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

Question

2 $\lambda_1$

$\frac{1}{2}$$\lambda_1$

$\frac{1}{\sqrt2}$$\lambda_1$

$\sqrt2~\lambda_1$

Solution

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