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 de Broglie wavelength of Ne at $+{727}^{\circ }C$. Assuming that Kinetic energy is proportional to absolute temperature, 'M' is

Answer 1

de Broglies's wavelength of a particle when kinetic energy (K.E) and mass (m) are given:
$\lambda =\frac{h}{\sqrt{2K.Em}}$
Given:
$\text{Mass of He atom}\left(m_{He}\right)=4amu$
$\text{Mass of Ne atom}\left(m_{Ne}\right)=20amu$
$\text{Temperature of He}\left(T_{He}\right)=-{73}^{\circ }C=200K$
$\text{Temperature of Ne}\left(T_{Ne}\right)=+{727}^{\circ }C=1000K$

We know that:
$K.E\alpha T$
$\frac{K.E_{He}}{K.E_{Ne}}=\frac{T_{He}}{T_{Ne}}=\frac{200}{1000}=\frac{1}{5}$

Now, ratio of de Broglies wavelengths of $Ne\text{and}He({\lambda }_{He},{\lambda }_{Ne})$:
$\frac{{\lambda }_{He}}{{\lambda }_{Ne}}=\frac{\sqrt{2K.E_{Ne}{m}_{Ne}}}{\sqrt{2K.E_{He}{m}_{He}}}=\sqrt{\frac{5}{1}\times \frac{20}{4}}=5\therefore {\lambda }_{He}=5\times {\lambda }_{Ne}$
So, value of M is 5