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

The de Broglie wavelength is: $$\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mKE}}$$

For gas molecules, the average KE is: $$KE = \frac{3}{2}k_BT$$

So: $$\lambda = \frac{h}{\sqrt{3mk_BT}} \propto \frac{1}{\sqrt{T}}$$

$$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{600}} = \frac{1}{\sqrt{2}}$$

$$\lambda_2 = \frac{\lambda_1}{\sqrt{2}}$$

The correct answer is Option 3.

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