Question

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium Which are at temperature ${27}^{o}C$ and ${127}^{o}C$ respectively is

• A. $\frac{1}{2}$
• B. $\sqrt{\frac{3}{8}}$
• C. $\sqrt{\frac{8}{3}}$
• D. $1$

Answer 1

Answer: (C)

$\sqrt{\frac{8}{3}}$

As we know,

de-Broglie wavelength $\lambda =\frac{h}{m{v}_{rms}},rms$ velocity of a gas particle at the given temperature $\left(T\right)$ is given as

$\frac{1}{2}m{v}_{rms}^2=\frac{3}{2}kT\Rightarrow v_{rms}=\sqrt{\frac{3kT}{m}}\Rightarrow m{v}_{rms}=\sqrt{3mkT}$

$\therefore \lambda =\frac{h}{m{v}_{rms}}=\frac{h}{\sqrt{3mkT}}$

$\Rightarrow \frac{{\lambda }_H}{{\lambda }_{He}}=\sqrt{\frac{m_{He}{T}_{He}}{m_H{T}_H}}=\sqrt{\frac{4(273+127)}{2(273+27)}}=\sqrt{\frac{8}{3}}$