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. $(5)$
Solution:
- PLAN: $KE=\dfrac{1}{2} m v^{2}=\dfrac{3}{2} R T$
$\therefore \quad m^{2} v^{2}=2 m KE \quad \therefore m v=\sqrt{2 m KE}$
$\lambda \text { (wavelength) }=\dfrac{h}{m v}=\dfrac{h}{\sqrt{2 m KE}} \propto \dfrac{h}{\sqrt{2 m(T)}}$
where, $\quad T=$ Temperature in Kelvin
$ \begin{aligned} \lambda\left(He \text { at }-73^{\circ} C=200 K\right) & =\dfrac{h}{\sqrt{2 \times 4 \times 200}} \\ \lambda\left(Ne \text { at } 727^{\circ} C=1000 K\right) & =\dfrac{h}{\sqrt{2 \times 20 \times 1000}} \\ \therefore \quad \dfrac{\lambda(He)}{\lambda(Ne)} & =M=\sqrt{\dfrac{2 \times 20 \times 1000}{2 \times 4 \times 200}}=5 \end{aligned} $
Thus, $\quad M=5$
BETA
