The atomic masses of He and Ne are 4 and 20 | vedclass

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2m(KE)}}$.Since $KE = \frac{3}{2}kT$,we have $\lambda \propto \frac{1}{\sqrt{mT}}$

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2m(KE)}}$.Since $KE = \frac{3}{2}kT$,we have $\lambda \propto \frac{1}{\sqrt{mT}}$.For $He$ at $-73^{\circ} C$ $(200 \ K)$: $m_{He} = 4$,$T_{He} = 200 \ K$.For $Ne$ at $727^{\circ} C$ $(1000 \ K)$: $m_{Ne} = 20$,$T_{Ne} = 1000 \ K$.$\frac{\lambda_{He}}{\lambda_{Ne}} = \sqrt{\frac{m_{Ne} T_{Ne}}{m_{He} T_{He}}} = \sqrt{\frac{20 	imes 1000}{4 	imes 200}} = \sqrt{\frac{20000}{800}} = \sqrt{25} = 5$.Thus,$M = 5$.

The atomic masses of $He$ and $Ne$ are $4$ and $20 \ amu$,respectively. The value of the de Broglie wavelength of $He$ gas at $-73^{\circ} C$ is $M$ times that of the de Broglie wavelength of $Ne$ at $727^{\circ} C$. $M$ is

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