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  • The ratio of de-Broglie wavelength of molecules of hydrogen and helium in two gas jars kept separately at temperature <img alt="27^{\circ} \mathrm{C}" src="https://entrancecorner.oncodecogs.com/gif

The ratio of de-Broglie wavelength of molecules of hydrogen and helium in two gas jars kept separately at temperature 27^{\circ} \mathrm{C} and 127^{\circ} \mathrm{C} respectively is

Option: 1

\frac{2}{\sqrt{3}}


Option: 2

2: 3


Option: 3

\frac{\sqrt{3}}{4}


Option: 4

\sqrt{\frac{8}{3}}


Answers (1)

best_answer

de-Broglie wavelength \lambda=\frac{\mathrm{h}}{ {mv}}

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

\frac{1}{2} \mathrm{mv}^2=\frac{3}{2} \mathrm{KT}

\Rightarrow \quad v=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}} \text { where } \mathrm{K}=\text { Boltzmann's constant and } \mathrm{m}=\text { mass }

\begin{aligned} & \text { of the gas particle and } \mathrm{T}=\text { temperature of the gas in } \mathrm{K} \\ & \Rightarrow \quad {mv}=\sqrt{3 \mathrm{mKT}} \end{aligned}

\Rightarrow \quad \lambda=\frac{\mathrm{h}}{ {mv}}=\frac{\mathrm{h}}{\sqrt{3 \mathrm{~m} \mathrm{KT}}}

\therefore \quad \frac{\lambda_{\mathrm{H}}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}} \mathrm{T}_{\mathrm{H}}}}

=\sqrt{\frac{(4 \mathrm{amu})(273+127)^{\circ} \mathrm{K}}{(2 \mathrm{amu})(273+27)^{\circ} \mathrm{K}}}=\sqrt{\frac{8}{3}}

 

Posted by

manish

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