Question

The atomic masses of $He$ and $Ne$ are $4$ and $20amu$, respectively. The value of the de-Broglie wavelength of $He$ gas at $-73°C$ is ‘$M$’ times that of the de-Broglie wavelength of $Ne$ at $727°C$. $M$ is:

• A. $25$
• B. $6$
• C. $5$
• D. $36$

Answer 1

The correct option is C $5$

According to the formula of Kinetic energy:
$K.E.=\frac{1}{2}m{v}^2=\frac{3}{2}RT$
$m^2{v}^2=2mK.E$
($mv=\sqrt{2mK.E.}$
Wavelength ($\lambda )=\frac{h}{mv}=\frac{h}{\sqrt{(}2mK.E.)}\propto \frac{h}{\sqrt{\left(2m\right(T)}}\propto \frac{1}{\sqrt{mT}}$
where, $T$ = Temperature in Kelvin;
$\lambda$ ($He$ at $-73°C=200K)=\frac{1}{\sqrt{(4\times 200)}}$
$\lambda$ ($Ne$ at $727°C=1000K)=\frac{1}{\sqrt{(20\times 1000}}$
∴ $\frac{{\lambda }_{\left(He\right)}}{{\lambda }_{\left(Ne\right)}}=M=\sqrt{\frac{20\times 1000}{(4\times 200}}$ = $5$
Thus, $M=5$