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The atomic masses of $H e$ and $N e$ are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of $H e$ gas at $- 73^{°} C$ is “M” times that of the de-Broglie wavelength of $N e$ at $727^{°} C$ . M is

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answer is 5.

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Detailed Solution

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

$\begin{array}{l} ∴ m^{2} v^{2} = 2 m K E \\ ∴ m v = \sqrt{2 m K E} \end{array}$

$λ (w a v e l e n g t h) = \frac{h}{m v} = \frac{h}{\sqrt{2 m K E}} = \frac{h}{\sqrt{2 m (T)}}$

Where, $T =$ Temperature in Kelvin

$λ (H e a t - 73^{°} C = 200 K) = \frac{h}{\sqrt{2 \times 4 \times 200}}$

$λ (N e a t 727^{°} C = 1000 K) = \frac{h}{\sqrt{2 \times 20 \times 1000}}$

$∴ \frac{λ (H e)}{λ (N e)} = M = \sqrt{\frac{2 \times 20 \times 1000}{2 \times 4 \times 200}} = 5$

Thus, $M = 5$

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