Atomic Structure - Result Question 93
####69. 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 the de-Broglie wavelength of $Ne$ at $727^{\circ} C . M$ is
(2013 Adv.)
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Answer:
Correct Answer: 69. $(4.14 eV)$
Solution:
- PLAN $KE=\frac{1}{2} m v^{2}=\frac{3}{2} R T$
$$ \therefore \quad m^{2} v^{2}=2 m KE \quad \therefore m v=\sqrt{2 m KE} $$
$$ \lambda \text { (wavelength) }=\frac{h}{m v}=\frac{h}{\sqrt{2 m KE}} \propto \frac{h}{\sqrt{2 m(T)}} $$
where, $\quad T=$ Temperature in Kelvin
$$ \begin{aligned} \lambda\left(He \text { at }-73^{\circ} C=200 K\right) & =\frac{h}{\sqrt{2 \times 4 \times 200}} \ \lambda\left(Ne \text { at } 727^{\circ} C=1000 K\right) & =\frac{h}{\sqrt{2 \times 20 \times 1000}} \ \therefore \quad \frac{\lambda(He)}{\lambda(Ne)} & =M=\sqrt{\frac{2 \times 20 \times 1000}{2 \times 4 \times 200}}=5 \end{aligned} $$
Thus, $\quad M=5$
