Question

Making use of Eq.(6.4g), find at `T=0`:
(a) the velocity distribution of free electrons,
(b) the ratio of the mean velocity of free electron to their maximum velocity.

Answer 1

(a)From
`dn(E )=(sqrt(2)m^(3//2))/(pi^(2)ħ^(3))E^(1//2)dE`
we get on using `E=(1)/(2)mv^(2), dn(E )=dn(V)`
`dn(v)=(sqrt(2)m^(3//2))/(pi^(2)ħ^(3))(1)/(sqrt(2))m^(1//2) vmvdv=(m^(3))/(pi^(2)ħ^(3))v^(2)dv`
This holds for `0 lt v lt V_(F)` where `(1)/(2)mV_(F)^(2)=E_(F)`
and `dn(v)=0 for v gt v_(F)`.
(b)Mean velocity
`lt v gt = int_(0)^(v_(F))v^(3)dv//int_(0)^(v_(F)) v^(2) dv=(3)/(4)V_(F)`
`:. (lt v gt)/(V_(F))=(3)/(4)`