The de Broglie wavelength of a molecule in a gas at room...

The root mean squared velocity of a gas is given by

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

$⇒ v ∝ \sqrt{T}$

Let

${T}_{1} = 300K{T}_{2} = 600K$

Taking velocity ratios at the given temperatures,

$\dfrac{{v}_{1}}{{v}_{2}} = \sqrt{\dfrac{{T}_{1}}{{T}_{2}}} = \sqrt{\dfrac{300}{600}} = \sqrt{\dfrac{1}{2}}$

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

$\lambda ∝ \dfrac{1}{v}$

The ratio of the wavelengths is

$\dfrac{{\lambda}_{1}}{{\lambda}_{2}} = \left(\dfrac{{v}_{2}}{{v}_{1}}\right) = \dfrac{\sqrt{2}}{1}$

$⇒ {\lambda}_{2} = \left(\dfrac{1}{\sqrt{2}}{\lambda}_{1}\right)$