At room temperature (27^{circ}C) and 1 atmosp | vedclass

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{3mk_BT}}$,where $m$ is the mass of a helium atom.$m = \frac{4 	imes 10^{-3} 	ext

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$r = 50 \lambda$

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{3mk_BT}}$,where $m$ is the mass of a helium atom.$m = \frac{4 	imes 10^{-3} 	ext{ kg}}{6.023 	imes 10^{23}} \approx 0.664 	imes 10^{-26} 	ext{ kg}$ and $T = 273 + 27 = 300 	ext{ K}$.Substituting the values: $\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{3(0.664 	imes 10^{-26})(1.38 	imes 10^{-23})(300)}} \approx 7.3 	imes 10^{-11} 	ext{ m}$.If $r$ is the average distance between atoms,then $r \approx (V/N)^{1/3}$.From the ideal gas law $PV = N k_B T$,we have $V/N = k_B T / P$.Thus,$r = (k_B T / P)^{1/3} = \left( \frac{1.38 	imes 10^{-23} 	imes 300}{1.01 	imes 10^5} ight)^{1/3} \approx 3.4 	imes 10^{-9} 	ext{ m}$.Comparing the two: $\frac{r}{\lambda} = \frac{3.4 	imes 10^{-9}}{7.3 	imes 10^{-11}} \approx 46.5 \approx 50$.Therefore,$r \approx 50 \lambda$.

At room temperature $(27^{\circ}C)$ and $1$ atmospheric pressure,the de Broglie wavelength associated with a helium atom is compared with the average distance between two atoms. Which of the following relations is correct?

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