MCQEasyJEE Main 2023 · 10 April, Shift 1de Broglie Relation

Physics Question from JEE Main 2023 · 10 April, Shift 1

The de Broglie wavelength of a molecule in a gas at room temperature (300K300 \, \text{K}) is λ1\lambda_1. If the temperature of the gas is increased to 600K600 \, \text{K}, the de Broglie wavelength becomes:

  • A

    2λ12\lambda_1

  • B

    λ1/2\lambda_1/\sqrt{2}

  • C

    2λ1\sqrt{2}\lambda_1

  • D

    λ1/2\lambda_1/2

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