SATHEE: Atomic Structure - Result Question 93

Atomic Structure - Result Question 93

####69. The atomic masses of $H e$ and $N e$ are 4 and $20 a m u$ , respectively. The value of the de-Broglie wavelength of $H e$ gas at $- 73^{\circ} C$ is ’ $M$ ’ times that of the de-Broglie wavelength of $N e$ at $727^{\circ} C . M$ is

(2013 Adv.)

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Answer:

Correct Answer: 69. $(4.14 e V)$

Solution:

PLAN $K E = \frac{1}{2} m v^{2} = \frac{3}{2} R T$

$∴ m^{2} v^{2} = 2 m K E ∴ m v = \sqrt{2 m K E}$

$λ (wavelength) = \frac{h}{m v} = \frac{h}{\sqrt{2 m K E}} \propto \frac{h}{\sqrt{2 m (T)}}$

where, $T =$ Temperature in Kelvin

$\begin{aligned} λ (H e at - 73^{\circ} C = 200 K) & = \frac{h}{\sqrt{2 \times 4 \times 200}} λ (N e at 727^{\circ} C = 1000 K) & = \frac{h}{\sqrt{2 \times 20 \times 1000}} ∴ \frac{λ (H e)}{λ (N e)} & = M = \sqrt{\frac{2 \times 20 \times 1000}{2 \times 4 \times 200}} = 5 \end{aligned}$

Thus, $M = 5$

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Atomic Structure - Result Question 93

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