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)

$\lambda_{H e}(200 K)=M \times \lambda_{N e}(1000 K)\left(\lambda=\frac{h}{m \cdot v}\right)$
$\frac{h}{4 a m u \times C_{m p}(200)}=M \times \frac{h}{20 a m u \times C_{m p}(1000)}\left(C_{m p}=\sqrt{\frac{2 K T}{m}}\right)$
$\Rightarrow \frac{1}{4 \times \sqrt{\frac{2 K \times 200}{4}}}=\frac{M}{20 \times \sqrt{\frac{2 K \times 1000}{20}}} $
$\Rightarrow \frac{1}{4 \times \sqrt{50}}=\frac{M}{20 \times \sqrt{50}}$
$ \Rightarrow M=5$