The de Broglie wavelength of a molecule in a gas at room temperature 300   K is λ 1 . If the…

The root mean squared velocity of a gas is given by 

$v=\sqrt{\frac{3RT}{M}}$

$\Rightarrow v\propto \sqrt{T}$

Let

 $$\begin{gathered} T_1=300 \mathrm{K} \\ {\mathrm{T}}_2=600 \mathrm{K} \end{gathered}$$

Taking velocity ratios at the given temperatures,

$\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{300}{600}}=\sqrt{\frac{1}{2}}$

The de Broglie wavelength is given by $\lambda =\frac{h}{mv}$. So,

$\lambda \propto \frac{1}{v}$

The ratio of the wavelengths is 

$\frac{{\lambda }_1}{{\lambda }_2}=\left(\frac{v_2}{v_1}\right)=\frac{\sqrt{2}}{1}$

$\Rightarrow {\lambda }_2=\left(\frac{1}{\sqrt{2}}{\lambda }_1\right)$