ,  z2,  me 4, ., , , ,  n2,  8  2 ch 3, ,  0, ,  z2, Z2, Z2,  ch, , , R, ch, , , 13, ., 6, eV,  n2, n2, n2, , , me 4, = Rydberg's constant = 1.09  107 per metre, 8  02 ch 3, , Each Bohr orbit has a definite energy, , 13 .6, eV, n2,  The state with n = 1 has the lowest (most negative) energy. For hydrogen atom it is E1 = – 13.6, eV.,  Rch = Rydberg's energy ~– 2 .17  10 18 J ~– 31 .6 eV .,  For hydrogen atom (Z = 1)  En  , , U, ., 2, (iv) Ionisation energy and potential : The energy required to ionise an atom is called ionisation, energy. It is the energy required to make the electron jump from the present orbit to the infinite orbit.,  E  K , , , 13 .6 Z 2, Z2 , eV, Hence Eionisation  E  En  0    13 .6 2   , n2, n , ,  13 .6(1)2,  13 .6 eV, n2, The potential through which an electron need to be accelerated so that it acquires energy equal to the, E, ionisation energy is called ionisation potential., Vionisation  ionisation, e, (v) Excitation energy and potential : When the electron is given energy from external source, it, jumps to higher energy level. This phenomenon is called excitation., , For H2-atom in the ground state E ionisation , , The minimum energy required to excite an atom is called excitation energy of the particular excited state, and corresponding potential is called exciting potential., EExcitation  EFinal  EInitial and VExcitation , , Eexcitation, e, , (vi) Binding energy (B.E.) : Binding energy of a system is defined as the energy released when it's, constituents are brought from infinity to form the system. It may also be defined as the energy needed to, separate it's constituents to large distances. If an electron and a proton are initially at rest and brought from

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genius PHYSICS, Atomic structure 5, large distances to form a hydrogen atom, 13.6 eV energy will be released. The binding energy of a hydrogen, atom is therefore 13.6 eV., , Note : , , 13 . 6, ., (B.E.), , For hydrogen atom principle quantum number n , , (5) Energy level diagram, The diagrammatic description of the energy of the electron in different orbits around the nucleus is called, energy level diagram., Energy level diagram of hydrogen/hydrogen like atom, , Note : , , n=, n=4, n=3, n=2, , Infinite, Fourth, Third, Second, , Infinite, Third, Second, First, , E = 0 eV, E4 = – 0.85 eV, E3 = – 1.51 eV, E2 = – 3.4 eV, , 0 eV, – 0.85 Z 2, – 1.51 Z 2, – 3.4 Z 2, , 0 eV, + 0.85 eV, + 1.51 eV, + 3.4 eV, , n=1, , First, , Ground, , E1 = – 13.6 eV, , – 13.6 Z 2, , + 13.6 eV, , Principle, quantum, number, , Orbit, , Excited, state, , Energy for H2, , Energy for H2, , Ionisation energy, , – atom, , – like atom, , from this level (for, H2 – atom), , In hydrogen atom excitation energy to excite electron, , from ground state to first, , excited state will be  3.4  (13 .6)  10 .2 eV ., and from ground state to second excited state it is [  1.51  (13 .6)  12 .09 eV ].,  In an H 2 atom when e  makes a transition from an excited state to the ground state it’s kinetic, energy increases while potential and total energy decreases., (6) Transition of electron, When an electron makes transition from higher energy level having energy E2(n2) to a lower energy level, having energy E1 (n1) then a photon of frequency  is emitted, (i) Energy of emitted radiation, E2, n2,  1, ,  Rc h Z 2  Rch Z 2 , 1, 2,   13 .6 Z , , E,,, E  E 2  E1 ,   , 2, ,  n2  n2 , n 22, n, 1, 2 , , ,  1, E1, n1, (ii) Frequency of emitted radiation, Emission, , , E E 2  E1, 1, 1, ,  Rc Z 2  2  2 , E  h   , h, h,  n1 n 2 , (iii) Wave number/wavelength, 1 , Wave, number, is, the, number, of, waves, in, unit, length,   ,  c, 2,  1, 1, 1  13 .6 Z  1, 1 ,   RZ 2  2  2  ,  2 , 2, , , hc  n1 n 2 ,  n1 n 2 , (iv) Number of spectral lines : If an electron jumps from higher energy orbit to lower energy orbit it, emits raidations with various spectral lines., If electron falls from orbit n2 to n1 then the number of spectral lines emitted is given by, (n  n1  1)(n 2  n1 ), NE  2, 2, If electron falls from nth orbit to ground state (i.e. n2 = n and n1 = 1) then number of spectral lines emitted, n (n  1), NE , 2

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genius PHYSICS, , 6 Atomic Structure, , Note : , , Absorption spectrum is obtained only for the transition from lowest energy level to, higher energy levels. Hence the number of absorption spectral lines will be (n – 1)., (v) Recoiling of an atom : Due to the transition of electron, photon is emitted and the atom is recoiled,  1, h, 1 , Recoil momentum of atom = momentum of photon   hRZ 2  2  2 , ,  n1 n 2 , 2, 2, h, p, Also recoil energy of atom , (where m = mass of recoil atom), , 2 m 2m  2, (7) Drawbacks of Bohr's atomic model, (i) It is valid only for one electron atoms, e.g. : H, He+, Li+2, Na+1 etc., (ii) Orbits were taken as circular but according to Sommerfield these are elliptical., (iii) Intensity of spectral lines could not be explained., (iv) Nucleus was taken as stationary but it also rotates on its own axis., (v) It could not be explained the minute structure in spectrum line., (vi) This does not explain the Zeeman effect (splitting up of spectral lines in magnetic field) and Stark, effect (splitting up in electric field), (vii) This does not explain the doublets in the spectrum of some of the atoms like sodium (5890Å &, 5896Å), , Hydrogen Spectrum and Spectral Series., When hydrogen atom is excited, it returns to its normal unexcited (or ground state) state by emitting the, energy it had absorbed earlier. This energy is given out by the atom in the form of radiations of different, wavelengths as the electron jumps down from a higher to a lower orbit. Transition from different orbits cause, different wavelengths, these constitute spectral series which are characteristic of the atom emitting them., When observed through a spectroscope, these radiations are imaged as sharp and straight vertical lines of a, single colour., Photon of, wavelength , +, , +, , Spectrum, , +, , Emission spectra, , Spectral series, The spectral lines arising from the transition of electron forms a spectra series., (i) Mainly there are five series and each series is named after it's discover as Lymen series, Balmer series,, Paschen series, Bracket series and Pfund series., (ii) According to the Bohr's theory the wavelength of the radiations emitted from hydrogen atom is given, by, 1, 1, 1 ,  R 2  2 , ,  n1 n 2 , where n2 = outer orbit (electron jumps from this orbit), n1 = inner orbit (electron falls in this orbit), , (iii) First line of the series is called first member, for this line wavelength is maximum (max)

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genius PHYSICS, Atomic structure 7, (iv) Last line of the series (n2 = ) is called series limit, for this line wavelength is minimum (min), Spectral, series, , Transition, , Wavelength () , , n12 n 22, , 2 , ,  1  n1  R, , n 22 , , Minimum, wavelength, n 2  , n1  n, , (n 22  n12 )R, , Maximum, wavelength, n1  n and n 2  n  1, , n12, , , ,  max, (n  1) 2, ,  min, (2n  1), , Region, , max , , n 2 (n  1)2, (2n  1)R, , min , ,  max , , (1)2 (1  1)2, 4, , (2  1  1)R 3 R, , n1 = n = 1, 1, min , R, , 4, 3, , Ultraviolet, region, , 4, R, , 9, 5, , Visible, region, , n2, R, , 1., Lymen, series, , n2 = 2, 3, 4 …, n1 = 1, , 2.Balmer, series, , n2 = 3, 4, 5 …, n1 = 2, , n1 = n = 2, n2 = 2 + 1 = 3, 36, max , 5R, , 3., Paschen, series, , n2 = 4, 5, 6 …, n1 = 3, , n1 = n = 3, n2 = 3 + 1 = 4, 144, max , 7R, , n1 = n = 3, 9, min , R, , 16, 7, , Infrared, region, , 4., Bracket, series, , n2 = 5, 6, 7 … , n1 = 4, , n1 = n = 4, n2 = 4 + 1 = 5, 400, max , 9R, , n1 = n = 4, 16, min , R, , 25, 9, , Infrared, region, , 5., Pfund, series, , n2 = 6, 7, 8 … , n1 = 5, , n1 =  = 5, n2 = 5 + 1 = 6, 900, max , 11 R, , 25, R, , 36, 11, , Infrared, region, ,  min , , min , , Quantum Numbers., An atom contains large number of shells and subshells. These are distinguished from one another on the, basis of their size, shape and orientation (direction) in space. The parameters are expressed in terms of, different numbers called quantum number., Quantum numbers may be defined as a set of four number with the help of which we can get complete, information about all the electrons in an atom. It tells us the address of the electron i.e. location, energy, the, type of orbital occupied and orientation of that orbital., (1) Principal Quantum number (n) : This quantum number determines the main energy level or shell, in which the electron is present. The average distance of the electron from the nucleus and the energy of the, 1, electron depends on it., (in H-atom), E n  2 and rn  n 2, n, The principal quantum number takes whole number values, n = 1, 2, 3, 4,….. , (2) Orbital quantum number (l) or azimuthal quantum number (l), This represents the number of subshells present in the main shell. These subsidiary orbits within a shell, will be denoted as 1, 2, 3, 4 … or s, p, d, f … This tells the shape of the subshells., The orbital angular momentum of the electron is given as L  l(l  1), , h, 2, , (for a particular value of n)., , For a given value of n the possible values of l are l = 0, 1, 2, ….. upto (n – 1), (3) Magnetic quantum number (ml) : An electron due to it's angular motion around the nucleus, generates an electric field. This electric field is expected to produce a magnetic field. Under the influence of, external magnetic field, the electrons of a subshell can orient themselves in certain preferred regions of space, around the nucleus called orbitals.

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genius PHYSICS, , 8 Atomic Structure, , The magnetic quantum number determines the number of preferred orientations of the electron present, in a subshell., The angular momentum quantum number m can assume all integral value between – l to +l including, zero. Thus ml can be – 1, 0, + 1 for l = 1. Total values of ml associated with a particular value of l is given by (2l +, 1)., (4) Spin (magnetic) quantum number (ms) : An electron in atom not only revolves around the, nucleus but also spins about its own axis. Since an electron can spin either in clockwise direction or in, anticlockwise direction. Therefore for any particular value of magnetic quantum number, spin quantum, 1, 1, number can have two values, i.e., or, (Spin down), ms  , m s  (Spin up), 2, 2, This quantum number helps to explain the magnetic properties of the substance., , Electronic Configurations of Atoms., The distribution of electrons in different orbitals of an atom is called the electronic configuration of the, atom. The filling of electrons in orbitals is governed by the following rules., (1) Pauli's exclusion principle, "It states that no two electrons in an atom can have all the four quantum number (n, l, ml and ms) the, same.", It means each quantum state of an electron must have a different set of quantum numbers n, l, ml and ms., This principle sets an upper limit on the number of electrons that can occupy a shell., N max in one shell = 2n2; Thus Nmax in K, L, M, N …. shells are 2, 8, 18, 32,, , Note : , , The maximum number of electrons in a subshell with orbital quantum number l is 2(2l +, , 1)., (2) Aufbau principle, Electrons enter the orbitals of lowest energy first., As a general rule, a new electron enters an empty orbital for which (n + l ) is minimum. In case the value, (n  l) is equal for two orbitals, the one with lower value of n is filled first., Thus the electrons are filled in subshells in the following order (memorize), 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, ……, (3) Hund's Rule, When electrons are added to a subshell where more than one orbital of the same energy is available, their, spins remain parallel. They occupy different orbitals until each one of them has at least one electron. Pairing, starts only when all orbitals are filled up., Pairing takes place only after filling 3, 5 and 7 electrons in p, d and f orbitals, respectively.

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genius PHYSICS, Atomic structure 9, Concepts, , , , With the increase in principal quantum number the energy difference between the two successive energy level decreases,, while wavelength of spectral line increases., E'  E' '  E' ' ', , E, , , '  ' '  ' ' ', , E, , , E  E ' E ' ' E ' ' ', , 1, , , , , , , , E, , , n=4, n=3, n=2, , 1, 1, 1, , , ' ' ' ' ' ', , E, , n=1, , Rydberg constant is different for different elements, R( =1.09  107 m–1) is the value of Rydberg constant when the nucleus is considered to be infinitely massive as compared to, the revolving electron. In other words, the nucleus is considered to be stationary., In case, the nucleus is not infinitely massive or stationary, then the value of Rydberg constant is given as R ' , , R, 1, , m, M, , where m is the mass of electron and M is the mass of nucleus., , , , Atomic spectrum is a line spectrum, Each atom has it's own characteristic allowed orbits depending upon the electronic configuration. Therefore photons, emitted during transition of electrons from one allowed orbit to inner allowed orbit are of some definite energy only. They, do not have a continuous graduation of energy. Therefore the spectrum of the emitted light has only some definite lines, and therefore atomic spectrum is line spectrum., , , , Just as dots of light of only three colours combine to form almost every conceivable colour on T.V. screen, only about 100, distinct kinds of atoms combine to form all the materials in the universe., , B, a, 92U, 235, , 92U, 236, , K, r

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genius PHYSICS, , 10 Atomic Structure, , Rutherford's -scattering experiment established that the mass of atom is concentrated with small positively, charged region at the centre which is called 'nucleus'., Nuclei are made up of proton and neutron. The number of protons in a nucleus, (called the atomic number or proton number) is represented by the symbol Z. The, number of neutrons (neutron number) is represented by N. The total number of, neutrons and protons in a nucleus is called it's mass number A so A = Z + N., , e–, , Neutrons and proton, when described collectively are called nucleons., , e–, , Nucleus contains two types of particles : Protons and neutrons, Nuclides are represented as, , Z, , e–, , X A ; where X denotes the chemical symbol of the, , element., , Neutron., Neutron is a fundamental particle which is essential constituent of all nuclei except that of hydrogen atom. It was, discovered by Chadwick., (1) The charge of neutron : It is neutral, (2) The mass of neutron : 1.6750  10–27 kg, (3) It's spin angular momentum :, , 1  h, , 2  2, , A free neutron outside the nucleus is unstable, and decays into proton and electron., , , J -s, , , 0n, , 1, , , , (4) It's magnetic moment : 9.57  10–27 J/Tesla, , 1, 1H, Proton, , 0, 1 , Electron, , , , , , , , Antinutrin o, , (5) It's half life : 12 minutes, (6) Penetration power : High, (7) Types : Neutrons are of two types slow neutron and fast neutron, both are fully capable of penetrating a nucleus, and causing artificial disintegration., Thermal neutrons, Fast neutrons can be converted into slow neutrons by certain materials called moderator's (Paraffin wax, heavy water, graphite) when fast moving, neutrons pass through a moderator, they collide with the molecules of the moderator, as a result of this, the energy of moving neutron decreases while, that of the molecules of the moderator increases. After sometime they both attains same energy. The neutrons are then in thermal equilibrium with the, molecules of the moderator and are called thermal neutrons., , Note : , , Energy of thermal neutron is about 0.025 eV and speed is about 2.2 km/s., , Nucleus., (1) Different types of nuclei, The nuclei have been classified on the basis of the number of protons (atomic number) or the total number of, nucleons (mass number) as follows, (i) Isotopes : The atoms of element having same atomic number but different mass number are called isotopes. All, isotopes have the same chemical properties. The isotopes of some elements are the following, 1, , H 1, 1H 2, 1H 3, , 8O, , 16, , , 8 O 17 , 8 O 18, , 2, , He 3 ,, , 2 He, , 4, , 17 Cl, , 35, , ,, , 17 Cl, , 37, , 92 U, , 235, , ,, , 92 U, , 238, , (ii) Isobars : The nuclei which have the same mass number (A) but different atomic number (Z) are called isobars., Isobars occupy different positions in periodic table so all isobars have different chemical properties. Some of the examples, of isobars are, 1, , H 3 and, , 2 He, , 3, , ,, , 6C, , 14, , and, , 7N, , 14, , , 8 O 17 and, , 9F, , 17, , (iii) Isotones : The nuclei having equal number of neutrons are called isotones. For them both the atomic number, (Z) and mass number (A) are different, but the value of (A – Z) is same. Some examples are, 4, , Be 9 and, , 5B, , 10, , , 6 C 13 and, , 7N, , 14, , , 8 O 18 and, , 9F, , 19, , , 3 Li 7 and, , 4 Be, , 8, , , 1 H 3 and, , 2 He, , 4

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genius PHYSICS, Atomic structure 11, (iv) Mirror nuclei : Nuclei having the same mass number A but with the proton number (Z) and neutron number, (A – Z) interchanged (or whose atomic number differ by 1) are called mirror nuclei for example., 1H, , 3, , and 2 He 3 , 3 Li 7 and, , 4 Be, , 7, , (2) Size of nucleus, (i) Nuclear radius : Experimental results indicates that the nuclear radius is proportional to A1/3, where A is the mass, number of nucleus i.e., , Note : , , R  A1 / 3, ,  R  R 0 A 1 / 3 , where R0 = 1.2  10–15 m = 1.2 fm., , Heavier nuclei are bigger in size than lighter nuclei., , (ii) Nuclear volume : The volume of nucleus is given by V , , 4, 4,  R 3   R 03 A  V  A, 3, 3, , (iii) Nuclear density : Mass per unit volume of a nucleus is called nuclear density., , Nuclear density( ) , , Massof nucleus, mA, , Volume of nucleus 4,  (R 0 A 1 / 3 )3, 3, , where m = Average of mass of a nucleon (= mass of proton + mass of neutron = 1.66  10–27 kg), and mA = Mass of nucleus,   , , 3m,  2 . 38  10 17 kg / m 3, 3, 4R 0, , Note : , ,  is independent of A, it means  is same of all atoms., ,  Density of a nucleus is maximum at it's centre and decreases as we move outwards from the nucleus., , (3) Nuclear force, Forces that keep the nucleons bound in the nucleus are called nuclear forces., (i) Nuclear forces are short range forces. These do not exist at large distances, greater than 10–15 m., (ii) Nuclear forces are the strongest forces in nature., , (iii) These are attractive force and causes stability of the nucleus., , At low speeds, electromagnetic, repulsion prevents the collision of, nuclei, , (iv) These forces are charge independent., (v) Nuclear forces are non-central force., Nuclear forces are exchange forces, According to scientist Yukawa the nuclear force between the two nucleons is the, result of the exchange of particles called mesons between the nucleons., , At high speeds, nuclei come close, enough for the strong force to bind, them together., ,  - mesons are of three types – Positive  meson (+), negative  meson ( –), neutral  meson (0), The force between neutron and proton is due to exchange of charged meson between them i.e., , p     n,, , n  p  , , The forces between a pair of neutrons or a pair of protons are the result of the exchange of neutral meson ( o), between them i.e., , p  p ' 0 and, , n  n' 0, , Thus exchange of  meson between nucleons keeps the nucleons bound together. It is responsible for the nuclear, forces.

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genius PHYSICS, , 12 Atomic Structure, , Dog-Bone analogy, The above interactions can be explained with the dog bone analogy according to which we, consider the two interacting nucleons to be two dogs having a common bone clenched in between, their teeth very firmly. Each one of these dogs wants to take the bone and hence they cannot be, separated easily. They seem to be bound to each other with a strong attractive force (which is the, bone) though the dogs themselves are strong enemies. The meson plays the same role of the, common bone in between two nucleons., (4) Atomic mass unit (amu), The unit in which atomic and nuclear masses are measured is called atomic mass unit (amu), 1 amu (or 1u) =, , 1, th of mass of 6 C 12 atom = 1.66  10–27 kg, 12, , Masses of electron, proton and neutrons, Mass of electron (me) = 9.1  10–31 kg = 0.0005486 amu, Mass of proton (mp) = 1.6726  10–27 kg = 1.007276 amu, Mass of neutron (mn) = 1.6750  10–27 kg = 1.00865 amu, Mass of hydrogen atom (me + mp) = 1.6729  10–27 kg = 1.0078 amu, Mass-energy equivalence, According to Einstein, mass and energy are inter convertible. The Einstein's mass energy relationship is given by E  mc 2, If m = 1 amu, c = 3  108 m/sec then E = 931 MeV i.e. 1 amu is equivalent to 931 MeV or 1 amu (or 1 u) = 931 MeV, (5) Pair production and pair-annihilation, When an energetic -ray photon falls on a heavy substance. It is absorbed by some nucleus of the substance and an, electron and a positron are produced. This phenomenon is called pair production and may be represented by the following, equation, , h, , (  photon), , , , 0, 1, (Positron), , , , 0, 1 , (Electron), , , , +1 0, , The rest-mass energy of each of positron and electron is, E0 = m0c2 = (9.1  10–31 kg)  (3.0  108 m/s)2, , h, , +Ze, , -photon, , Nucleus, , , , = 8.2  10–14 J = 0.51 MeV, , –1 0, , Hence, for pair-production it is essential that the energy of -photon must be at least 2  0.51 = 1.02 MeV. If the, energy of -photon is less than this, it would cause photo-electric effect or Compton effect on striking the matter., The converse phenomenon pair-annihilation is also possible. Whenever an electron and a positron come very close, to each other, they annihilate each other by combining together and two -photons (energy) are produced. This, phenomenon is called pair annihilation and is represented by the following equation., 0, 1 , (Positron), , 0, 1 , (Electron), , , , , , h, , ( - photon ), , , , h, , ( - photon ), , (6) Nuclear stability, Among about 1500 known nuclides, less than 260 are stable. The others are unstable that decay to form other, nuclides by emitting , -particles and  - EM waves. (This process is called radioactivity). The stability of nucleus is, determined by many factors. Few such factors are given below :, , N, , Ratio , Z, , , (i) Neutron-proton ratio , , The chemical properties of an atom are governed entirely by the number of protons (Z) in the nucleus, the stability of, an atom appears to depend on both the number of protons and the number of neutrons.

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genius PHYSICS, Atomic structure 13, For lighter nuclei, the greatest stability is achieved when the number of protons and neutrons are approximately, equal (N  Z) i.e., , N, 1, Z, , Neutron, number (N), , Heavy nuclei are stable only when they have more neutrons than protons. Thus heavy nuclei are neutron rich, compared to lighter nuclei (for heavy nuclei, more is the number of protons in the nucleus, greater is the electrical, repulsive force between them. Therefore more neutrons are added to provide the strong attractive forces necessary to keep, the nucleus stable.), Stable nuclei, (Line of, stability), N=Z, 10 20 30405060708090, Proton number, (Z), , Figure shows a plot of N verses Z for the stable nuclei. For mass number upto about A = 40. For larger value of Z the, nuclear force is unable to hold the nucleus together against the electrical repulsion of the protons unless the number of, neutrons exceeds the number of protons. At Bi (Z = 83, A = 209), the neutron excess in N – Z = 43. There are no stable, nuclides with Z > 83., , Note : , , The nuclide, , 83, , Bi 209 is the heaviest stable nucleus., , ,  A nuclide above the line of stability i.e. having excess neutrons, decay through  emission (neutron, , emission,, , N, ratio decreases., Z, , A nuclide below the line of stability have excess number of protons. It decays, by , the, , , , emission, results in decreasing Z and increasing N. In , , , , emission,, , N, ratio increases., Z, , Neutron, Number (N), , , changes into proton). Thus increasing atomic number Z and decreasing neutron number N. In , , –, , N=Z, , +, Proton number (Z), , (ii) Even or odd numbers of Z or N : The stability of a nuclide is also determined by the consideration whether it, contains an even or odd number of protons and neutrons., It is found that an even-even nucleus (even Z and even N) is more stable (60% of stable nuclide have even Z and even, N)., An even-odd nucleus (even Z and odd N) or odd-even nuclide (odd Z and even N) is found to be lesser sable while, the odd-odd nucleus is found to be less stable., 2, Only five stable odd-odd nuclides are known : 1 H ,, , 3 Li, , 6, , , 5 Be 10 , 7 N 14 and, , 75 Ta, , 180, , (iii) Binding energy per nucleon : The stability of a nucleus is determined by value of it's binding energy per nucleon., In general higher the value of binding energy per nucleon, more stable the nucleus is, , Mass Defect and Binding Energy., (1) Mass defect (m)

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genius PHYSICS, , 14 Atomic Structure, , It is found that the mass of a nucleus is always less than the sum of masses of it's constituent nucleons in free state., This difference in masses is called mass defect. Hence mass defect, m = Sum of masses of nucleons – Mass of nucleus, ,  Zm p  ( A  Z)m n   M  Zm p  Zm e  ( A  Z)m z   M ', where mp = Mass of proton, mn = Mass of each neutron, me = Mass of each electron, M = Mass of nucleus, Z = Atomic number, A = Mass number, M = Mass of atom as a whole., , Note : , , The mass of a typical nucleus is about 1% less than the sum of masses of nucleons., , (2) Packing fraction, Mass defect per nucleon is called packing fraction, Packing fraction (f ) , , m M  A, , A, A, , where M = Mass of nucleus, A = Mass number, , Packing fraction measures the stability of a nucleus. Smaller the value, of packing fraction, larger is the stability of the nucleus., (i) Packing fraction may be of positive, negative or zero value., , 40, 30, 20, 10, 0, , (iii) At A = 16, f  Zero, (3) Binding energy (B.E.), , – 10, – 20, , A > 240, , Mass, number (A), , The neutrons and protons in a stable nucleus are held together by, nuclear forces and energy is needed to pull them infinitely apart (or the, same energy is released during the formation of the nucleus). This energy is called the binding energy of the nucleus., or, The binding energy of a nucleus may be defined as the energy equivalent to the mass defect of the nucleus., If m is mass defect then according to Einstein's mass energy relation, Binding energy = m  c2 = [{mpZ + mn(A – Z)} – M] c2, (This binding energy is expressed in joule, because m is measured in kg), If m is measured in amu then binding energy = m amu = [{mpZ + mn(A – Z)} – M] amu = m  931 MeV, (4) Binding energy per nucleon, The average energy required to release a nucleon from the nucleus is called binding energy per nucleon., , Binding energy per nucleon , , Total bind ing energy, m  931 MeV, , Mass number (i.e . total number of nucleons), A, Nucleon, , Binding energy per nucleon  Stability of nucleus, , Binding Energy Curve., , Binding energy per, nucleon (MeV), , It is the graph between binding energy per nucleon and total number of nucleons (i.e. mass number A), 8.0 He, , 26Fe56, , 6.0, 4.0, , Li, , 2.0 H2, 0, 56, , 100, 150, Mass number A, , 200, , (1) Some nuclei with mass number A < 20 have large binding energy per nucleon than their neighbour nuclei. For, example, , 2, , He 4 , 4 Be 8 , 6 C 12 , 8 O 16 and, , 10, , Ne 20 . These nuclei are more stable than their neighbours.

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genius PHYSICS, Atomic structure 15, 56, (2) The binding energy per nucleon is maximum for nuclei of mass number A = 56 ( 26 Fe ) . It's value is 8.8 MeV, , per nucleon., (3) For nuclei having A > 56, binding energy per nucleon gradually decreases for uranium (A = 238), the value of, binding energy per nucleon drops to 7.5 MeV., , Note : , , When a heavy nucleus splits up into lighter nuclei, then binding energy per nucleon of lighter, , nuclei is more than that of the original heavy nucleus. Thus a large amount of energy is liberated in this, process (nuclear fission).,  When two very light nuclei combines to form a relatively heavy nucleus, then binding energy per nucleon, increases. Thus, energy is released in this process (nuclear fusion)., B. E., A, , +, , Fusion, , Fission, , +, , Nuclear Reactions., , A, , The process by which the identity of a nucleus is changed when it is bombarded by an energetic particle is called, nuclear reaction. The general expression for the nuclear reaction is as follows., , X, , (Parent nucleus), , , , a, , (Incident particle), , , , C, , (Com pound nucleus), , , , Y, , (Com pound nucleus), , , , b, ,  Q, , (Product particles), , (Energy ), , Here X and a are known as reactants and Y and b are known as products. This reaction is known as (a, b) reaction, and can be represented as X(a, b) Y, (1) Q value or energy of nuclear reaction, The energy absorbed or released during nuclear reaction is known as Q-value of nuclear reaction., Q-value = (Mass of reactants – mass of products)c2 Joules, = (Mass of reactants – mass of products) amu, If Q < 0, The nuclear reaction is known as endothermic. (The energy is absorbed in the reaction), If Q > 0, The nuclear reaction is known as exothermic (The energy is released in the reaction), (2) Law of conservation in nuclear reactions, (i) Conservation of mass number and charge number : In the following nuclear reaction, 2, , He 4  7 N 14  8 O 17  1 H 1, , Mass number (A) , , Before the reaction, 4 +14 = 18, , Charge number (Z) , , 2+7=9, , After the reaction, 17 + 1 = 18, 8+1=9, , (ii) Conservation of momentum : Linear momentum/angular momentum of particles before the reaction is equal to, the linear/angular momentum of the particles after the reaction. That is p = 0, (iii) Conservation of energy : Total energy before the reaction is equal to total energy after the reaction. Term Q is, added to balance the total energy of the reaction., (3) Common nuclear reactions, The nuclear reactions lead to artificial transmutation of nuclei. Rutherford was the first to carry out artificial, transmutation of nitrogen to oxygen in the year 1919., 2, , He 4  7 N 14  9 F 18  8 O 17  1 H 1

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genius PHYSICS, , 16 Atomic Structure, , It is called (, p) reaction. Some other nuclear reactions are given as follows., (p, n) reaction , , 1, , H 1  5 B 11  6 C 12  6 C 11  0 n 1, , (p, ) reaction , , 1, , H 1  3 Li 11  4 Be 8  2 He 4  2 He 4, , (p, ) reaction , , 1H, ,  6 C 12  7 N 13  7 N 13  , , 1, ,  7 N 14  7 N 15  6 C 14  1 H 1, , (n, p) reaction , , 0n, , (, n) reaction , ,   1 H 2  1 H 1  0 n1, , 1, , Nuclear Fission and Fusion., Nuclear fission, The process of splitting of a heavy nucleus into two lighter nuclei of comparable masses (after bombardment with a, energetic particle) with liberation of energy is called nuclear fission., The phenomenon of nuclear fission was discovered by scientist Ottohann and F. Strassman and was explained by N., Bohr and J.A. Wheeler on the basis of liquid drop model of nucleus., (1) Fission reaction of U235, (i) Nuclear reaction :, 92 U, , 235, ,  0 n1 , , 236, 92 U, (unstable nucleus), , , , 56 Ba, , 141, , , , 36 Kr, , 92, ,  3 0 n1  Q, , (ii) The energy released in U235 fission is about 200 MeV or 0.8 MeV per nucleon., (iii) By fission of, , 92 U, , 235, , , on an average 2.5 neutrons are liberated. These neutrons are called fast neutrons and their, , energy is about 2 MeV (for each). These fast neutrons can escape from the reaction so as to proceed the chain reaction they, are need to slow down., (iv) Fission of U235 occurs by slow neutrons only (of energy about 1eV) or even by thermal neutrons (of energy about, 0.025 eV)., (v) 50 kg of U235 on fission will release  4 × 1015 J of energy. This is equivalence to 20,000 tones of TNT explosion., The nuclear bomb dropped at Hiroshima had this much explosion power., (vi) The mass of the compound nucleus must be greater than the sum of masses of fission products., (vii) The, , Binding energy, of compound nucleus must be less than that of the fission products., A, , (viii) It may be pointed out that it is not necessary that in each fission of uranium, the two fragments, 36, , Kr are formed but they may be any stable isotopes of middle weight atoms., Same other U 235 fission reactions are, 92 U, , 235, ,  0 n 1  54 Xe 140 ,  57 La 148 , , 38 Sr, , 35 Br, , 94, , 85, ,  2 0 n1, ,  3 0 n1, ,  Many more, (ix) The neutrons released during the fission process are called prompt neutrons., (x) Most of energy released appears in the form of kinetic energy of fission fragments., Ba, Energy, , Slow, Neutron, , (2) Chain reaction, , Energy, 92U235, , 92U236, , Energy, Energy, Kr, , 56, , Ba and

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genius PHYSICS, Atomic structure 17, In nuclear fission, three neutrons are produced along with the release of large energy. Under favourable conditions,, these neutrons can cause further fission of other nuclei, producing large number of neutrons. Thus a chain of nuclear, fissions is established which continues until the whole of the uranium is consumed., Kr, U, , Kr, n, , U, , Ba, Kr, , Ba, , U, Ba, , In the chain reaction, the number of nuclei undergoing fission increases very fast. So, the energy produced takes a, tremendous magnitude very soon., Difficulties in chain reaction, (i) Absorption of neutrons by U 238 , the major part in natural uranium is the isotope U238 (99.3%), the isotope U 235, is very little (0.7%). It is found that U 238 is fissionable with fast neutrons, whereas U 235 is fissionable with slow neutrons., 238, Due to the large percentage of U , there is more possibility of collision of neutrons with U 238 . It is found that the, , 238, neutrons get slowed on coliding with U , as a result of it further fission of U238 is not possible (Because they are slow, , and they are absorbed by U238). This stops the chain reaction., , , , Removal : (i) To sustain chain reaction, 92 U, , 235, , 92 U, , 235, , is separated from the ordinary uranium. Uranium so obtained, ,  is known as enriched uranium, which is fissionable with the fast and slow neutrons and hence chain reaction can, , be sustained., (ii) If neutrons are slowed down by any method to an energy of about 0.3 eV, then the probability of their absorption, 238, , by U, becomes very low, while the probability of their fissioning U 235 becomes high. This job is done by moderators., Which reduce the speed of neutron rapidly graphite and heavy water are the example of moderators., (iii) Critical size : The neutrons emitted during fission are very fast and they travel a large distance before being, slowed down. If the size of the fissionable material is small, the neutrons emitted will escape the fissionable material, before they are slowed down. Hence chain reaction cannot be sustained., Removal : The size of the fissionable material should be large than a critical size., The chain reaction once started will remain steady, accelerate or retard depending upon, a factor called neutron, reproduction factor (k). It is defined as follows., , k, , Rate of production of neutrons, Rate of loss of neutrons, ,  If k = 1, the chain reaction will be steady. The size of the fissionable material used is said to be the critical size and, it's mass, the critical mass.,  If k > 1, the chain reaction accelerates, resulting in an explosion. The size of the material in this case is super, critical. (Atom bomb),  If k < 1, the chain reaction gradually comes to a halt. The size of the material used us said to be sub-critical., Types of chain reaction : Chain reactions are of following two types, Controlled chain reaction, , Uncontrolled chain reaction, , Controlled by artificial method, , No control over this type of nuclear reaction, , All neurons are absorbed except one, , More than one neutron takes part into reaction, , It's rate is slow, , Fast rate

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genius PHYSICS, , 18 Atomic Structure, , Reproduction factor k = 1, , Reproduction factor k > 1, , Energy liberated in this type of reaction is always less, than explosive energy, , A large amount of energy is liberated in this type of, reaction, , Chain reaction is the principle of nuclear reactors, , Uncontrolled chain reaction is the principle of atom, bomb., , Note : , , The energy released in the explosion of an atom bomb is equal to the energy released by 2000, , tonn of TNT and the temperature at the place of explosion is of the order of 10 7 oC., , Nuclear Reactor., A nuclear reactor is a device in which nuclear fission can be carried out through a sustained and a controlled chain reaction. It is also called an atomic, pile. It is thus a source of controlled energy which is utilised for many useful purposes., Cadmium rods, , Core, , Coolant, Coolant out, , Turbine, , Concrete, wall, , To electric, generator, Condenser, , Moderator, , Water, Coolant in, , (1) Parts of nuclear reactor, , Heat, exchanger, , Fuel elements, , (i) Fissionable material (Fuel) : The fissionable material used in the reactor is called the fuel of the reactor., Uranium isotope (U235) Thorium isotope (Th232) and Plutonium isotopes (Pu239, Pu240 and Pu241) are the most commonly, used fuels in the reactor., (ii) Moderator : Moderator is used to slow down the fast moving neutrons. Most commonly used moderators are, graphite and heavy water (D2O)., (iii) Control Material : Control material is used to control the chain reaction and to maintain a stable rate of, reaction. This material controls the number of neutrons available for the fission. For example, cadmium rods are inserted, into the core of the reactor because they can absorb the neutrons. The neutrons available for fission are controlled by, moving the cadmium rods in or out of the core of the reactor., (iv) Coolant : Coolant is a cooling material which removes the heat generated due to fission in the reactor., Commonly used coolants are water, CO2 nitrogen etc., (v) Protective shield : A protective shield in the form a concrete thick wall surrounds the core of the reactor to, save the persons working around the reactor from the hazardous radiations., , Note : , , It may be noted that Plutonium is the best fuel as compared to other fissionable material. It is, , because fission in Plutonium can be initiated by both slow and fast neutrons. Moreover it can be obtained, from U 238 .,  Nuclear reactor is firstly devised by fermi., (2) Uses of nuclear reactor, , , , Apsara was the first Indian nuclear reactor., , (i) In electric power generation., , (ii) To produce radioactive isotopes for their use in medical science, agriculture and industry., (iii) In manufacturing of PU 239 which is used in atom bomb., (iv) They are used to produce neutron beam of high intensity which is used in the treatment of cancer and nuclear, research., , Note : , Nuclear fusion, , A type of reactor that can produce more fissile fuel than it consumes is the breeder reactor.

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genius PHYSICS, Atomic structure 19, In nuclear fusion two or more than two lighter nuclei combine to form a single heavy nucleus. The mass of single, nucleus so formed is less than the sum of the masses of parent nuclei. This difference in mass results in the release of, tremendous amount of energy, , or, , 1, , H 2  1 H 2  1 H 3  1 H 1  4 MeV, , 1, , H 3  1 H 2  2 He 4  0 n 1  17 .6 MeV, , 1, , H 2  1 H 2  2 He 4  24 MeV, , For fusion high pressure ( 106 atm) and high temperature (of the order of 10 7 K to 108 K) is required and so the, reaction is called thermonuclear reaction., Fusion energy is greater then fission energy fission of one uranium atom releases about 200 MeV of energy. But the, 2, 3, fusion of a deutron (1 H ) and triton (1 H ) releases about 17.6 MeV of energy. However the energy released per nucleon, , in fission is about 0.85 MeV but that in fusion is 4.4 MeV. So for the same mass of the fuel, the energy released in fusion is, much larger than in fission., Plasma : The temperature of the order of 108 K required for thermonuclear reactions leads to the complete, ionisation of the atom of light elements. The combination of base nuclei and electron cloud is called plasma. The, enormous gravitational field of the sun confines the plasma in the interior of the sun., The main problem to carryout nuclear fusion in the laboratory is to contain the plasma at a temperature of 108K. No, solid container can tolerate this much temperature. If this problem of containing plasma is solved, then the large quantity, of deuterium present in sea water would be able to serve as in-exhaustible source of energy., , Note : , , To achieve fusion in laboratory a device is used to confine the plasma, called Tokamak., , Stellar Energy, Stellar energy is the energy obtained continuously from the sun and the stars. Sun radiates energy at the rate of, about 1026 joules per second., Scientist Hans Bethe suggested that the fusion of hydrogen to form helium (thermo nuclear reaction) is continuously, taking place in the sun (or in the other stars) and it is the source of sun's (star's) energy., The stellar energy is explained by two cycles, Proton-proton cycle, , Carbon-nitrogen cycle, , 1H, , 1, ,  1 H  1 H  1 e  Q1, , 1H, , 1H, , 2, ,  1 H 1  2 He 3  Q 2, , 7, , 2, , 1, , 2, , 0, , He 3  2 He 3  2 He 4  2 1 H 1  Q 3, , 4 1 H  2 He  2, 1, , 4, , 1 e, , 0, ,  2  26 .7 MeV, , 1, ,  6 C 12  7 N 13  Q1, , N 13  6 C 13 , , 0, 1 e, 14, , 1H, , 1, ,  6 C 13  7 N, , 1H, , 1, , 7 N, , 8, , O, , 1H, , 15, 1, , 14, , 7N, ,  8O, 15, , 15, ,  Q2, ,  Q3, ,  1e  Q4, 0, ,  7 N 15  6 C 12  2 He 4, , 4 1 H 1  2 He 4  2 1 e 0  24 .7 MeV, , About 90% of the mass of the sun consists of hydrogen and helium., , Nuclear Bomb., , Based on uncontrolled nuclear reactions., Atom bomb, , Hydrogen bomb, , Based on fission process it involves the fission of U235, , Based on fusion process. Mixture of deutron and, tritium is used in it, , In this critical size is important, , There is no limit to critical size, , Explosion is possible at normal temperature and, pressure, , High temperature and pressure are required, , Less energy is released compared to hydrogen bomb, , More energy is released as compared to atom bomb so, it is more dangerous than atom bomb, , Concepts, , , , A test tube full of base nuclei will weight heavier than the earth.

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genius PHYSICS, , 20 Atomic Structure, , , , The nucleus of hydrogen contains only one proton. Therefore we may say that the proton is the nucleus of hydrogen, atom., , , , If the relative abundance of isotopes in an element has a ratio n1 : n2 whose atomic masses are m1 and m2 then atomic, n m  n2 m 2, mass of the element is M  1 1, n1  n 2, , Examples, Example: 1, , A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio 8 : 1. The ratio, of radii of the fragments is, (a) 1 : 2, , (b) 1 : 4, , Solution : (a), , v1, , (c) 4 : 1, , (d) 2 : 1, , By conservation of momentum m1v1 = m2v2, , m1, , , , M, m2, , v1 8 m 2,  , v 2 1 m1, , …… (i), 1/3, , Also from r  A1 / 3 , , v2, Example: 2, , The ratio of radii of nuclei, , 27, 13, , Al and, , (a) 6 : 10, , (b) 13 : 52, r1  A1, , r2  A2, , 1/3, , , , , , 125, 52 Te, , r1  A1 , , , r2  A2 , , 1/3, , 1,  , 8, , , , 1, 2, , is approximately, , [J & K CET 2000], , (c) 40 : 177, 1/3, ,  27 , , ,  125 , , , , (d) 14 : 7, , 8, 6, , 5 10, , Solution : (a), , By using r  A1 / 3 , , Example: 3, , If Avogadro’s number is 6  1023 then the number of protons, neutrons and electrons in 14 g of 6C14 are, respectively, , Solution : (a), , Example: 4, , (a) 36  1023, 48  1023, 36  1023, , (b) 36  1023, 36  1023, 36  1021, , (c) 48  1023, 36  1023, 48  1021, , (d) 48  1023, 48  1023, 36  1021, , Since the number of protons, neutrons and electrons in an atom of, gm of, , 14, , of 6 C, , are, , 6C, 14, , 14, , are 6, 8 and 6 respectively. As 14, , contains 6  1023 atoms, therefore the numbers of protons, neutrons and electrons in 14 gm, , 6  6  10 23  36  10 23 ,, , 8  6  10 23  48  10 23 , 6  6  10 23  36  10 23 ., , Two Cu64 nuclei touch each other. The electrostatics repulsive energy of the system will be, (a) 0.788 MeV, , Solution : (c), , 6C, , (b) 7.88 MeV, , Radius of each nucleus R  R0 ( A), , 1/3, , (c) 126.15 MeV, ,  1 .2 (64 ), , 1/3, , (d) 788 MeV, ,  4 .8 fm, , Distance between two nuclei (r) = 2R, So potential energy U , Example: 5, , When, , 92 U, , 235, , undergoes fission. 0.1% of its original mass is changed into energy. How much energy is, , released if 1 kg of, (a) 9  1010 J, , k q2, 9  10 9  (1 .6  10 19  29 ) 2, ,  126 .15 MeV ., r, 2  4 .8  10 15  1 .6  10 19, , 92 U, , 235, , undergoes fission, (b) 9  1011 J, , [MP PET 1994; MP PMT/PET 1998; BHU 2001; BVP 2003], , (c) 9  1012 J, , (d) 9  1013 J, , Solution : (d), ,  0 .1, ,  1  (3  10 8 )2  9  10 13 J, By using E  m  c 2  E  , 100, , , , Example: 6, , 1 g of hydrogen is converted into 0.993 g of helium in a thermonuclear reaction. The energy released is, [EAMCET (Med.) 1995; CPMT 1999], , (a) 63  107 J, Solution : (b), , (b) 63  1010 J, , m = 1 – 0.993 = 0.007 gm,  E = mc2 = 0.007  10–3  (3  108)2 = 63  1010 J, , (c) 63  1014 J, , (d) 63  1020 J

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genius PHYSICS, Atomic structure 21, Example: 7, , The binding energy per nucleon of deuteron (12 H ) and helium nucleus (42 He ) is 1.1 MeV and 7 MeV, respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is, [MP PMT 1992; Roorkee 1994; IIT-JEE 1996; AIIMS 1997; Haryana PMT 2000; Pb PMT 2001; CPMT 2001; AIEEE 2004], , (a) 13.9 MeV, Solution : (c), , 1H, , 2, , (b) 26.9 MeV, , (c) 23.6 MeV, , (d) 19.2 MeV, ,  1 H  2 He  Q, 2, , 4, , Total binding energy of helium nucleus = 4  7 = 28 MeV, Total binding energy of each deutron = 2  1.1 = 2.2 MeV, Hence energy released = 28 – 2  2.2 = 23.6 MeV, Example: 8, , The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and, protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu. The binding energy of, the helium nucleus will be [1 amu= 931 MeV], [CPMT 1986; MP PMT 1995; CBSE 2003], (a) 28.4 MeV, , Solution : (a), , (b) 20.8 MeV, , (c) 27.3 MeV, , (d) 14.2 MeV, , Helium nucleus consist of two neutrons and two protons., So binding energy E = m amu = m  931 MeV, , Example: 9, ,  E = (2  mp + 2mn – M)  931 MeV = (2  1.0073 + 2  1.0087 – 4.0015)  931 = 28.4 MeV, A atomic power reactor furnace can deliver 300 MW. The energy released due to fission of each of, uranium atom U 238 is 170 MeV. The number of uranium atoms fissioned per hour will be, (a) 5  1015, , Solution : (c), , By using P , , (b) 10  1020, W n E, , t, t, , where n = Number of uranium atom fissioned and E = Energy released due to, , 300  10 6 , , (a) 3.52, Solution : (c), , (d) 30  1025, , n  170  10 6  1.6  10 19,  n = 40  1021, 3600, The binding energy per nucleon of O16 is 7.97 MeV and that of O17 is 7.75 MeV. The energy (in MeV), required to remove a neutron from O17 is, [IIT-JEE 1995], each fission so, , Example: 10, , (c) 40  1021, , (b) 3.64, , (c) 4.23, , (d) 7.86, , O17  O16  0 n1, ,  Energy required = Binding of O17 – binding energy of O16 = 17  7.75 – 16  7.97 = 4.23 MeV, Example: 11, , A gamma ray photon creates an electron-positron pair. If the rest mass energy of an electron is 0.5 MeV, and the total kinetic energy of the electron-positron pair is 0.78 MeV, then the energy of the gamma ray, photon must be, [MP PMT 1991], (a) 0.78 MeV, , (b) 1.78 MeV, , (c) 1.28 MeV, , Solution : (b), , Energy of -rays photon = 0.5 + 0.5 +0.78 = 1.78 MeV, , Example: 12, , What is the mass of one Curie of U234, (a) 3.7  1010 gm, , Solution : (c), , (b) 2.348  1023 gm, , [MNR 1985], , (c) 1.48  10–11 gm, , (d) 6.25  10–34 gm, , 1 curie = 3.71  1010 disintegration/sec and mass of 6.02  1023 atoms of U 234  234 gm,  Mass of 3.71  1010 atoms , , Example: 13, , (d) 0.28 MeV, , In the nuclear fusion reaction, , 234  3 .71  10 10, 6 .02  10 23, 2, 1H, ,  1 .48  10 11 gm, , 13 H 42 He  n, given that the repulsive potential energy between the, , two nuclei is  7.7  10 14 J , the temperature at which the gases must be heated to initiate the reaction is, nearly [Boltzmann’s constant k  1 .38  10 23 J /K ], [AIEEE 2003], (a) 10 9 K, Solution : (a), , (c) 10 5 K, , (d) 10 3 K, , Kinetic energy of molecules of a gas at a temperature T is 3/2 kT,  To initiate the reaction, , Example: 14, , (b) 107 K, , 3, kT  7 . 7  10 14 J  T = 3.7  109 K., 2, , A nucleus with mass number 220 initially at rest emits an -particle. If the Q value of the reaction is 5.5, MeV. Calculate the kinetic energy of the -particle, [IIT-JEE (Screening) 2003]

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genius PHYSICS, , 22 Atomic Structure, , (a) 4.4 MeV, , (b) 5.4 MeV, , (c) 5.6 MeV, , Solution : (b), , (d) 6.5 MeV, , k1, , k2, , , p1, , , p2, m2 = 4, , m1 = 216, , M = 220, , Q-value of the reaction is 5.5 eV i.e. k 1  k 2  5.5 MeV, By conservation of linear momentum p 1  p 2 , , ……(i), , 2(216 )k 1 , , 2(4 )k 2  k2 = 54 k1, , ……(ii), , On solving equation (i) and (ii) we get k2 = 5.4 MeV., Example: 15, , Let m p be the mass of a proton, m n the mass of a neutron, M1 the mass of a, 40, 20, , Ne nucleus and M 2 the mass of a, , Ca nucleus. Then, , (a) M 2  2M1, Solution : (c, d), , 20, 10, , [IIT 1998; DPMT 2000], , (b) M 2  2M1, , (c), , M 2  2M1, , (d) M 1  10(m n  m p ), , Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus, is always less then the sum of masses of it's constituent particles, neutrons. Therefore, mass of, , 20, 10, , 20, 10, , Ne is made up of 10 protons plus 10, , Ne nucleus M 1  10 (m p  m n ), , Also heavier the nucleus, more is he mass defect thus 20 (mn  m p )  M 2  10(m p  mn )  M1, or, , 10 (m p  m n )  M 2  M 1, , , , M 2  M1  10 (m p  mn )  M 2  M 1  M 1  M 2  2M 1, , Tricky example: 1, , (a) Y  2Z, (b) W  X + Z, (c) W  2Y, , Binding energy, nucleon in, MeV, , Binding energy per nucleon vs mass number curve for nuclei is shown in the figure. W, X, Y and Z, are four nuclei indicated on the curve. The process that would release energy is, [IIT-JEE 1999], , 5.0, , (d) X  Y + Z, Solution : (c), , Y, , 8.5, 8.0, 7.5, , X, , W, , Z, 30, 60 90 120, Mass number of nuclei, , Energy is released in a process when total binding energy of the nucleus (= binding energy per, nucleon  number of nucleon) is increased or we can say, when total binding energy of products, is more than the reactants. By calculation we can see that only in case of option (c) this happens., Given, , W  2Y, , Binding energy of reactants = 120  7.5 = 900 MeV, and binding energy of products = 2 (60  8.5) = 1020 MeV > 900 MeV, , Radioactivity., The phenomenon of spontaneous emission of radiatons by heavy elements is called radioactivity. The elements which shows this phenomenon are called, radioactive elements., , (1) Radioactivity was discovered by Henery Becquerel in uranium salt in the year 1896., (2) After the discovery of radioactivity in uranium, Piere Curie and Madame Curie discovered a new radioactive, element called radium (which is 106 times more radioactive than uranium), (3) Some examples of radio active substances are : Uranium, Radium, Thorium, Polonium, Neptunium etc., (4) Radioactivity of a sample cannot be controlled by any physical (pressure, temperature, electric or magnetic field), or chemical changes., (5) All the elements with atomic number (Z ) > 82 are naturally radioactive., (6) The conversion of lighter elements into radioactive elements by the bombardment of fast moving particles is, called artificial or induced radioactivity., (7) Radioactivity is a nuclear event and not atomic. Hence electronic configuration of atom don't have any, relationship with radioactivity., Nuclear radiatons

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genius PHYSICS, Atomic structure 23, According to Rutherford's experiment when a sample of radioactive substance is put in a lead box and allow the, emission of radiation through a small hole only. When the radiation enters into the external electric field, they splits into, three parts, –, –, –, –, –, ,  -rays, , rays, , rays, , ,  -  -rays, , rays, , , , , , , , , +, +, +, +, +, , , , , , , , Magnetic, , , , ,  field, rays , , (i) Radiations which deflects towards negative plate are called -rays (stream of positively charged particles), (ii) Radiations which deflects towards positive plate are called  particles (stream of negatively charged particles), (iii) Radiations which are undeflected called -rays. (E.M. waves or photons), , Note : , , Exactly same results were obtained when these radiations were subjected to magnetic field., ,  No radioactive substance emits both  and  particles simultaneously. Also -rays are emitted after the, emission of  or -particles.,  -particles are not orbital electrons they come from nucleus. The neutron in the nucleus decays into, proton and an electron. This electron is emitted out of the nucleus in the form of -rays., Properties of ,  and -rays, , - particles, , Features, 1. Identity, ,  - particles, Fast, , 2. Charge, , + 2e, , –e, , Zero, , 3. Mass 4 mp (mp = mass of, proton = 1.87  10–27, , 4 mp, , me, , Massless, , 4. Speed, ,  107 m/s, , 1% to 99% of speed of light, , Speed of light, , 5. Range of kinetic energy, , 4 MeV to 9 MeV, , All possible values between, a minimum certain value to, 1.2 MeV, , Between a minimum, value to 2.23 MeV, , 6. Penetration power (, ,, ), , 1, , 100, , 10,000, , (Stopped by a paper), , (100 times of ), , (100 times of  upto 30, cm of iron (or Pb) sheet, , 7. Ionisation power ( >  > ), , 10,000, , 100, , 1, , 8. Effect of, magnetic field, , Deflected, , Deflected, , Not deflected, , 9. Energy spectrum, , Line and discrete, , Continuous, , Line and discrete, , 10. Mutual interaction with, matter, , Produces heat, , Produces heat, , Produces, photo-electric, effect, Compton effect,, pair production, , 11. Equation of decay, , or, , Z, ,  decay, X A , , Z 2 Y, , A 4, ,  2 He 4, , electron, , Photons (E.M. waves), , Helium nucleus or, doubly ionised helium, atom (2He4), , electric, , moving, ,  - rays, , –, , (  or  ), 0, , Z, , XA , , Z, , , X A , , , n, , Z 1 Y, , A, , , , Z' X, , 1 e, , A, , 0, , , , z, , X A  zXa 

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genius PHYSICS, , 24 Atomic Structure, , , X A , , , n, , Z, ,  nα , , Z' Y, ,  n β  (2 n α  Z  Z'), , A', , A'  A, 4, , Radioactive Disintegration., (1) Law of radioactive disintegration, According to Rutherford and Soddy law for radioactive decay is as follows., "At any instant the rate of decay of radioactive atoms is proportional to the number of atoms present at that instant", i.e. , , dN, N, dt, , , , dN,  N . It can be proved that N = N0e–t, dt, , This equation can also be written in terms of mass i.e. M = M0e–t, where N = Number of atoms remains undecayed after time t, N0 = Number of atoms present initially (i.e. at t = 0), M =, Mass of radioactive nuclei at time t, M0 = Mass of radioactive nuclei at time t = 0, N0 – N = Number of disintegrated nucleus in, time t, , dN, = rate of decay,  = Decay constant or disintegration constant or radioactivity constant or Rutherford Soddy's, dt, , constant or the probability of decay per unit time of a nucleus., , Note : , ,  depends only on the nature of substance. It is independent of time and any physical or chemical, N0, , changes., , N, , N = N0e–t, , (2) Activity, 0, t, It is defined as the rate of disintegration (or count rate) of the substance (or the number of atoms of any material, decaying per second) i.e. A  , , dN,  N  N 0 e t  A0 e t, dt, , where A0 = Activity of t = 0, A = Activity after time t, , Units of activity (Radioactivity), It's units are Becqueral (Bq), Curie (Ci) and Rutherford (Rd), 1 Becquerel = 1 disintegration/sec,, 1 Rutherford = 106 dis/sec,, , Note : , , 1 Curie = 3.7  1011 dis/sec, , Activity per gm of a substance is known as specific activity. The specific activity of 1 gm of radium –, , 226 is 1 Curie.,  1 millicurie = 37 Rutherford,  The activity of a radioactive substance decreases as the number of undecayed nuclei decreases with time., , , Activity , , 1, Half life, , (3) Half life (T1/2), , Time interval in which the mass of a radioactive substance or the number of it's atom reduces to half of it's initial value is called the half life of the, substance., N, , N0, then t  T1 / 2, 2,  t, Hence from N  N 0 e, N0, log e 2 0.693,  (T1 / 2 ),  T1 / 2 ,  N0e, , 2, , , i.e., , if N , , N0, , Half life = T, N0/2, N0/4, 0, , Time (t), t=0, t = T1/2, , Number of undecayed atoms (N), (N0 = Number of initial atoms), N0, N0, 2, , Remaining fraction of, active atoms (N/N0), probability of survival, 1, (100%), 1, 2, , (50%), , 1, , 2T, , 3T, , t, , Fraction of atoms, decayed (N0 – N) /N0, probability of decay, 0, 1, 2, , (50%)

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genius PHYSICS, Atomic structure 25, t = 2(T1/2), , N, 1 N0, ,  02, 2, 2, (2), N, 1 N0, ,  0, 2 (2) (2)3, , t = 3(T1/2), t = 10 (T1/2), , N0, 10, , (2), t = n (N1/2), , N, (2), , 2, , 1, 4, , (25%), , 3, 4, , (75%), , 1, 8, , (12.5%), , 7, 8, , (87.5%), , 1,  , 2, , 10, , 1,  , 2, , n, ,  99.9%, ,  0 .1 %, ,   1  n , 1    ,   2  , , Useful relation, n, , t/T, , 1,  1  1/2, After n half-lives, number of undecayed atoms N  N 0    N 0  , 2, 2, (4) Mean (or average) life (), The time for which a radioactive material remains active is defined as mean (average) life of that material., , Other definitions, (i) It is defined as the sum of lives of all atoms divided by the total number of atoms, , Sum of the lives of all the atoms 1, , Total number of atoms, , N, ln, N0,  t, (ii) From N  N 0 e ,   slope of the line shown in the graph, t, N, i.e. the magnitude of inverse of slope of ln, vs t curve is known as mean life ()., N0, i.e.  , , ln, , N, N0, , t, ,  t, (iii) From N  N 0 e, , If t , , Slope = – , , 1,    N  N 0 e 1  N 0    0 .37 N 0  37 % of N0., , e , , 1, , i.e. mean life is the time interval in which number of undecayed atoms (N) becomes, original number of atoms., , or, , , , , It is the time in which number of decayed atoms (N0 – N) becomes  1 , , 1, times or 0.37 times or 37% of, e, , 1,  times or 0.63 times or 63% of original, e, , number of atoms., (iv) From T1 / 2 , , 0.693, , , , , , 1, , , ,  , , 1, . (t1 / 2 )  1.44 (T1 / 2 ), 0.693, , i.e. mean life is about 44% more than that of half life. Which gives us  > T(1/2), , Note : , , Half life and mean life of a substance doesn't change with time or with pressure, temperature etc., , Radioactive Series., If the isotope that results from a radioactive decay is itself radioactive then it will also decay and so on., , The sequence of decays is known as radioactive decay series. Most of the radio-nuclides found in nature are, members of four radioactive series. These are as follows, Mass number Series (Nature), Parent Stable, and Integer, Number, of, lost, product, n, particles, 4n, Thorium (natural), 52, Th 232, Pb 208,  = 6,  = 4, 90, , 4n + 1, , Neptunium, (Artificial), , 93, , 82, , Np, , 237, 83, , Bi 209, , 52, ,  = 8,  = 5

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genius PHYSICS, , 26 Atomic Structure, 4n + 2, , Uranium (Natural), , 4n + 3, , Actinium (Natural), , Note : , , 92 U, 89, , 238, 82, , Ac, , 227, 82, , The 4n + 1 series starts from, , 94, , Pb 206, , 51, ,  = 8,  = 6, , 207, , 51, ,  = 7,  = 4, , Pb, , PU 241 but commonly known as neptunium series because, , neptunium is the longest lived member of the series.,  The 4n + 3 series actually starts from 92 U 235 ., , Successive Disintegration and Radioactive Equilibrium., Suppose a radioactive element A disintegrates to form another radioactive element B which intern disintegrates to, still another element C; such decays are called successive disintegration., 1, , A, , 2, , B, , C, , dN 1,  1 N 1 (which is also the rate of formation of B), dt, dN 2, Rate of disintegration of B ,   2 N 2, dt, Rate of disintegration of A , ,  Net rate of formation of B = Rate of disintegration of A – Rate of disintegration of B, = 1N1 – 2N2, Equilibrium, In radioactive equilibrium, the rate of decay of any radioactive product is just equal to it's rate of production from the previous member., , 1N1 = 2N2, , i.e., , Note : , , , , (T ), 1 N 2  2, , ,  1/2,  2 N 2  1 (T1 / 2 )1, , In successive disintegration if N0 is the initial number of nuclei of A at t = 0 then number of nuclei, , of product B at time t is given by N 2 , , 1 N 0, (e 1 t  e 2 t ) where 12 – decay constant of A and B., ( 2  1 ), , Uses of radioactive isotopes, (1) In medicine, (i) For testing blood-chromium - 51, (ii) For testing blood circulation - Na - 24, (iii) For detecting brain tumor- Radio mercury - 203, (iv) For detecting fault in thyroid gland - Radio iodine - 131, (v) For cancer - cobalt – 60, (vi) For blood - Gold - 189, (vii) For skin diseases - Phospohorous - 31, (2) In Archaeology, (i) For determining age of archaeological sample (carbon dating) C 14, (ii) For determining age of meteorites - K 40, (3) In agriculture, , (iii) For determining age of earth-Lead isotopes, , (i) For protecting potato crop from earthworm- CO 60, (ii) For artificial rains - AgI, (iii) As fertilizers - P 32, (4) As tracers - (Tracer) : Very small quantity of radioisotopes present in a mixture is known as tracer, (i) Tracer technique is used for studying biochemical reaction in tracer and animals., (5) In industries, (i) For detecting leakage in oil or water pipe lines (ii) For determining the age of planets., Concept, , , , If a nuclide can decay simultaneously by two different process which have decay constant 1 and 2, half life T1 and T2 and, mean lives 1 and 2 respectively then,   = 1 + 2, 1, T1, 1, 1, TT, , T  1 2, , T, 1  T2, T, 2, , , , Example: 16, , When, , 90 Th, , (a) 8, 7, , 228, , transforms to, , 83 Bi, , 2, T2, 2, , 212, , (b) 4, 7, , , ,  , ,  1 2, 1   2, , , then the number of the emitted –and –particles is, respectively, (c) 4, 4, , (d) 4, 1, , [MP PET 2002]

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genius PHYSICS, Atomic structure 27, Solution : (d), , Z  90 Th, , A  228, , , , Z '  83 Bi, , A '  212, , A  A ' 228  212, , 4, 4, 4,  2n  Z  Z '  2  4  90  83  1 ., , Number of -particles emitted n , Number of -particles emitted n , Example: 17, , A radioactive substance decays to 1/16th of its initial activity in 40 days. The half-life of the radioactive, substance expressed in days is, (a) 2.5, (b) 5, (c) 10, (d) 20, t/T, , Solution : (c), Example: 18, , 40 / T, , 1/2, N, 1, 1,  1  1/2, ,  , By using N  N 0  , ,  T1 / 2  10 days., N0, 16  2 , 2, A sample of radioactive element has a mass of 10 gm at an instant t = 0. The approximate mass of this, element in the sample after two mean lives is, [CBSE PMT 2003], (a) 2.50 gm, (b) 3.70 gm, (c) 6.30 gm, (d) 1.35 gm, ,  t, ,  (2 ), , 2,   ,  10 e   , , 2, , 1,  10    1.359 gm, e , , Solution : (d), , By using M  M 0 e, , Example: 19, , The half-life of 215 At is 100 s. The time taken for the radioactivity of a sample of 215 At to decay to 1/16th, of its initial value is, [IIT-JEE (Screening) 2002], (a) 400 s, (b) 6.3 s, (c) 40 s, (d) 300 s, ,  M  10 e, , t/T, , n, , Solution : (a), Example: 20, , Solution : (c), , t / 100, , N, 1,  1  1/2, 1, 1,  ,  , By using N  N 0   , ,  t = 400  sec., N, 2, 16, 2,  , 2,  , 0, The mean lives of a radioactive substance for  and  emissions are 1620 years and 405 years respectively., After how much time will the activity be reduced to one fourth, [RPET 1999], (a) 405 year, (b) 1620 year, (c) 449 year, (d) None of these, 1, 1,  , per year and   , per year and it is given that the fraction of the remained activity, 405, 1620, A, 1, , A0, 4, 1, 1, 1, , , per year, 1620 405, 324, A, 1, 1, 2, We know that A  A 0 e t  t  log e 0  t  loge 4  loge 2 = 324  2  0.693 = 449 years., , A, , , At any instant the ratio of the amount of radioactive substances is 2 : 1. If their half lives be, respectively 12 and 16 hours, then after two days, what will be the ratio of the substances, (a) 1 : 1, (b) 2 : 1, (c) 1 : 2, (d) 1 : 4, , Total decay constant       , , Example: 21, , 2 24, , Solution : (a), , Example: 22, ,  1  12,  , n, n1, N 1 ( N 0 )1 (1 / 2), 2 2, 1, 1, , , , , , By using N  N 0   , n2, 2, , 24, 2, N, (, N, ), 1, 1, (1 / 2),  , 2, 0 2,  1  16,  , 2, From a newly formed radioactive substance (Half-life 2 hours), the intensity of radiation is 64 times the, permissible safe level. The minimum time after which work can be done safely from this source is, [IIT 1983; SCRA 1996], , (a) 6 hours, , (b) 12 hours, 0, , n, , Solution : (b), , Example: 23, , (c) 24 hours, , A, 1, 1, 1, 1, ,    , By using A  A 0   , 2, A, 64, 2,  ,  , 2, 0, t, ,  6  t  6  2  12 hours., T1 / 2, , (d) 128 hours, , n, , n=6, , nucleus of mass number A, originally at rest, emits an -particle with speed v. The daughter nucleus, recoils with a speed, [DCE 2000; AIIMS 2004], (a) 2v /( A  4), (b) 4v /( A  4), (c) 4v /( A  4), (d) 2v /( A  4), m, , Solution : (c), A, Rest, , v, , A–4, , m, +, , 4, , v

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genius PHYSICS, , 28 Atomic Structure, , 4v, ., A4, The counting rate observed from a radioactive source at t = 0 second was 1600 counts per s econd and, at t = 8 seconds it was 100 counts per second. The counting rate observed as counts per second at t =, 6 seconds will be, [MP PET 1996; UPSEAT 2000], (a) 400, (b) 300, (c) 200, (d) 150, , According to conservation of momentum 4v  ( A  4)v'  v ' , Example: 24, , n, , 8 / T1 / 2, , 1, 1, By using A  A0    100  1600  , 2, 2, , Example: 25, , 1,  200 ., Again by using the same relation the count rate at t = 6 sec will be A  1600  , 2, The kinetic energy of a neutron beam is 0.0837 eV. The half-life of neutrons is 693s and the mass of, neutrons is 1.675  10 27 kg. The fraction of decay in travelling a distance of 40m will be, (a) 10 3, (b) 10 4, (c) 10 5, (d) 10 6, , Solution : (c), , v, , , , 1 1,  , 16  2 , , 8 / T1 / 2, , Solution : (c), ,  T1 / 2  2 sec, 6/2, , 2E, , m, , 2  0 . 0837  1 . 6  10 19, 1 . 675  10  27, , = 4  103 m/sec, ,  Time taken by neutrons to travel a distance of 40 m t' , , , 40, 4  10, , 3, ,  10  2 sec, , dN, dN, N ,   dt, dt, N, , N, 0 . 693, 0 . 693,   t , t ,  10  2  10  5, N, T, 693, The fraction of atoms of radioactive element that decays in 6 days is 7/8. The fraction that decays in 10 days, will be, (a) 77/80, (b) 71/80, (c) 31/32, (d) 15/16, N , ,  N0 ,  log e 0 , T1 / 2 log e , , t / T1 / 2, N 1, N, t, 1, ,  N , By using N  N 0  ,  t,  t  log e 0  1 , N, log e (2), t2 , N , 2,  log e 0 , N 2, , N, N, log e (8 / 1), 10, 6, log e (8 )  log e 32  0  32 ., Hence,  log e 0 , , N, 6, N, 10 log e ( N 0 / N ), ,  Fraction of neutrons decayed in t sec in, Example: 26, , Solution : (c), , So fraction that decays  1 , , 1, 31, , ., 32 32, , Tricky example: 2, , Solution : (b), , Half-life of a substance is 20 minutes. What is the time between 33% decay and 67% decay [AIIMS 2000], (a) 40 minutes, (b) 20 minutes, (c) 30 minutes, (d) 25 minutes, Let N0 be the number of nuclei at beginning,  Number of undecayed nuclei after 33% decay = 0.67 N0, and number of undecayed nuclei after 67% of decay = 0.33 N0, 0 . 67 N 0,  0.33 N0 ~, and in the half-life time the number of undecayed nuclei becomes half., 2, , Example, s, Example: 1, , The ratio of areas within the electron orbits for the first excited state to the ground state for hydrogen, atom is, (a) 16 : 1, , Solution : (a), , For a hydrogen atom, , (b) 18 : 1, , (c) 4 : 1, , (d) 2 : 1

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genius PHYSICS, Atomic structure 29, Radius r  n 2 , Example: 2, , r12, r22, , , , n14, , , , n 24, , r12, r22, , , , n14, n 24, , , , A1, n4, A, 24, 16,  14  4  16  1 , A2, A2, 1, n2, 1, , The electric potential between a proton and an electron is given by V  V0 ln, , r, , where r0 is a constant., r0, , Assuming Bohr’s model to be applicable, write variation of rn with n, n being the principal quantum, number, [IIT-JEE (Screening) 2003], , (a) rn  n, Solution : (a), , (b) rn  1 / n, , Potential energy U  eV  eV 0 ln,  Force F  , , (c) rn  n, , (d) rn  1 / n 2, , 2, , r, r0, , dU eV0, . The force will provide the necessary centripetal force. Hence, , dr, r, , eV, mv 2,  0 v, r, r, , eV 0, m, , …..(i), , mvr , , and, , nh, 2, , …..(ii), ,  nh  m, Dividing equation (ii) by (i) we have mr  , or r  n, ,  2  eV 0, Example: 3, , The innermost orbit of the hydrogen atom has a diameter 1.06 Å. The diameter of tenth orbit is, [UPSEAT 2002], , (a) 5.3 Å, , (b) 10.6 Å, r2  n 2 , , , r1  n 1 , , 2, , (c) 53 Å, 2, , (d) 106 Å, , d 2  n2 , d2,  10 ,  ,  ,     d = 106 Å, d 1  n1 , 1 . 06  1 , 2, , Solution : (d), , Using r  n 2 , , Example: 4, , Energy of the electron in nth orbit of hydrogen atom is given by En  , , or, , 13 .6, eV . The amount of energy, n2, , needed to transfer electron from first orbit to third orbit is, (a) 13.6 eV, Solution : (c), , Using E  , , (b) 3.4 eV, , 13 .6, n2, , For n = 1 , E1 , , (c) 12.09 eV, , (d) 1.51 eV, , eV, , 13 .6, 1, , 2, ,  13 .6 eV and for n = 3 E 3  , , 13 .6, 32, ,  1 .51 eV, , So required energy  E 3  E1  1.51  (13.6)  12.09 eV, Example: 5, , If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the, electron from the first excited state of Li   is, [AIEEE 2003], (a) 122.4 eV, (b) 30.6 eV, (c) 13.6 eV, (d) 3.4 eV, , Solution : (b), , Using E n  , , 13 .6  Z 2, n2, , eV, , For first excited state n = 2 and for Li   , Z = 3,  E, , 13 .6, 2, , 2, ,  32  , , 13 .6  9,  30 .6 eV . Hence, remove the electron from the first excited state of Li   be 30.6, 4, , eV, Example: 6, , The ratio of the wavelengths for 2  1 transition in Li++, He+ and H is, (a) 1 : 2 : 3, (b) 1 : 4 : 9, (c) 4 : 9 : 36, , [UPSEAT 2003], , (d) 3 : 2 : 1, ,  1, 1 1 1, 1 , 1,  RZ 2  2  2    ,   Li :  He  :  H  : :  4 : 9 : 36, 2, , , 9 4 1, , n2 , Z,  n1, 1, , Solution : (c), , Using, , Example: 7, , Energy E of a hydrogen atom with principal quantum number n is given by E , , 13 .6, eV . The energy of a, n2, , photon ejected when the electron jumps n = 3 state to n = 2 state of hydrogen is approximately

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genius PHYSICS, , 30 Atomic Structure, , [CBSE PMT/PDT Screening 2004], , (a) 1.9 eV, Solution : (a), Example: 8, , (b) 1.5 eV, , (d) 3.4 eV, , 1 , 5,  1, E  13 .6  2  2   13 .6 ,  1 .9 eV, 36, 2, 3, , , In the Bohr model of the hydrogen atom, let R, v and E represent the radius of the orbit, the speed of, electron and the total energy of the electron respectively. Which of the following quantity is proportional, to the quantum number n, [KCET 2002], (a) R/E, , Solution : (d), , (c) 0.85 eV, , (b) E/v, , Rydberg constant R , Velocity v , , 0n h, 2, , (c) RE, , (d) vR, , 2, , mZe 2, , mZ 2 e 4, Ze 2, and energy E   2 2 2, 2 0 nh, 8 0 n h, , Now, it is clear from above expressions R.v  n, Example: 9, , The energy of hydrogen atom in nth orbit is En, then the energy in nth orbit of singly ionised helium atom, will be, (a) 4En, , (b) En/4, , 13 .6 Z 2, , , , (c) 2En, ,  Z, EH,   H, E He,  Z He, , 2, , (d) En/2, , , 1,      E He  4 E n ., 2, , 2, , Solution : (a), , By using E  , , Example: 10, , The wavelength of radiation emitted is 0 when an electron jumps from the third to the second orbit of, hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom, the, wavelength of radiation emitted will be, [SCRA 1998; MP PET 2001], (a), , Solution : (b), , n2, , 16, 0, 25, , (b), , 20, 0, 27, , (c), , 27, 0, 20, , (d), , Wavelength of radiation in hydrogen atom is given by,  1, 1 , 1, 1 , 5,  1, 1 1 ,  R 2  2  ,  R 2  2   R   , R, , 0, n 2 , 3 ,  4 9  36, 2,  n 1, , …..(i), , 1, 1 , 1  3R,  1, 1,  R 2  2   R   , , 4 ,  4 16  16, 2, , …..(ii), , 1, , and, , From equation (i) and (ii), Example: 11, , (b) 256, , 1, sin 4 ( / 2), , , , (d) 108, , N2, N1, ,  1,  sin ,  2, ,   2,  sin ,   2, , 4, , ,   90   ,  , ,  sin , N, ,  2 ,  2 ,   60   , N1, ,  , ,  sin ,  ,   2 , , 4, ,  sin 45  , , ,  sin 30  , , 4, ,  4  N 2  4 N 1  4  56  224, , (b) 3.32  10 34 J-s, , (c) 1.05  10 34 J-s, , (d) 2.08  10 34 J-s, , Change in angular momentum, , L  L 2  L1 , Example: 13, , (c) 98, , When an electron in hydrogen atom is excited, from its 4 th to 5th stationary orbit, the change in angular, momentum of electron is (Planck’s constant: h  6.6  10 34 Js ), [AFMC 2000], (a) 4.16  10 34 J- s, , Solution : (c), , [RPMT 2000], , Using Scattering formula, , N, , Example: 12, ,   5 R 16 20, 20,   , , , , 0, , 36 3 R 27, 27, , If scattering particles are 56 for 90 o angle then this will be at 60 o angle, (a) 224, , Solution : (a), , 25, 0, 16, , n 2 h n1 h, h, 6.6  10 34,  L , (n 2  n1 ) , (5  4 )  1.05  10 34 J-s, , 2, 2  3.14, 2, 2, , In hydrogen atom, if the difference in the energy of the electron in n = 2 and n = 3 orbits is E, the, ionization energy of hydrogen atom is, (a) 13.2 E, , (b) 7.2 E, , (c) 5.6 E, , (d) 3.2 E

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genius PHYSICS, Atomic structure 31, Solution : (b), , Energy difference between n = 2 and n = 3;, , 1 , 5,  1, 1 1, E  K 2  2   K   , K, 3 ,  4 9  36, 2, , 1 ,  1, Ionization energy of hydrogen atom n1  1 and n 2   ; E   K  2  2   K,  , 1, From equation (i) and (ii) E  , Example: 14, , …..(i), …..(ii), , 36, E  7 .2 E, 5, , In Bohr model of hydrogen atom, the ratio of periods of revolution of an electron in n = 2 and n = 1 orbits is, [EAMCET (Engg.) 2000], , (a) 2 : 1, Solution : (c), Example: 15, , (b) 4 : 1, , According to Bohr model time period of electron T  n 3 , , T2, , T1, , (d) 16 : 1, , n 23, n13, , , , 2, , 3, , 1, , 3, , , , 8,  T2  8T1 ., 1, , A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of, required radiation for emitting electron from first to third Bohr orbit in Li  will be (Ionisation energy of, hydrogen atom is 13.6 eV), (a) 182.51 Å, , Solution : (d), , (c) 8 : 1, , (b) 177.17 Å, , (c) 142.25 Å, , (d) 113.74 Å, , Energy of a electron in nth orbit of a hydrogen like atom is given by, , E n  13 .6, , Z2, n2, , eV , and Z = 3 for Li, , Required energy for said transition, , 1 ,  1, 8 , E  E 3  E1  13 .6 Z 2  2  2   13 .6  3 2    108 .8 eV  108 .8  1 .6  10 19 J, 3 , 9 , 1, Now using E , Example: 16, , hc, , , ,   , , The absorption transition between two energy states of hydrogen atom are 3. The emission transitions, between these states will be, (a) 3, , Solution : (d), , hc, 6 .6  10 34  3  10 8,  ,  0 .11374  10 7 m    113 .74 Å, 19, E, 108 .8  1 .6  10, , (b) 4, , (d) 6, , Number of absorption lines = (n – 1)  3 = (n – 1)  n = 4, Hence number of emitted lines , , Example: 17, , (c) 5, n(n  1) 4 (4  1), , 6, 2, 2, , The energy levels of a certain atom for 1st, 2nd and 3rd levels are E, 4E/3 and 2E respectively. A photon of, wavelength  is emitted for a transition 3  1. What will be the wavelength of emissions for transition 2 , 1, [CPMT 1996], , Solution : (d), , (a) /3, , (b) 4/3, , For transition 3  1, , E  2 E  E , , For transition 2  1, From equation (i) and (ii), Example: 18, , hc, , , ,  E, , 4E, hc, 3hc,  E, E, 3, , , , hc, , , , (d) 3, , …..(i), …..(ii), ,    3, , Hydrogen atom emits blue light when it changes from n = 4 energy level to n = 2 level. Which colour of, light would the atom emit when it changes from n = 5 level to n = 2 level, [KCET 1993], (a) Red, , Solution : (d), , (c) 3/4, , (b) Yellow, , (c) Green, , (d) Violet, , In the transition from orbits 5  2 more energy will be liberated as compared to transition from 4  2., So emitted photon would be of violet light.

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genius PHYSICS, , 32 Atomic Structure, Example: 19, , A single electron orbits a stationary nucleus of charge +Ze, where Z is a constant. It requires 47.2 eV to, excited electron from second Bohr orbit to third Bohr orbit. Find the value of Z, [IIT-JEE 1981], (a) 2, , Solution : (b), , (b) 5, , (c) 3, , (d) 4, , Excitation energy of hydrogen like atom for n 2  n1,  1, 47 . 2  36, 1 , 1 , 5 2,  1,  24 . 98 ~ 25, E  13 . 6 Z 2  2  2  eV  47 .2  13 .6 Z 2 ,  2   13 .6 , Z  Z2 , 2, n, , 13, ., 6, , 5, 36, n, 2, 3, , , 2 ,  1, , Z=5, Example: 20, , The first member of the Paschen series in hydrogen spectrum is of wavelength 18,800 Å. The short, wavelength limit of Paschen series is, [EAMCET (Med.) 2000], (a) 1215 Å, , Solution : (c), , (b) 6560 Å, , First member of Paschen series mean it's  max , , (c) 8225 Å, , 144, 7R, , Short wavelength of Paschen series means  m in , Hence, Example: 21, , (d) 12850 Å, , 9, R, ,  max 16, 7, 7,   m ax ,  18 ,800  8225 Å .,   m in , , 16, 16,  min, 7, , Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is, [EAMCET (Engg.) 1995; MP PMT 1997], , (a) 1 : 3, Solution : (c), , (b) 27 : 5, , For Lyman series, , For Balmer series, , 1, ,  L1, 1, ,  B1, , 1  3R, 1,  R 2  2 , 4, 2 , 1, , …..(i), , 1  5R,  1,  R 2  2 , 3  36, 2, , …..(ii), , From equation (i) and (ii), Example: 22, ,  L1,  B1, , , , (d) 4 : 9, , 5, ., 27, , The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The, ground state energy of an electron of this ion will be, [RPET 1997], (a) 3.4 eV, , Solution : (c), , (c) 5 : 27, , Using, , , (b) 13.6 eV, , (c) 54.4 eV, , (d) 122.4 eV, ,  1, 1 , 1, 1 ,  1,  RZ 2  2  2  ,  1 .1  10 7  Z 2  2  2 , , 9, , , , n2 , 108 .5  10, 5 , 2,  n1, 1, , 1, 108 .5  10, , 9, ,  1 .1  10 7  Z 2 , , 21, 100,  Z2 , 4 Z=2, , 9, 100, 108 .5  10  1 .1  10 7  21, , Now Energy in ground state E  13 .6 Z 2 eV  13 .6  2 2 eV  54 .4 eV, Example: 23, , Solution : (a), , Hydrogen (H), deuterium (D), singly ionized helium (He  ) and doubly ionized lithium (Li   ) all have one, electron around the nucleus. Consider n  2 to n  1 transition. The wavelengths of emitted radiations are, [KCET 1994], 1 , 2 , 3 and 4 respectively. Then approximately, (a) 1  2  4 3  94, , (b) 4 1  22  23  4, , Using E  Z 2, , (∵ n 1 and n 2 are same), , , , hc, , , , (c) 1  22  2 2 3  3 2 4 (d), , 1  2  23  34, ,  Z 2  Z 2  constant  1 Z12   2 Z 22   3 Z 32   4 Z 4  1  1   2  1 2   3  2 2   4  3 3, ,  1   2  4 3  94 ., Example: 24, , Hydrogen atom in its ground state is excited by radiation of wavelength 975 Å. How many lines will be, there in the emission spectrum, (a) 2, , (b) 4, , (c) 6, , (d) 8

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genius PHYSICS, Atomic structure 33, Solution : (c), ,  1, 1 , 1, 1 ,  1,  R 2  2  ,  1 .097  10 7  2  2   n = 4, 10, , n 2 , 975  10, n , 1,  n 1, 1, , Using, , Now number of spectral lines N , Example: 25, , A photon of energy 12.4 eV is completely absorbed by a hydrogen atom initially in the ground state so that, it is excited. The quantum number of the excited state is, (b) n= 3, , (a) n =1, Solution : (c), , (d) n = , , (c) n = 4, , Let electron absorbing the photon energy reaches to the excited state n. Then using energy conservation,  , , 13 .6, n, , Example: 26, , n(n  1) 4 (4  1), , 6., 2, 2, , 2, ,  13 .6  12 .4  , , 13 .6, n, , 2, , 13 . 6,  12  n = 3.46 ≃ 4, 1 .2, ,  1 .2  n 2 , , The wave number of the energy emitted when electron comes from fourth orbit to second orbit in, hydrogen is 20,397 cm–1. The wave number of the energy for the same transition in He  is, (a) 5,099 cm–1, , (b) 20,497 cm–1, , (c) 40,994 cm–1, , (d) 81,998 cm–1, , 2,  1, Z , , 1 , Z,    RZ 2  2  2     Z 2  2   2      4   2    4  81588 cm 1 ., n, ,  1  Z1 , n 2 , 1,  1, 2, , 1, , Solution : (d), , Using, , Example: 27, , In an atom, the two electrons move round the nucleus in circular orbits of radii R and 4R. the ratio of the, time taken by them to complete one revolution is, (a) 1/4, , Solution : (d), , (b) 4/1, , Time period T , , (c) 8/1, , n3, Z2, , For a given atom (Z = constant) So T  n 3 …..(i),  From equation (i) and (ii) T  R 3 / 2 , Example: 28, , radius R  n 2, , and, , T1  R 1, , T 2  R 2, , , , , , 3/2, ,  R , , ,  4R , , 3/2, , , , …..(ii), , 1, ., 8, , Ionisation energy for hydrogen atom in the ground state is E. What is the ionisation energy of Li  atom in the 2nd excited, state, (a) E, , Solution : (a), , (d) 1/8, , (b) 3E, , Ionisation energy of atom in nth state E n , , (c) 6E, , (d) 9E, , Z2, n2, , For hydrogen atom in ground state (n = 1) and Z = 1  E  E0, For Li   atom in 2nd excited state n = 3 and Z = 3, hence E  , , E0, 32, , …..(i), ,  3 2  E0, , …..(ii), , From equation (i) and (ii) E  E ., Example: 29, , An electron jumps from n = 4 to n = 1 state in H-atom. The recoil momentum of H-atom (in eV/C) is, (a) 12.75, , Solution : (a), , (b) 6.75, , (d) 0.85, , The H-atom before the transition was at rest. Therefore from conservation of momentum, Photon, Precoil, , Example: 30, , (c) 14.45, , momentum, =, Recoil, E  E1, 0 .85 eV  (13 .6 eV ), h, eV, ,  4, ,  12 . 75, c, c, c, c, , momentum, , of, , H-atom, , or, , If elements with principal quantum number n > 4 were not allowed in nature, the number of possible, elements would be, [IIT-JEE 1983; CBSE PMT 1991, 93; MP PET 1999; RPET 1993, 2001; RPMT 1999, 2003; J & K CET 2004], , (a) 60, Solution : (a), , Maximum value of n = 4, , (b) 32, , (c) 4, , (d) 64

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genius PHYSICS, , 34 Atomic Structure, , So possible (maximum) no. of elements, , N  2  1 2  2  2 2  2  3 2  2  4 2  2  8  18  32  60 ., , Tricky example: 1, If the atom, , 100 Fm, , 257, , follows the Bohr model and the radius of, , 100 Fm, , 257, , is n times the Bohr radius, then find, , n, [IIT-JEE (Screening) 2003], , (a) 100, , Solution : (d), , (b) 200, , m2, (rm )  ,  Z, , , (d) 1/4, , 2, ,  (0 . 53 Å)  (n  0 . 53 Å)  m  n, , Z, , , m = 5 for, , 100, , , , n, , Tricky example: 2, , (c) 4, , Fm 257 (the outermost shell) and z = 100, , (5) 2, 1, , 100, 4, , An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy (in eV), required to remove both the electrons from a neutral helium atom is, (a) 79.0, , Solution : (a), , (b) 51.8, , (c) 49.2, , (d) 38.2, , After the removal of first electron remaining atom will be hydrogen like atom., So energy required to remove second electron from the atom E  13 .6 ,  Total energy required = 24.6 + 54.4 = 79 eV, , 22,  54 .4 eV, 1, , [IIT