De Broglie Wavelength Equivalence of Electron and Thermal Neutron

To find the value of $\alpha$, we first determine the de Broglie wavelength of the electron in the specified orbit and equate it to the de Broglie wavelength of the thermal neutron.

Step 1: De Broglie Wavelength of the Electron

According to Bohr's quantization condition for angular momentum, the circumference of the $n^{\text{th}}$ orbit is an integral multiple of the electron's de Broglie wavelength: $2\pi r_n = n \lambda_e \implies \lambda_e = \frac{2\pi r_n}{n}$

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

For the $n = 3$ orbit: $r_3 = \frac{3^2 a_0}{Z} = \frac{9 a_0}{Z}$

Substituting $r_3$ into the de Broglie wavelength equation for $n = 3$: $\lambda_e = \frac{2\pi}{3} \left( \frac{9 a_0}{Z} \right) = \frac{6 \pi a_0}{Z}$

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Step 2: De Broglie Wavelength of the Neutron

The thermal kinetic energy of the neutron at temperature $T$ is given as $E_k = k_{\text{B}}T$.

The momentum of the neutron of mass $m_{\text{N}}$ is: $p_{\text{N}} = \sqrt{2 m_{\text{N}} E_k} = \sqrt{2 m_{\text{N}} k_{\text{B}} T}$

Thus, the de Broglie wavelength of the neutron is: $\lambda_{\text{N}} = \frac{h}{p_{\text{N}}} = \frac{h}{\sqrt{2 m_{\text{N}} k_{\text{B}} T}}$

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Step 3: Equating the Wavelengths and Finding $\alpha$

Given that $\lambda_{\text{N}} = \lambda_e$: $\frac{h}{\sqrt{2 m_{\text{N}} k_{\text{B}} T}} = \frac{6 \pi a_0}{Z}$

Squaring both sides of the equation: $\frac{h^2}{2 m_{\text{N}} k_{\text{B}} T} = \frac{36 \pi^2 a_0^2}{Z^2}$

Rearranging the expression to solve for temperature $T$: $T = \frac{Z^2 h^2}{72 \pi^2 a_0^2 m_{\text{N}} k_{\text{B}}}$

Comparing this derived temperature equation with the given formula: $T = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_{\text{N}} k_{\text{B}}}$

We find that: $\alpha = 72$