$(a)\;\large\frac{2}{\sqrt{3}}\qquad(b)\;2:3 \qquad(c)\;\large\frac{\sqrt{3}}{4}\qquad(d)\;\sqrt{\large\frac{8}{3}}$

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Answer : (d) $\;\sqrt{\large\frac{8}{3}}$

Explanation :

De-Broglie wavelength $\;\lambda=\large\frac{h}{mV}$

Where speed (r.m.s ) of a gas particle at given Temperature(T) is given as

$\large\frac{1}{2}mV^2=\large\frac{3}{2}\;KT$

$V=\sqrt{\large\frac{3KT}{m}}\qquad \; $ where K = Boltzmann's constant and m=mass of gas particle

T=temp of gas in K

$mV=\sqrt{3mKT}$

$\lambda=\large\frac{h}{mV}=\large\frac{h}{\sqrt{3 m KT}}$

$\large\frac{\lambda_{H}}{\lambda_{He}}=\sqrt{\large\frac{m_{He} T_{He}}{m_{H}T_{H}}}$

$=\sqrt{\large\frac{(4 amu)\;(273+127)^{0}K}{(2 amu)\;(273+27)_{0}K}}$

$=\sqrt{\large\frac{8}{3}}\;.$

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