Consider an electron in the n=3 orbit of a hy | vedclass

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(A) The velocity of an electron in the $n$-th orbit is given by $v = \frac{Z e^2}{2 \epsilon_0 n h}$.For an electron,the de Broglie wavelength is $\la

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(A) The velocity of an electron in the $n$-th orbit is given by $v = \frac{Z e^2}{2 \epsilon_0 n h}$.For an electron,the de Broglie wavelength is $\lambda_e = \frac{h}{m v} = \frac{2 \epsilon_0 n h^2}{m Z e^2}$.For a neutron with thermal energy $E = k_B T$,the de Broglie wavelength is $\lambda_n = \frac{h}{\sqrt{2 m_N k_B T}}$.Equating $\lambda_e = \lambda_n$,we get $\frac{h}{m v} = \frac{h}{\sqrt{2 m_N k_B T}}$,which implies $m^2 v^2 = 2 m_N k_B T$.Substituting $v = \frac{Z e^2}{2 \epsilon_0 n h}$,we get $T = \frac{m^2 Z^2 e^4}{8 \epsilon_0^2 n^2 h^2 m_N k_B}$.Using $n=3$,$T = \frac{m^2 Z^2 e^4}{72 \epsilon_0^2 h^2 m_N k_B}$.Since $a_0 = \frac{h^2 \epsilon_0}{\pi m e^2}$,we have $a_0^2 = \frac{h^4 \epsilon_0^2}{\pi^2 m^2 e^4}$.Substituting $\frac{m^2 e^4}{\epsilon_0^2} = \frac{h^4}{\pi^2 a_0^2}$,we get $T = \frac{Z^2 h^2}{72 \pi^2 a_0^2 m_N k_B}$.Comparing this with the given expression $T = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$,we find $\alpha = 72$.

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