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JEE Main 2023 (10 Apr, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2023 (10 Apr, Shift 1)

Q71JEE Main 2023 · 10 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength

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

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Answer: (c)

Thermal kinetic energy $\propto T$ , so $p = \sqrt{3mkT}$ and $\lambda = \dfrac{h}{\sqrt{3mkT}} \propto \dfrac{1}{\sqrt{T}}$ .
$\dfrac{\lambda_2}{\lambda_1} = \sqrt{\dfrac{300}{600}} = \dfrac{1}{\sqrt{2}}$
So the answer is (c) $\dfrac{1}{\sqrt{2}}\lambda_1$ .

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