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JEE Advanced Physics Atomic Physics 2025 JEE Advanced 2025 (Paper 1)

JEE Advanced Physics Question (2025) — Solution

Question

Consider an electron in the n=3 orbit of a hydrogen-like atom with atomic number Z. At absolute temperature T, a neutron having thermal energy k_ B T has the same de Broglie wavelength as that of this electron. If this temperature is given by T= Z^2 h^2 ^2 a_0^2 m_N k_B , (where h is the Planck's constant, k_B is the Boltzmann constant, m_ N is the mass of the neutron and a_0 is the first Bohr radius of hydrogen atom) then the value of is ________

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

mv ^2 r = KZe ^2 r ^2 mv ^2 r = 1 4 _0 Ze ^2 mvr = nh 2 (1)/(2) gives v = Ze ^2 4 _0 nh 2 = Ze ^2 2 _0 nh aligned & h m v = h 2 m_N K_B T \\ & T= m^2 Z^2 e^4 8 _0^2 n^2 h^2 m_N K_B \\ & n=3 T= m^2 Z^2 e^4 72 _0^2 h^2 m_N K_B \\ & (1) (2)^2 1 m r = Z e^2 4 _0 n^2 h^2 4 ^2 aligned aligned & r = n ^2 ~h ^2 _0 Ze ^2 ~m a _0= h ^2 _0 e ^2 ~m \\ & a _0^2= h ^4 _0^2 ^2 e ^4 ~m ^2 \\ & Ta _0^2= m ^2 Z ^2 e ^4 72 _0 ~h ^2 ~m _ N k _ B h ^4 _0^2 ^2 e ^4 ~m ^2 \\ & ~T = h ^2 Z ^2 72 ^2 a _0^2 ~m _ N k _ B =72 aligned

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Related: Physics — Atomic Physics · All PYQ Banks