Question

The atomic masses of $He$ and $Ne$ are $4$ and $20$ amu, respectively. The value of the de-Broglie wavelength of $He$ gas at $-73^{\circ} C$ is ' $M$ ' times that of the de-Broglie wavelength of $Ne$ at $727^{\circ} C . M$ is

Answer 1

Atomic mass of helium and neon is $4 amu$ and $20 amu$
$ m _{ He }=4 amu \text { and } m _{ Ne }=20 amu $
$ T _{ He }=-73^{\circ} C =(273+(-73)) K =200 K $
$ T _{ Ne }=727^{\circ} C =(273+727) K =1000 K $
As we know that the de Broglie wavelength of gas molecules is given by$\lambda=\frac{ h }{\sqrt{3 mkT }}$
where, $k =$ Boltzmann constant $ \lambda_{ He }= M \lambda_{ Ne } \text { (Given) } $
$\Rightarrow \frac{ h }{\sqrt{3 \times 4 \times k \times 200}}$= M $\frac{ h }{\sqrt{3 \times 20 \times k \times 1000}} $
$\Rightarrow M =\sqrt{25}=5$
Hence, the value of de Broglie wavelength of helium is 5 times the de Broglie wavelength of neon.