Find the de-Broglie wavelength of an electron in a metal at `12.^(@)C` and compare it with the mean separation between two electrons in a metal which is about `2Å`, Given `h=6.6xx10^(-34) Js, m_(e)=9.1xx10^(-31) kg`. Boltzmann constant `k=1.38xx10^(-23) JK^(-1)`

Answer

`5.4 nm, 27`

Explanation

Here, T=127+273=400K.
Energy of electron `=1/2mv^(2)=3/2kT`
or `mv=sqrt(3mkT)`
De-broglie wavelength of electron is
`lambda=h/(mv)=h/(sqrt(3mkT))`
`=(6.63xx10^(-34))/(sqrt(3xx(9.1xx10^(-31))xx(1.38xx10^(-23))xx400))`
`=(6.63xx10^(-34))/(12.276xx10^(-26))=5.4xx10^(-9)m`
Mean separation between two electrons in metal is `r=2xx10^(-10)m :. lambda/r=(5.4xx10^(-9))/(2xx10^(-10))=27`