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 \(k_{\mathrm{B}} T\) has the same de Broglie wavelength as that of this electron. If this temperature is given by \(T=\frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}\), (where \(h\) is the Planck's constant, \(k_B\) is the Boltzmann constant, \(m_{\mathrm{N}}\) is the mass of the neutron and \(a_0\) is the first Bohr radius of hydrogen atom) then the value of \(\alpha\) is ________

\(\frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{\mathrm{KZe}^2}{\mathrm{r}^2}\)
\(\mathrm{mv}^2 \mathrm{r}=\frac{1}{4 \pi \epsilon_0} \mathrm{Ze}^2\)
\(\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}\)
(1)/(2) gives
\(\mathrm{v}=\frac{\frac{\mathrm{Ze}^2}{4 \pi \epsilon_0}}{\frac{\mathrm{nh}}{2 \pi}}=\frac{\mathrm{Ze}^2}{2 \epsilon_0 \mathrm{nh}}\)
\(\begin{aligned} & \frac{h}{m v}=\frac{h}{\sqrt{2 m_N \cdot K_B T}} \\ & T=\frac{m^2 Z^2 e^4}{8 \epsilon_0^2 n^2 h^2 m_N K_B} \\ & n=3 \Rightarrow T=\frac{m^2 Z^2 e^4}{72 \epsilon_0^2 h^2 m_N K_B} \\ & \frac{(1)}{(2)^2} \Rightarrow \frac{1}{m r}=\frac{\frac{Z e^2}{4 \pi \epsilon_0}}{\frac{n^2 h^2}{4 \pi^2}}\end{aligned}\)
\(\begin{aligned} & \mathrm{r}=\frac{\mathrm{n}^2 \mathrm{~h}^2 \epsilon_0}{\pi \mathrm{Ze}^2 \cdot \mathrm{~m}} \Rightarrow \mathrm{a}_0=\frac{\mathrm{h}^2 \epsilon_0}{\pi \mathrm{e}^2 \mathrm{~m}} \\ & \mathrm{a}_0^2=\frac{\mathrm{h}^4 \epsilon_0^2}{\pi^2 \mathrm{e}^4 \mathrm{~m}^2} \\ & \mathrm{Ta}_0^2=\frac{\mathrm{m}^2 \mathrm{Z}^2 \mathrm{e}^4}{72 \epsilon_0 \mathrm{~h}^2 \mathrm{~m}_{\mathrm{N}} \mathrm{k}_{\mathrm{B}}} \cdot \frac{\mathrm{h}^4 \epsilon_0^2}{\pi^2 \mathrm{e}^4 \mathrm{~m}^2} \\ & \mathrm{~T}=\frac{\mathrm{h}^2 \mathrm{Z}^2}{72 \pi^2 \mathrm{a}_0^2 \mathrm{~m}_{\mathrm{N}} \mathrm{k}_{\mathrm{B}}} \Rightarrow \alpha=72\end{aligned}\)