The de Broglie wavelength of an electron in a | vedclass

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{3mk_BT}}$.Here,$m = 9.11 	imes 10^{-31} \ kg$,$k_B = 1.38 	imes 10^{-23} \ J/K$,

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$0.03$

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{3mk_BT}}$.Here,$m = 9.11 	imes 10^{-31} \ kg$,$k_B = 1.38 	imes 10^{-23} \ J/K$,and $T = 27 + 273 = 300 \ K$.Substituting these values:$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{3 	imes (9.11 	imes 10^{-31}) 	imes (1.38 	imes 10^{-23}) 	imes 300}}$$\lambda \approx 6.2 	imes 10^{-9} \ m$.The mean distance between two electrons is $r = 2 	imes 10^{-10} \ m$.The ratio is $\frac{r}{\lambda} = \frac{2 	imes 10^{-10}}{6.2 	imes 10^{-9}} \approx 0.03$.

The de Broglie wavelength of an electron in a metal at $27^{\circ}C$ is compared with the given mean distance between two electrons in the metal,which is $2 \times 10^{-10} \ m$. The ratio of the mean distance to the de Broglie wavelength is approximately:

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