The ratio of de-Broglie wavelengths of molecu | vedclass

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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m v_{rms}}$.For a gas molecule,the root mean square velocity is $v_{rms} =

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

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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m v_{rms}}$.For a gas molecule,the root mean square velocity is $v_{rms} = \sqrt{\frac{3kT}{m}}$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{3mkT}}$.Given $T_H = 27^oC = 300 K$ and $T_{He} = 127^oC = 400 K$.The masses are $m_H = 2$ units and $m_{He} = 4$ units.The ratio is $\frac{\lambda_H}{\lambda_{He}} = \sqrt{\frac{m_{He} T_{He}}{m_H T_H}} = \sqrt{\frac{4 	imes 400}{2 	imes 300}} = \sqrt{\frac{1600}{600}} = \sqrt{\frac{8}{3}}$.

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperatures $27^oC$ and $127^oC$ respectively is

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