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 $-{73}^{0}C$ is $"M"$ times that of the de Broglie wavelength of $Ne$ at ${727}^{0}C$. $M$ is:

• A. $5$
• B. $0.2$
• C. $2$
• D. $0.5$

Answer 1

Answer: (A)

$5$

We have de broglie relation $\lambda =\frac{h}{mv}$
Average kinetic energy $\frac{1}{2}m{v}^2=\frac{3}{2}KT$
Hence, $\lambda =\frac{h}{mv}=\frac{h}{m}\sqrt{\frac{m}{3KT}}=\frac{h}{\sqrt{3mKT}}$
Thus, $\lambda \propto \frac{1}{{\left(mT\right)}^{\frac{1}{2}}}$
Hence,$\frac{\lambda He}{\lambda Ne}={\left[\frac{\left(200mu\right)\left(100K\right)}{\left(4mu\right)\left(200K\right)}\right]}^{\frac{1}{2}}=5$
Therefore, the value of $m$ is $5$.