The ratio of de Broglie wavelength of molecul | vedclass

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.For a gas molecule at temperature $T$,the root mean square speed is $v = \sqrt{\fra

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$\sqrt{8/3}$

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.For a gas molecule at temperature $T$,the root mean square speed is $v = \sqrt{\frac{3kT}{m}}$,where $k$ is the Boltzmann constant and $m$ is the mass of the molecule.Substituting $v$ into the wavelength formula: $\lambda = \frac{h}{m\sqrt{\frac{3kT}{m}}} = \frac{h}{\sqrt{3mkT}}$.For hydrogen $(H_2)$ at $T_1 = 27\,^oC = 300\,K$ and helium $(He)$ at $T_2 = 127\,^oC = 400\,K$,the masses are $m_H = 2\,amu$ and $m_{He} = 4\,amu$.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 wavelength of molecules of hydrogen and helium in two gas jars kept separately at temperatures of $27\,^oC$ and $127\,^oC$ respectively is

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