Q 12-11-099JEE MainJEE Main 2023 (10 Apr, Shift 1)Easy
The de Broglie wavelength of a molecule in a gas at room temperature $300$ K is $\lambda_1$. If the temperature of the gas is increased to $600$ K, then the de Broglie wavelength of the same gas molecule becomes
Answer: (C) $\dfrac1{\sqrt2}\lambda_1$
$\lambda=\dfrac{h}{\sqrt{3mkT}}\propto\dfrac1{\sqrt T}$: doubling $T$ gives $\dfrac{\lambda_1}{\sqrt2}$.
Solution by Sreeraj P, M.Sc Physics