Step-by-step derivation:

1. Determine the de Broglie wavelength of the electron ($\\lambda_e$).

   According to the Bohr model, for an electron in a stable orbit, the circumference of the orbit is an integral multiple of its de Broglie wavelength. This is given by the relation: $2 \pi r_n = n \lambda_e$ where $r_n$ is the radius of the $n$-th orbit and $n$ is the principal quantum number.

   The de Broglie wavelength of the electron is therefore: $\lambda_e = \frac{2 \pi r_n}{n}$

   The radius of the $n$-th orbit in a hydrogen-like atom with atomic number $Z$ is given by: $r_n = \frac{n^2 a_0}{Z}$ where $a_0$ is the first Bohr radius of the hydrogen atom.

   Substituting the expression for $r_n$ into the equation for $\lambda_e$: $\lambda_e = \frac{2 \pi}{n} \left( \frac{n^2 a_0}{Z} \right) = \frac{2 \pi n a_0}{Z}$

   The problem states that the electron is in the $n=3$ orbit. So, we substitute $n=3$: $\lambda_e = \frac{2 \pi (3) a_0}{Z} = \frac{6 \pi a_0}{Z}$

2. Determine the de Broglie wavelength of the neutron ($\\lambda_N$).

   The problem states that the neutron has thermal energy equal to $k_{\mathrm{B}} T$. This is the kinetic energy ($K_N$) of the neutron. $K_N = k_{\mathrm{B}} T$

   The de Broglie wavelength of a particle is related to its momentum $p$ by $\lambda = h/p$. The momentum is related to kinetic energy by $K = p^2 / (2m)$, so $p = \sqrt{2mK}$.

   For the neutron, its momentum is $p_N = \sqrt{2 m_{\mathrm{N}} K_N}$, where $m_{\mathrm{N}}$ is the mass of the neutron. $p_N = \sqrt{2 m_{\mathrm{N}} k_{\mathrm{B}} T}$

   The de Broglie wavelength of the neutron is: $\lambda_N = \frac{h}{p_N} = \frac{h}{\sqrt{2 m_{\mathrm{N}} k_{\mathrm{B}} T}}$

3. Equate the two wavelengths and solve for the temperature T.

   The problem states that the de Broglie wavelengths of the electron and the neutron are the same: $\lambda_e = \lambda_N$ $\frac{6 \pi a_0}{Z} = \frac{h}{\sqrt{2 m_{\mathrm{N}} k_{\mathrm{B}} T}}$

   To solve for $T$, we first square both sides of the equation: $\left( \frac{6 \pi a_0}{Z} \right)^2 = \left( \frac{h}{\sqrt{2 m_{\mathrm{N}} k_{\mathrm{B}} T}} \right)^2$ $\frac{36 \pi^2 a_0^2}{Z^2} = \frac{h^2}{2 m_{\mathrm{N}} k_{\mathrm{B}} T}$

   Now, we rearrange the equation to isolate $T$: $T = \frac{h^2 Z^2}{36 \pi^2 a_0^2 \cdot 2 m_{\mathrm{N}} k_{\mathrm{B}}}$ $T = \frac{Z^2 h^2}{72 \pi^2 a_0^2 m_{\mathrm{N}} k_{\mathrm{B}}}$

4. Compare the derived expression for T with the given expression to find $\\alpha$.

   The problem provides the following expression for the temperature: $T = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_{\mathrm{N}} k_{\mathrm{B}}}$

   Comparing our derived expression for $T$ with the given expression: $\frac{Z^2 h^2}{72 \pi^2 a_0^2 m_{\mathrm{N}} k_{\mathrm{B}}} = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_{\mathrm{N}} k_{\mathrm{B}}}$

   By equating the denominators, we find the value of $\alpha$: $72 \pi^2 a_0^2 m_{\mathrm{N}} k_{\mathrm{B}} = \alpha \pi^2 a_0^2 m_{\mathrm{N}} k_{\mathrm{B}}$ $\alpha = 72$

Thus, the value of $\alpha$ is 72.