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JEE Advanced Physics Question (2024) — Solution

Question

A particle of mass m is moving in a circular orbit under the influence of the central force F(r)=-k r, corresponding to the potential energy V(r)=k r^2 / 2, where k is a positive force constant and r is the radial distance from the origin. According to the Bohr's quantization rule, the angular momentum of the particle is given by L=n , where =h /(2 ), h is the Planck's constant, and n a positive integer. If v and E are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?

Options

  1. A. r^2=n 1 m k
  2. B. v^2=n k m^3
  3. C. L m r^2 = k m
  4. D. E= n 2 k m

Answer

C. L m r^2 = k m

Step-by-step solution

The central force will provide necessary centripetal force kr = mv ^2 r or, kr ^2= mv ^2 ...(1) By quantisation rule n = mvr or, n r = mv ...(2) aligned & (1) (2)^2 kr ^2 n ^2 ^2 r ^2 = mv ^2 ~m ^2 v ^2 \\ & k n ^2 ^2 r ^4= 1 ~m \\ & r = ( n ^2 ^2 ~km )^ 1 4 r ^2= n mk aligned aligned & (2) Using (1), K n mk = mv ^2 \\ & v ^2= n k m ^3 aligned (3) L mr ^2 = mvr mr ^2 = v r = k m from (1) aligned & (4) E = 1 2 mv ^2+ 1 2 kr ^2= n 2 k m + 1 2 k n mk \\ & E = n k m aligned

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