PhysicsMedium145×since 2002Q6586The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is λ1\lambda_1λ1. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomesA2 λ1\lambda_1λ1B12\frac{1}{2}21λ1\lambda_1λ1C12\frac{1}{\sqrt2}21λ1\lambda_1λ1D2 λ1\sqrt2~\lambda_12 λ1Check answerSkip