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

Answer: 5

Atomic mass of helium and neon is $4amu$ and $20amu$
$m_{He}=4amu\text{ and }{m}_{Ne}=20amu$
$T_{He}=-73^{\circ }C=(273+(-73))K=200K$
$T_{Ne}=727^{\circ }C=(273+727)K=1000K$
As we know that the de Broglie wavelength of gas molecules is given by$\lambda =\frac{h}{\sqrt{3mkT}}$
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.