Ratio of De Broglie Wavelength of Helium and Neon Gases

To find the value of $M$, we use the relation for the thermal de Broglie wavelength $\lambda$ of a gas molecule at temperature $T$:

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m E_k}} = \frac{h}{\sqrt{3 m k_B T}}$

where:

• $h$ is Planck's constant
• $m$ is the mass of the gas atom
• $k_B$ is the Boltzmann constant
• $T$ is the absolute temperature in Kelvin

From this relation, the thermal de Broglie wavelength is inversely proportional to $\sqrt{m T}$: $\lambda \propto \frac{1}{\sqrt{m T}}$

Given:

• For Helium ($\text{He}$):

  • $m_{\text{He}} = 4\text{ a.m.u.}$
  • $T_{\text{He}} = -73\text{ }^{\circ}\text{C} = (-73 + 273)\text{ K} = 200\text{ K}$

• For Neon ($\text{Ne}$):

  • $m_{\text{Ne}} = 20\text{ a.m.u.}$
  • $T_{\text{Ne}} = 727\text{ }^{\circ}\text{C} = (727 + 273)\text{ K} = 1000\text{ K}$

The ratio of their de Broglie wavelengths is: $\frac{\lambda_{\text{He}}}{\lambda_{\text{Ne}}} = \sqrt{\frac{m_{\text{Ne}} \cdot T_{\text{Ne}}}{m_{\text{He}} \cdot T_{\text{He}}}}$

Substitute the given values into the expression: $M = \sqrt{\frac{20 \times 1000}{4 \times 200}} = \sqrt{\frac{20000}{800}} = \sqrt{25} = 5$

Thus, the value of $M$ is $5$.