# Modern Physics-02-Objective Solved Problems

Educator en Study Innovations
26 de May de 2023
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### Modern Physics-02-Objective Solved Problems

• 1. S SO OL LV VE ED D O OB BJ JE EC CT TI IV VE E P PR RO OB BL LE EM MS S Problem 1. The ratio of minimum to maximum wavelengths of radiation that an excited electron in a hydrogen atom can emit while going to the ground state is (a) 1/2 (b) Zero (c) 3/4 (d) 27/32. Ans. (c) Solution: Energy of radiation that corresponds to the energy difference between two energy levels 1 n and 2 n is given as 2 2 1 2 1 1 13.6 E eV n n         E is minimum when 1 1 n  & 2 2 n   min 1 1 3 13.6 13.6 1 4 4 E eV eV           E is maximum when 1 1 n  & 2 n   (the atom is ionized, that is known as ionization energy)  max 1 13.6 1 13.6 E eV           .  min max 3 4 E E   max min / 3 / 4 hc hc     min max 3 4    Problem 2. The wavelength of α K X-ray produced by an X-ray tube is 0.76 Å. The atomic number of anticathode material is (a) 82 (b) 41 (c) 20 (d) 10. Ans. (b) Solution: For K X-ray line, 2 2 2 2 1 1 1 1 ( 1) ( 1) 1 4 1 2 R Z R Z                      2 1 3 ( 1) 4 R Z     …(i) With reference to given data, 0.76    Å = 0.76 10 10  m 7 1.097 10 R   m Putting these values in equation (i) 2 10 7 4 1 ( 1) 3 0.76 10 1.097 10 Z       1600   1 40 Z    41 Z  . Problem 3. If the stationary proton and α - particle are accelerated through same potential difference, the ratio of de Broglie’s wavelength will be (a) 2 (b) 1 (c) 2 2 (d) none of these.
• 2. Ans. (c) Solution: The gain in K.E. of a charged particle after moving through a potential difference of V is given by eV, that is also equal to 2 1 2 mv where v is the velocity of the charged particle. Disregarding the relativistic effect, 2 1 2 mv qV   2qV v m   2 mv mqV   de Broglie wavelength 2 h h mv mqV      p p p p m q V m q V        Putting (4)(2) , 2 2 (1)(1) p p V V        . Problem 4. There are two radioactive substances A and B . Decay constant of B is two times that of A . Initially, both have equal number of nuclei. After n half lives of A , rate of disintegration of both are equal. The value of n is (a) 4 (b) 2 (c) 1 (d) 5. Ans. (c) Solution: Let A     2 B    If 0 N is total no. of atoms in A and B at 0 t  , then initial rate of disintegration of 0 A N   , and initial rate of disintegration of 0 2 B N   As 2 B a     1 2 B A T T  i.e. half life of B is half the half life of A . After one half life of A 0 2 A N dN dt          Equivalently, after two half lives of B 0 0 2 4 2 B N N dN dt            Clearly, A B dN dN dt dt                , after 1 n  , i.e., one half life of A . Problem 5. A hydrogen atom is in an excited state of principal quantum number ( n) it emits a photon of wavelength ( λ ), when it returns to the ground state. The value of n is (a) λR λR - 1 (b) (λR - 1) λR (c) λ(R - 1) (d) λR λR - 1 . Ans. (d)
• 3. Solution: As 2 2 1 2 1 1 1 R n n           2 2 1 1 1 1 R n          Multiply both sides by  2 1 1 1 R n          2 1 1 1 R n    or 2 1 1 1 1 R R R n        1 R n R     . Problem 6. The weight based ratio of 238 U and 226 Pb in a sample of rock is 4 : 3 . If the half life of 238 U is 9 4.5× 10 years, then the age of rock is (a) 9 9.0×10 years (b) 9 6.3×10 years (c) 9 4.5× 10 years (d) 9 3.78× 10 years. Ans. (d) Solution: Let initial no. of U-atoms = 0 N After time t , (age of rock), let no. of atoms remaining undecayed N  .  0 238 4 26( ) 3 N N N    0 1.79 N N  0 log / log 2 T N N t  9 4.5 10 log1.79 0.301    = 9 3.78 10  years. Problem 7. 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 (a) Y  2Z (b) W  X + Z (c) W  2Y (d) X  Y + Z. Binding Energy/Nucleon (in eV) Mass number of nuclei 5.0 7.5 8.0 8.5 Z Y X W 0 30 60 90 120 Ans. (c) Solution: If it mass A 2 Y Z  Reactant: 60 8.5 510 R    MeV Product: 2 30 5 300 P     MeV 210 E    MeV ENDOTHERMIC If it was B
• 4. W X Z   120 7.5 900 R    MeV 90 8 30 5 870 P      MeV 30 E    MeV ENDOTHERMIC If it was C 2 W Y  120 7.5 900 R    MeV 2 60 8.5 1020 P     MeV 120 E   Mev EXOTHERMIC If it was D X Y Z   90 8.0 720 R    MeV 60 8.5 30 5.0 660 P      MeV 60 E    MeV ENDOTHERMIC Problem 8. In the sun about 4 billion kg of matter is converted to energy each second. The power output of the sun in watt is (a) 26 3.6 ×10 (b) 26 0.36 ×10 (c) 26 36 ×10 (d) 26 0.036 × 10 . Ans. (b) Solution: 8 4 10 m t   kgs-1 2 E mc   2 E m c t t         8 16 4 10 9 10 E t      25 3.6 10 E t   Js-1  25 3.6 10 E W t   . Problem 9. A star initially has 40 10 deutrons. It produces energy via the processes  2 2 3 1 1 1 H + H H + p and  2 3 4 1 1 2 H + H He + n . If the average power radiated by the star is 16 10 W , the deuteron supply of the star is exhausted in a time of the order of (a) 6 10 s (b) 8 10 s (c) 12 10 s (d) 16 10 s . (The masses of nuclei are: 2 m(H ) = 2.014amu, m(p) = 1.007amu , 4 m(n) = 1.0084amu, m(He ) = 4.001amu ) Ans. (c) Solution: 2 2 3 1 1 1 H H H p    2 3 4 1 1 2 H H He n     2 4 1 2 3 H He p n    4 2 2 1 ( ) ( ) ( ) 3 ( ) m m He m p m n m H       [4.001 1.007 1.008 3(2.014)] m amu     
• 5.  0.026 m amu     2 | | | | E c m     16 27 (9 10 )(0.026 1.67 10 ) E        (931.5)(0.026) E   MeV  12 3.87 10 J E     As each reaction involves 3 deuterons, so total number of reactions involved in the process 40 10 3  . If each reaction produces an energy E  , then 40 28 10 1.29 10 J 3 total E E     total E Pt  Time of exhaustion of the star 28 16 1.29 10 10 t    12 1.29 10 t s   . Problem 10. Consider the following reaction  2 2 4 1 1 2 H + H He + Q . If 2 1 m( H ) = 2.0141amu ; 4 2 m( He ) = 4.0024amu . The energy Q released (in MeV) in this fusion reaction is (a) 12 (b) 6 (c) 24 (d) 48. Ans. (c) Solution: 2 2 4 1 1 2 H H He Q     4 2 2 1 ( ) 2 ( ) m m He m H     4.0024 2(2.0141) m     0.0258 m amu    Since, 2 Q c m    (0.0258)(931.5) Q  MeV  24 Q MeV.