00:01
For part a of the given problem, according to free electron theory of metals, fermi energy is given by ef, that is equal to h squared divided by two times of a mass of electrons times 3 in e divided by 8 pi, power to 2 over 3.
00:21
The ne is the special concentration of free electrons.
00:25
Then we see this e -fermi energy is proportional to the n -e, n -e to the power 2 over 3.
00:41
For part b of the problem, we substitute with the values of the constants into the equation that we wrote above.
00:52
So we get the fermi energy, e -fermi energy, will be 6 .6 .6.
00:58
To 6 times 10 to the power minus 34 joules second, taking square of this, divided by 2 times 9 .1, 1 times 10 to power minus 31, is the mass of the electrons in a kg.
01:16
We multiply this with a constant that is 1 .60 times 10 to the power minus 19 joules per electron volts, then times 3 over 8 pi, to the power 3, 2 over 3 times ne 2 over 3.
01:34
So the fermi energy we get here is 3 .65 times 10 to the power minus 19 times ne 2 over 3rd power...