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Atomic mass of He and Ne are 4 and 20 amu respectively. The value of the de Broglie wavelength of He gas -73°C is “M” times that of the de-Broglie wavelength of Ne at 727°C. M is

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Detailed Solution

The de-Broglie wavelength of gas molecules is given as,

$λ = \frac{h}{\sqrt{3 mkT}}$ where k = Boltzmann constant

$λ_{Hc} = {Mλ}_{Nc}$

$\frac{h}{\sqrt{3 \times 4 \times k \times 200}} = M \frac{h}{\sqrt{3 \times 20 \times k \times 1000}}$

$M = \frac{\sqrt{3 \times 20 \times k \times 1000}}{\sqrt{3 \times 4 \times k \times 200}} = \sqrt{\frac{20 \times 1000}{4 \times 200}} = \sqrt{25} = 5$

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