Saved Bookmarks
| 1. |
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» Solution :(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)` |
|