The ratio of de-Broglie wavelength of molecul | vedclass

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(B) The root mean square velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3kT}{m}}$.The de-Broglie wavelength is $\lambda = \frac{h}{mv_{r

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

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(B) The root mean square velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3kT}{m}}$.The de-Broglie wavelength is $\lambda = \frac{h}{mv_{rms}} = \frac{h}{m} \sqrt{\frac{m}{3kT}} = \frac{h}{\sqrt{3mkT}}$.For hydrogen $(H_2)$,$m_H = 2$ units and $T_H = 27 + 273 = 300	ext{ K}$.For helium $(He)$,$m_{He} = 4$ units and $T_{He} = 127 + 273 = 400	ext{ K}$.The ratio is $\frac{\lambda_H}{\lambda_{He}} = \sqrt{\frac{m_{He} T_{He}}{m_H T_H}}$.Substituting the values: $\frac{\lambda_H}{\lambda_{He}} = \sqrt{\frac{4 	imes 400}{2 	imes 300}} = \sqrt{\frac{1600}{600}} = \sqrt{\frac{16}{6}} = \sqrt{\frac{8}{3}}$.

The ratio of de-Broglie wavelength of molecules of hydrogen and helium in two jars kept separately at temperatures of $27\,^{\circ}\text{C}$ and $127\,^{\circ}\text{C}$ respectively is

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