The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelengt — Atomic Structure Chemistry Question

💡 Solution & Explanation

The thermal de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{3mkT}}$, which means $\lambda \propto \frac{1}{\sqrt{mT}}$. For He: $T_{He} = 273 - 73 = 200$ K, $m_{He} = 4$. For Ne: $T_{Ne} = 273 + 727 = 1000$ K, $m_{Ne} = 20$. The ratio is $M = \frac{\lambda_{He}}{\lambda_{Ne}} = \sqrt{\frac{m_{Ne} T_{Ne}}{m_{He} T_{He}}} = \sqrt{\frac{20 \times 1000}{4 \times 200}} = \sqrt{\frac{20000}{800}} = \sqrt{25} = 5$. Therefore, correct answer is 0005.