Compute the typical de-Broglie wavelength of an electrons in a metal at `27^(@)C` and compare it with the mean separation between two electrons in a metal which is given to be about `2xx10^(-10)m`.

Explanation

Given , mea inter- electron separation , `r_0=2xx10^(10)m`
Temperature , T = 300 K
The de Broglie wavelength is given by
`lamda=h/(sqrt(3mkT))`
`=(6.62xx10^(-34))/(sqrt(3xx9.1xx10^(-31)xx1.38xx10^(-23)xx300))`
`lamda=6.2xx10^(-9)m`
Then , `lamda/r_0=(6.2xx10^(-9))/(2xx10^(-10))=31`
i.e. `lamda` is 31 times the inter electron separation `(r_0)` in a metal.