Question

The atomic masses of helium and neon are $4.0$ and $20.0$ amu respectively. The value of the de Broglie wavelength of helium gas at $-73^{\circ }C$ is $M$ times the de Broglie wavelength of neon at $727^{\circ }C$. The value of $M$ is

• A. $5$
• B. $25$
• C. $\frac{1}{5}$
• D. $\frac{1}{25}$

Answer 1

Answer: (A)

${\lambda }_{He}(200K)=M\times {\lambda }_{Ne}(1000K)\left(\lambda =\frac{h}{m\cdot v}\right)$
$\frac{h}{4amu\times C_{mp}(200)}=M\times \frac{h}{20amu\times C_{mp}(1000)}\left(C_{mp}=\sqrt{\frac{2KT}{m}}\right)$
$\Rightarrow \frac{1}{4\times \sqrt{\frac{2K\times 200}{4}}}=\frac{M}{20\times \sqrt{\frac{2K\times 1000}{20}}}$
$\Rightarrow \frac{1}{4\times \sqrt{50}}=\frac{M}{20\times \sqrt{50}}$
$\Rightarrow M=5$