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 $$k_B T$$ 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, $$m_N$$ is the mass of the neutron and $$a_0$$ is the first Bohr radius of hydrogen atom) then the value of $$\alpha$$ is ________.

The de Broglie wavelength $$\lambda$$ of a particle is given by $$\lambda = \dfrac{h}{p}$$, where $$p$$ is the linear momentum.

Electron in the $$n = 3$$ Bohr orbit
For a hydrogen-like atom (nuclear charge $$Z$$):

• Radius of the $$n^{\text{th}}$$ orbit $$r_n = \dfrac{n^2 a_0}{Z}$$

• In Bohr’s model, exactly $$n$$ de Broglie wavelengths fit along the circumference, so $$n \lambda_e = 2\pi r_n \;\; \Longrightarrow \;\; \lambda_e = \dfrac{2\pi r_n}{n}$$

Putting $$r_n$$ from above and $$n = 3$$,

$$\lambda_e = \dfrac{2\pi}{3}\,\dfrac{(3)^2 a_0}{Z} = \dfrac{2\pi \cdot 3 a_0}{Z} = \dfrac{6\pi a_0}{Z} \;-(1)$$

Thermal neutron
For a neutron of mass $$m_N$$ in thermal equilibrium at temperature $$T$$ (non-relativistic), the average kinetic energy is $$\dfrac{p_N^2}{2m_N} = k_B T$$, so

$$p_N = \sqrt{2 m_N k_B T}$$

Hence its de Broglie wavelength is

$$\lambda_N = \dfrac{h}{p_N} = \dfrac{h}{\sqrt{2 m_N k_B T}} \;-(2)$$

Given condition
The two wavelengths are equal: $$\lambda_e = \lambda_N$$.

Equating $$ (1) $$ and $$ (2) $$:

$$\dfrac{6\pi a_0}{Z} = \dfrac{h}{\sqrt{2 m_N k_B T}}$$

Rearrange to isolate $$T$$:

$$\sqrt{2 m_N k_B T} = \dfrac{hZ}{6\pi a_0}$$
Square both sides:

$$2 m_N k_B T = \dfrac{h^2 Z^2}{36 \pi^2 a_0^2}$$

Therefore

$$T = \dfrac{h^2 Z^2}{72 \pi^2 a_0^2 m_N k_B} \;-(3)$$

The temperature is given in the question as $$T = \dfrac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$$

Comparing this with expression $$ (3) $$ shows

$$\alpha = 72$$

Hence, the required value of $$\alpha$$ is 72.