The atomic masses of helium and neon are 4.0 | vedclass

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m \cdot v}$.For a gas,the most probable velocity is $C_{mp} = \sqrt{\frac{

Vedclass · Accepted Answer

$5$

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m \cdot v}$.For a gas,the most probable velocity is $C_{mp} = \sqrt{\frac{2KT}{m}}$.Thus,$\lambda = \frac{h}{m \sqrt{\frac{2KT}{m}}} = \frac{h}{\sqrt{2mKT}}$.Given $T_{He} = -73 + 273 = 200 \ K$ and $T_{Ne} = 727 + 273 = 1000 \ K$.$\lambda_{He} = M 	imes \lambda_{Ne}$ $\Rightarrow \frac{h}{\sqrt{2 	imes 4 	imes K 	imes 200}} = M 	imes \frac{h}{\sqrt{2 	imes 20 	imes K 	imes 1000}}$.$\frac{1}{\sqrt{1600K}} = M 	imes \frac{1}{\sqrt{40000K}}$.$\frac{1}{40\sqrt{K}} = M 	imes \frac{1}{200\sqrt{K}}$.$M = \frac{200}{40} = 5$.

The atomic masses of helium and neon are $4.0$ and $20.0 \ amu$ respectively. The value of the de Broglie wavelength of helium gas at $-73^{\circ} C$ is $M$ times the de Broglie wavelength of neon at $727^{\circ} C$. The value of $M$ is

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