00:01
Hi there.
00:01
So for this problem, we are told to show that from the fermi distribution that in a metal with the temperature equals to 0 kelvin, the average energy of an electron is equal to 3 divided by 5 times the fermi energy.
00:26
So we know that at the temperature equals to 0 kelvin, the number of particles in terms of the energy can be 1 or 0, 1 when the energy is between 0 but less or equal to the fermi energy and 0 when the energy is greater than the fermi energy.
01:00
Now, hence, the total number of particles is given by the following integral, that is the integral from zero to infinity of the product between the number of particles with the respect energy, well, n, which is the number of moles in this case, so times the number of particles, and this times the differential in energy.
01:31
So in this case, we will have that we are left in this case with the integral from zero to the fermi energy of the number of particles in function of the energy integrated over the energy.
01:50
So we know that in this case, the number of particles is going to be in function of the energy.
02:02
Of the volume and the energy.
02:06
So we can take out everything that is a constant that do not depend on the energy.
02:12
So we will have a times pi times the volume times two times the mass to the three and this elevated to one divided by two and this divided by plants constant to the three.
02:27
The integral from zero to the fermi energy of the square root of the energy integrated over the energy.
02:40
So we know that the integral of the square root of the energy is going to be the energy elevated to 3 divided by 2.
02:48
And a factor that comes from there, which is 2 divided by 3 plus 1, of course.
02:57
3 divided by 2 plus.
02:59
So that's the term that we need to divide this from that integral.
03:05
In this case for n is equal to 16 times pi times the volume times two times the mass to the three and that elevated to one divided by two and this divided by three times plums constant to the three times the fermi energy elevated to three divided by two and we obtained this after just evaluating this integral the result of that integral...