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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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The ratio of de-Broglie wavelength of moleculus of hydrogen and helium in two gas jars kept separately at temperatures $\;27^{0} C\;$ and $\;127^{0} C\;$ respectively is

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

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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
$mV=\sqrt{3mKT}$
$\lambda=\large\frac{h}{mV}=\large\frac{h}{\sqrt{3 m KT}}$
$\large\frac{\lambda_{H}}{\lambda_{He}}=\sqrt{\large\frac{m_{He} T_{He}}{m_{H}T_{H}}}$
$=\sqrt{\large\frac{(4 amu)\;(273+127)^{0}K}{(2 amu)\;(273+27)_{0}K}}$
$=\sqrt{\large\frac{8}{3}}\;.$
answered Feb 25, 2014 by yamini.v
 

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