Matter Waves: de Broglie Wavelength question from MHT CET 2026 (12 May, Shift 2)

Answer: (b)

1. Thermal de Broglie wavelength: $\lambda = \dfrac{h}{\sqrt{3mkT}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{mT}}$.
2. $T_{H_2} = 300\text{ K}$, $T_{He} = 400\text{ K}$, $m_{He} = 2m_{H_2}$.
3. $\dfrac{\lambda_{H_2}}{\lambda_{He}} = \sqrt{\dfrac{m_{He}T_{He}}{m_{H_2}T_{H_2}}} = \sqrt{\dfrac{2 \times 400}{300}} = \sqrt{\dfrac{8}{3}}$
4. So the answer is (b) $\sqrt{\frac{8}{3}}$.