Question

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

• A. 8
• B. 9
• C. 6
• D. 7

Answer 1

Answer: (A)

$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2m(KE)}}$

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

${\lambda }_{(He)}=\frac{h}{\sqrt{2\times 4\left(\frac{3}{2}R\times 100\right)}}$

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

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

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

$\Rightarrow m=8$