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Q. The value of the de Broglie wavelength of $He$ gas at $100 \,K$ is $5$ times that of the de-Broglie wavelength gas $X$, at $1250 \,K$. What is atomic mass of $X$?

Structure of Atom

Solution:

$\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m(K E)}}$

$=\frac{h}{\sqrt{2 m\left(\frac{3}{2} R T\right)}}$

$\lambda_{(H e)}=\frac{h}{\sqrt{2 \times 4\left(\frac{3}{2} R \times 100\right)}}$

$\lambda_{(X)}=\frac{h}{\sqrt{2 \times m\left(\frac{3}{2} R \times 1250\right)}}$

$\frac{\lambda_{( He )}}{\lambda_{(X)}}=\sqrt{\frac{m \times 1250}{4 \times 100}}=5$

$\therefore \frac{m \times 1250}{4 \times 100}=25$

$\Rightarrow m=8$