JEE Advanced 2025 Paper 1 · Q13 · Bohr Model

Question

Consider an electron in the $n = 3$ orbit of a hydrogen-like atom of atomic number $Z$. At absolute temperature $T$, a neutron with thermal energy $k_B T$ has the same de Broglie wavelength as that electron. If $T = Z^2 h^2/(\alpha\,\pi^2 a_0^2 m_N k_B)$, where $h$ is Planck's constant, $k_B$ Boltzmann's constant, $m_N$ the neutron mass, and $a_0$ the Bohr radius, find $\alpha$.

SolveKar AI · Accepted Answer

For a hydrogenic atom, $n$-th Bohr orbit radius $r_n = n^2 a_0/Z$. Electron momentum from quantization: $m_e v r = n \hbar \implies p_e = n\hbar/r_n = n\hbar Z/(n^2 a_0) = Z\hbar/(n a_0)$. de Broglie wavelength of electron: $\lambda_e = h/p_e = h n a_0/(Z\hbar) = 2\pi n a_0/Z$ (since $h/\hbar = 2\pi$). For $n = 3$: $\lambda_e = 6\pi a_0/Z$. Neutron de Broglie wavelength at thermal energy $E = k_B T$: $p_n = \sqrt{2 m_N k_B T}$, so $\lambda_n = h/\sqrt{2 m_N k_B T}$. Equate: $6\pi a_0/Z = h/\sqrt{2 m_N k_B T} \implies \sqrt{2 m_N k_B T} = h Z/(6\pi a_0)$. Square: $2 m_N k_B T = h^2 Z^2/(36\pi^2 a_0^2) \implies T = Z^2 h^2/(72\,\pi^2 a_0^2 m_N k_B)$. Hence $\alpha = 72$.

JEE Advanced 2025 Paper 1 · Q13 · Bohr Model

Solution