# Modern Physics-03- Subjective Solved Problems

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26. May 2023
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### Modern Physics-03- Subjective Solved Problems

• 1. S SO OL LV VE ED D S SU UB BJ JE EC CT TI IV VE E P PR RO OB BL LE EM MS S Problem 1. How may different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to states with principal quantum number n ? Solution: From the n th state, the atom may go to ( 1) n  th state, …., 2nd state or 1st state. So there are ( 1) n  th possible transitions staring from the n th sate. The atoms reaching ( 1) n  th state may make ( 2) n  different transitions. Similarly for other lower states. The total number of possible transitions is ( 1) ( 2) ( 3) .....2 1 n n n        ( 1) 2 n n   . Problem 2. A doubly ionized lithium atom is hydrogen-like with atomic number Z = 3. Find the wavelength of the radiation required to excite the electron in Li2 from the first to the third Bohr orbit. Given the ionization energy of hydrogen atom as 13.6 eV. Solution: The energy of th n orbit of a hydrogen-like atom is given as 2 2 13.6 n Z E n   Thus for 2 Li  atom, as Z = 3, the electron energies for the first and third Bohr orbits are : For n = 1, 2 1 2 13.6 (3) 1 E eV    122.4 eV   For n = 3, 2 3 2 13.6 (3) (3) E eV    13.6 eV   Thus the energy required to transfer an electron from 1 E level to 3 E level is, 1 3 E E E   13.6 ( 122.4) 108.8eV      Therefore the radiation needed to cause this transition should have photons of this energy. hv = 108.8 eV The wavelength of this radiation is  108.8 hc eV   or 34 8 19 (6.63 10 ) (3 10 ) 113.74 108.8 108.8 1.6 10 hc m eV            A. Problem 3. A moving hydrogen atom makes a head-on inelastic collision with a stationary hydrogen atom. Before collision both atoms are in the ground state and after collision they move together. What is the minimum velocity of the moving hydrogen atom if one of the atoms is to be given the minimum excitation energy after the collision? Solution: Let u be the velocity of the hydrogen atom before collision and  the velocity of the two atoms moving together after collision. By the principle of conservation of momentum, we have : 0 2 Mu M M     or 2 u   The loss in kinetic energy E  due to collision is given by 2 2 1 1 (2 ) 2 2 E Mu M    
• 2. As / 2 u   , we have 2 2 1 1 (2 ) ( / 2) 2 2 E Mu M u    2 2 1 1 2 4 Mu Mu   2 1 4 Mu  This loss in energy is due to the excitation of one of the hydrogen atoms. The ground state ( 1 n  ) energy of a hydrogen atom is : 1 13.6 E eV   The energy of the first excited level ( 2 n  ) is : 2 3.4 E eV   Thus the minimum energy required to excite a hydrogen atom from ground state to first excited state is : 2 1 [ 3.4 ( 13.6)] E E eV      10.2 eV  19 10.2 1.6 10    J 19 16.32 10 J    As per problem, the loss in kinetic energy in collision is due to the energy used up in exciting one of the atoms. Thus 2 1 E E E    or 2 19 1 16.32 10 4 Mu    or 19 2 4 16.32 10 u M     The mass of the hydrogen atom is 1.0078 amu or 1.0078 27 1.66 10   kg.  19 2 8 27 4 16.32 10 39.02 10 1.0078 1.66 10 u          or 4 -1 6.246 10 ms u   . Problem 4. A particle of charge equal to that of an electron, -e, and mass 208 times the mass of electron (called  - meson) moves in a circular orbit around a nucleus of charge +3e. (Take the mass of the nucleus to be infinite). Assuming that Bohr model of the atom is applicable to this system: (i) derive an expression for the radius of the nth Bohr orbit. (ii) find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom. (iii) find the wavelength of the radiation emitted when the  - meson jumps from the third orbit to the first orbit. (Rydberg’s constant = m 7 1 1.097 10   ) Solution: (i) We have the radius of the th n orbit given by 2 2 0 2 n n h r me z    Substituting 0 1 , 3 and 208 , 4 e k Z m m     we get 2 2 0 2 624 n e n h r m e    . (ii) The radius of first Bohr orbit for hydrogen is : 2 0 1 2 e h r m e    For n r ( -mesonic atom) = 1 r (hydrogen atom), we have
• 3. 2 2 2 0 0 2 2 624 e e n h h m e m e      or 2 624 n  or 25 n (iii) The energy for the th n orbit is given by 2 2 4 2 2 2 n mk Z e E n    Substituting 0 1 208 , 3, and 4 2 e h m e Z k        , we get 4 2 2 2 0 234 e n m e E n h    4 2 3 2 0 1872 8 e m e hc h c n          2 1872Rhc n   where 4 2 3 0 8 e m e R h c   is the Rydberg constant. When the  - meson jumps from the third orbit to the first orbit, the difference in energy is radiated as a photon of frequency v given by 3 1 hv E E   As , c v   we have 3 1 hc E E    2 2 1 1 1872 1 3 Rhc         or 1 1 1872 1 9 R          or 7 9 9 1872 8 1872 8 (1.097 10 ) R          10 0.5478 10  m 0.5478  Å. Problem 5. Suppose the potential energy between electron and proton at a distance r is given by ke r 2 3 3  . Use Bohr’s theory to obtain energy levels of such a hypothetical atom. Solution: As we know that negative of potential energy gradient is force for a conservative field. dU F dr   It is given that 2 3 3 ke U r   Hence, force 2 2 3 4 3 dU ke ke F dr r r            According to Bohr’s theory this force provides the necessary centripetal force for orbital motion.  2 2 4 ke m r r   …(i)
• 4. Also 2 nh m r    …(ii) Hence, 2 nh mr    …(iii) Substituting this value in Eq. (i), we get 2 2 2 2 2 2 4 4 mn h ke m r r r   or 2 2 2 2 4 e km r n h   Substituting this value or r in Eq. (iii), we get 3 3 2 2 2 8 n h km e    Total energy 2 2 3 1 2 3 ke E KE PE m r      2 3 3 3 2 2 2 3 2 2 2 2 2 3 8 4 m n h ke n h km e ke m                 6 2 2 ( ) 6( ) n ke m   where 2 h          . Problem 6. Consider a hypothetical hydrogen-like atom. The wavelength in A for the spectrial lines for transitions from n p  to n 1  are given by p p 2 2 1500 1    where p = 2, 3, 4, …. (a) find the wavelength of the least energetic and the most energetic photons in this series. (b) construct an energy level diagram for this element representing at least three energy levels. (c) what is the ionization potential of this element ? Solution: As we know energy of a photon is given by hc E   From the given condition 2 2 1500 1 p p    Hence, 10 2 1 1 10 1500 hc hc E p            J 10 19 2 1 1 10 (1500)(1.6 10 ) hc eV p            2 1 8.28 1 eV p         Hence energy of th n state is given by 2 8.28 . n E eV n  (a) Maximum energy is released for transition from p   to 1; p  hence wavelength of most energetic photon is 1500 Å. Least energy is released for transition from 2 n  to 1 n  transition. For 2 p  2000   Å (b) The energy level diagram is shown in the figure. (c) The ionization potential corresponds to energy required to liberate an electron from its ground state.
• 5. i.e., ionization energy = 8.28 eV Hence, ionization potential = 8.28 V n=3 n=2 n=1 -0.92 eV -2.07 eV -8.28 eV Problem 7. Calculate the wavelength of the emitted characteristic X-ray from a tungsten (Z = 74) target when an electron drops from an M shell to a vacancy in the K shell. Solution: Tungsten is a multielectron atom. Due to the shielding of the nuclear charge by the negative charge of the inner core electrons, each electron is subject to an effective nuclear charge eff Z which is different for different shells. Thus, the energy of an electron in the th n level of a multielectron atom is given by 2 2 13.6 eff n Z E eV n   For an electron in the K shell ( 1), ( 1) eff n Z Z    . Thus, the energy of the electron in the K shell is : 2 2 (74 1) 13.6 72500 1 K E eV      For an electron in the M shell ( 3 n  ), the nucleus is shielded by one electron of the 1 n  state and eight electrons of the 2 n  state, a total of nine electrons, so that 9 eff Z Z   . Thus the energy of an electron in the M shell is : 2 2 (74 9) 13.6 6380 3 M E eV      Therefore, the emitted X-ray photon has an energy given by 6380 ( 72500 ) 66100 M K hv E E eV eV eV        or 19 66100 1.6 10 hc      J  19 66100 1.6 10 hc      m 34 8 19 (6.63 10 ) (3 10 ) 66100 1.6 10         m 9 0.0188 10   m. Problem 8. If the short series limit of the Balmer series for hydrogen is 3646 Å, calculate the atomic no. of the element which gives X-ray wavelength down to 1.0 A. Identify the element. Solution: The short limit of the Balmer series is given by 2 2 1 1 1/ / 4 2 v R R              10 -1 4/ (4/ 3646) 10 m R     Further the wavelengths of the k series are given by the relation 2 2 2 1 1 1 ( 1) 1 v R Z n            The maximum wave number correspondence to n   and, therefore, we must have 2 1 ( 1) v R Z     or 10 2 10 1 3646 10 ( 1) 911.5 4 1 10 Z R          
• 6.  ( 1) 911.5 Z   30.2  or 30.2 31 Z   Thus the atomic number of the element concerned is 31. The element having atomic number Z = 31 is Gallium. Problem 9. The disintegration rate of a radioactive sample at a certain instant is 4750 disintegration per minute. Five minute later the rate becomes 2700 disintegration per minute. Calculate (i) the decay constant, and (ii) the half life of the sample. Given 10 log 1.76 0.2455  . Solution: The rate of disintegration R is given by 0 t R R e  where 0 R is the initial rate at t = 0. (i) This equation gives 0 t R e R   Taking the logarithm of both sides, we get 0 loge R t R   or 0 0 10 1 2.303 log log e R R t R t R    According to problem, 0 4750, 2700, 5 R R t    min  10 10 2.303 4750 2.303 log log (1.76) 5 2700 5    or 2.303 0.2455 0.1131 5     per min (ii) Half life is given by 1/ 2 log 2 0.693 6.13 0.1131 e T     min. Problem 10. The mean lives of a radioactive substance are 1620 year and 405 year for  - emission and  - emission respectively. Find the time during which three-fourth of a sample will decay if it is decaying both by  - emission and  - emission simultaneously. Solution: The decay constant  is the reciprocal of the mean life . Thus, 1 1620    per year and 1 405    per year  Total decay constant,        or 1 1 1 1620 405 324     per year When 3 4 th part of the sample has disintegrated, 0 / 4 N N   0 0 4 t N N e  or 4 t e  Taking logarithm of both sides, we get
• 7. log 4 e t   or 2 1 2 log 2 log 2 e e t     2 324 0.693 449     year. Problem 11. There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half life of neutron is 700 second, what fraction of neutrons will decay before they travel a distance of 10 m? Given mass of neutron = 27 1.675 10  kg. Solution: From the given kinetic energy of the neutrons we first calculate their velocity. Thus 2 19 1 0.0327 1.6 10 2 m       19 2 4 27 2 0.0327 1.6 10 625 10 1.675 10           or 2500 / m s   with this speed, the time taken by the neutrons to travel a distance of 10 m is, 3 10 4 10 2500 t s      The fraction of neutrons decayed in time t  second is, N t N    also, 1/ 2 0.693 T    3 6 1/ 2 0.693 0.693 (4 10 ) 3.96 10 700 N t N T           . Problem 12. In the chain analysis of a rock, the mass ratio of two radioactive isotopes is found to be 100 : 1. The mean lives of the two isotopes are 9 4 10  year and 9 2 10  year respectively. If it is assumed that at the time of formation of the rock, both isotopes were in equal proportion, calculate the age of the rock. Ratio of atomic weights of the two isotopes is 1.02 : 1. ( 10 log 1.02 0.0086  ). Solution: At the time of formation of the rock, both isotopes have the same number of nuclei 0 N . Let 1  and 2  be the decay constants of the two isotopes. If 1 N and 2 N are the number of their nuclei after a time t, we have 1 1 0 t N N e  and 2 2 0 t N N e   1 2 ( ) 1 2 t N e N     …(i) Let the masses of the two isotopes at time t be 1 m and 2 m and let their respective atomic weights be 1 M and 2 M . We have 1 1 1 m N M  and 2 2 2 m N M   1 1 2 2 2 1 . N m M N m M  …(ii) Substituting the value given in the problem, we get 1 2 100 1 100 . 1 1.02 1.02 N N   Let 1  and 2  be the mean lives of the two isotopes. Then 1 1 1    and 2 2 1   
• 8. Which gives 9 9 9 1 2 1 2 9 9 1 2 2 10 4 10 0.25 10 (2 10 ) (4 10 )                     Setting this value in Eqn. (i), we get 9 (0.25 10 ) 1 2 t N e N    or 9 1 100 log 1.02 0.25 10 e t          9 18.34 10   year. Problem 13. The nuclear reaction, n B Li He 10 7 4 5 3 2    is observed to occur even when very slow-moving neutrons ( n M u 1.0087  ) strike a boron atom at rest. For a particular reaction in which n K 0  , the helium ( He M u 4.0026  ) is observed to have a speed of m s 6 9.30 10 /  . Determine (a) the kinetic energy of the (lithium Li M u 7.0160  ), and (b) the Q-value of the reaction. Solution: (a) Since the neutron and boron are both initially at rest, the total momentum before the reaction is zero, and afterward is also zero. Therefore, Li Li He He M M    We solve this for Li  and substitute it into the equation for kinetic energy. We can use classical kinetic energy with little error, rather than relativistic formulas, because 6 9.30 10 He    m/s is not close to the speed of light c, and Li  will be even less since Li He M M  . Thus we can write: 2 2 2 2 1 1 2 2 2 He He He He Li Li Li Li Li Li M M K M M M M             We put in numbers, changing the mass in u to kg and recalling that 13 1.60 10  J = 1 MeV: 2 27 2 6 2 27 (4.0026 ) (1.66 10 / ) (9.30 10 m/s) 2(7.0160 )(1.66 10 / ) Li u kg u K u kg u       13 1.64 10 J=1.02MeV    (b) We are given the data 0, a X K K   so Li He Q K K   where 2 27 6 2 1 1 (4.0026 )(1.66 10 / )(9.30 10 m/s) 2 2 He He He K M u kg u       13 2.87 10 J=1.80MeV    Hence, 1.02MeV+1.80MeV=2.82 MeV Q  Problem 14. A radioactive source in the form of metal sphere of diameter 3 10 m emits beta of metal sphere of diameter 3 10 m emits beta particle at a constant rate of 10 6.25 10  particles per second. If the source is electrically insulated, how long will it take for its potential to rise by 1.0 volt, assuming that 80% of the emitted beta particles escape from the source? Solution: Let t  time for the potential of metal sphere to rise by one volt. Now   particles emitted in this time 11 (6.25 10 ) t    Number of  -particles escaped in this time 10 10 (80/100) (6.25 10 ) 5 10 t t       Charge acquired by the sphere in t sec. 10 19 (5 10 ) (1.6 10 ) Q t     
• 9. 19 8 10 t    coulomb …(i) ( emission of  -particle lends to a charge e on metal sphere) The capacitance C of a metal sphere is given by 0 4 C r    3 12 9 1 10 10 2 18 9 10                   farad …(ii) we know that Q C V   {Here 1 V  volt}  12 9 10 (8 10 ) 1 18 t            . Solving it for t , we get 6.95 t   sec. Problem 15 . A small quantity of solution containing 24 Na radionuclide (half life 15 hours) of activity 1.0 microcurie is injected into the blood of a person. A sample of the blood of volume 1 cm3 taken after 5 hours shows an activity of 296 disintegrations per minute. Determine the total volume of blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person. (1 curie 10 3.7 10   disintegrations per second). Solution: We know that 1/ 2 0.693 T   or 1/ 2 0.693 0.693 15 3600 T     = 5 1.283 10  sec. Now activity 0 dN A N dt    where 1 A  micro curie 4 1 3.7 10    4 3.7 10   disintegrations/sec  4 5 3.7 10 1.283 10    0 N  4 9 0 5 3.7 10 2.883 10 1.283 10 N       Let the number of radioactive nuclei present after 5 hours be 1 N in 1cc sample of blood. Then 1 dN N dt   or 1 296 0.693 60 15 3600 N   or 5 1 296 15 3600 3.844 10 60 0.693 N       Let 0 N  be the number of radioactive nuclei in per cc of sample, then / 1 0 1 2 t T N N         or / 0 1 (2)t T N N     5/15 1/ 3 5 0 1 (2) (2) 3.844 10 N N       5 1.269 3.844 10    [ 1/ 3 (2) 1.269  ] 5 4.878 10   Volume of blood 9 0 5 0 2.883 10 4.878 10 N V N      4 0.5910 10   cm3 = 5.91 litres.