The ratio of the de-Broglie wavelength of mol | vedclass

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Question

(B) The de-Broglie wavelength $\lambda$ of a gas molecule at temperature $T$ is given by $\lambda = \frac{h}{\sqrt{3mkT}}$,where $h$ is Planck's const

Vedclass · Accepted Answer

$2\sqrt{2}:\sqrt{3}$

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ of a gas molecule at temperature $T$ is given by $\lambda = \frac{h}{\sqrt{3mkT}}$,where $h$ is Planck's constant,$m$ is the mass of the molecule,$k$ is the Boltzmann constant,and $T$ is the absolute temperature.For hydrogen $(H_2)$,$m_1 = 2u$ and $T_1 = 27^{\circ} C = 300 \ K$.For helium $(He)$,$m_2 = 4u$ and $T_2 = 127^{\circ} C = 400 \ K$.The ratio is $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2 T_2}{m_1 T_1}} = \sqrt{\frac{4 	imes 400}{2 	imes 300}} = \sqrt{\frac{1600}{600}} = \sqrt{\frac{8}{3}} = \frac{2\sqrt{2}}{\sqrt{3}}$.Thus,the ratio is $2\sqrt{2}:\sqrt{3}$.

The ratio of the de-Broglie wavelength of molecules of hydrogen $(H_2)$ and helium $(He)$ which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is:

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