Question 13

For an electron in the $n^{th}$ Bohr orbit: De Broglie wavelength is related to the orbit circumference by $2\pi r_n = n\lambda_e$, where $r_n = \frac{n^2 a_0}{Z}$. For $n = 3$: $\lambda_e = \frac{2\pi r_3}{3} = \frac{2\pi (3^2 a_0 / Z)}{3} = \frac{2\pi (9 a_0 / Z)}{3} = \frac{6\pi a_0}{Z}$. For a neutron with thermal energy $E = k_B T$: The kinetic energy is $E = \frac{p^2}{2m_N} = k_B T \implies p = \sqrt{2m_N k_B T}$. The de Broglie wavelength of the neutron is $\lambda_N = \frac{h}{p} = \frac{h}{\sqrt{2m_N k_B T}}$. Given $\lambda_e = \lambda_N$: $\frac{6\pi a_0}{Z} = \frac{h}{\sqrt{2m_N k_B T}}$ Squaring both sides: $\frac{36\pi^2 a_0^2}{Z^2} = \frac{h^2}{2m_N k_B T}$ Rearranging for $T$: $T = \frac{Z^2 h^2}{72 \pi^2 a_0^2 m_N k_B}$ Comparing this with the given expression $T = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$, we find $\alpha = 72$.