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JEE Advanced 2025 · Paper 1, Physics question 13

Consider an electron in the n=3n = 3 orbit of a hydrogen-like atom with atomic number ZZ. At absolute temperature TT, a neutron having thermal energy kBTk_BT has the same de Broglie wavelength as that of this electron. If this temperature is given by T=Z2h2απ2a02mNkBT = \dfrac{Z^2h^2}{\alpha\pi^2a_0^2m_Nk_B}, (where hh is the Planck's constant, kBk_B is the Boltzmann constant, mNm_N is the mass of the neutron and a0a_0 is the first Bohr radius of hydrogen atom) then the value of α\alpha is _____.

Answer and worked solution

Official answer: 72

Electron. In the nn-th Bohr orbit 2πrn=nλ2\pi r_n=n\lambda, with rn=n2a0Zr_n=\frac{n^2a_0}{Z}. So
λe=2πna0Z=6πa0Z(n=3).\lambda_e=\frac{2\pi na_0}{Z}=\frac{6\pi a_0}{Z}\quad(n=3).

Neutron. With kinetic energy kBTk_BT, λN=h2mNkBT\lambda_N=\dfrac{h}{\sqrt{2m_Nk_BT}}.

Equate.
h22mNkBT=36π2a02Z2  ⟹  T=Z2h272π2a02mNkB,\frac{h^2}{2m_Nk_BT}=\frac{36\pi^2a_0^2}{Z^2}\implies T=\frac{Z^2h^2}{72\pi^2a_0^2m_Nk_B},
so α=72\alpha=72.