Fermi energy of simple 3D gas

[erogard]

Homework Statement

Consider a 3D gas of N non-interacting fermions in a volume V at temperature T << E_f / k.
Suppose that the particles in the energy range [0.25 Ef, 0.5 Ef] are suddenly removed.
Calculate the Fermi energy of the remaining particles after the system reaches its new thermal equilibrium, and the energy releases to its surrounding.

Homework Equations

So correct me if I'm wrong, but here the occupation number should be given by
$$<n_\epsilon> = \frac{1}{\exp[(\epsilon-\mu)kT]+1}$$
Now in the limit of low T (which I suppose is what is implied by the given inequality), this should be 1 for energies less than the Fermi energy, and 0 otherwise.

So, to compute the energy corresponding to the removed particles, should I integrate this equation times epsilon over that interval? ANd knowing what the new N is. how do I then compute the adjusted Fermi energy?

Any hint/explanation would be more than welcome. Thanks!

EDIT: Ok, so I think I got both the released energy and the new number of particles by integrating with the appropriate DoS function. Now wondering how to get the new Ef.

 

[Simon Bridge]

Well done, since you have the new number of particles, you can get the new fermi-energy by applying the definition of "fermi energy". So what is that definition?

 

[erogard]

Well done, since you have the new number of particles, you can get the new fermi-energy by applying the definition of "fermi energy". So what is that definition?

So using the new N can I directly apply the definition
$$E_f = (\frac{3}{8 \pi})^{2/3} \frac{h^2}{2m} (\frac{N}{V})^{2/3}$$
?

assuming the particles rearrange themselves in the expectated way, i.e. the one left with energies higher than E_max will loose energy to refill those lower empty states.

However, would I also need to include this "transition" energy into the energy released to the external environment?

 

[Simon Bridge]

assuming the particles rearrange themselves in the expectated way, i.e. the one left with energies higher than E_max will loose energy to refill those lower empty states.

You can see if this is a good assumption by checking the details of the problem.
Given that kT<<E_f and that a bunch of states E<E_f have suddenly been vacated, and that the system reaches a new equilibrium...

However, would I also need to include this "transition" energy into the energy released to the external environment?

You are asked for the energy released into the external environment.
Is the transition energy released into the external environment?
(Can it go anywhere else and still satisfy the assumptions?)

What other source of energy were you considering?

 

[erogard]

You can see if this is a good assumption by checking the details of the problem.
Given that kT<<E_f and that a bunch of states E<E_f have suddenly been vacated, and that the system reaches a new equilibrium...


You are asked for the energy released into the external environment.
Is the transition energy released into the external environment?
(Can it go anywhere else and still satisfy the assumptions?)

What other source of energy were you considering?

Originally thought of including the energy of the removed particles only, but yes I would assume that the transitional energies are to be included as well (kind of like when an electron drops to a lower orbit I guess)

 

[Simon Bridge]

Is there a relationship between the energies of the removed particles and the transition energy?

 

[unscientific]

Hint: Calculate the number of particles removed.

Now with a lower number of particles,what is the new fermi energy? (Look at the formula for fermi energy)

What happens to the difference in fermi energy?

 

[erogard]

Hint: Calculate the number of particles removed.

That would be
$$N' = N - C \int_{\epsilon_f / 4}^{\epsilon_f /2} \frac{\epsilon^{1/2} d\epsilon}{\exp[(\epsilon -\epsilon_f)/kT] +1}$$
correct? where C is a constant arising from the DoS. Don't know how to evaluate that integral though

Now with a lower number of particles,what is the new fermi energy? (Look at the formula for fermi energy)

Then that's just
$$\epsilon_f' = (\frac{3}{8 \pi})^{2/3} \frac{h^2}{2m} (\frac{N'}{V})^{2/3}$$

What happens to the difference in fermi energy?

Would that be the released energy, then? i.e. e_f-e_f' ? Or, rather, the difference in energy is obtained by integrating E all the way up to the initial fermi energy, minus E integrated up to the new fermi energy. COrrect me if I'm wrong.

 

[unscientific]

That would be
$$N' = N - C \int_{\epsilon_f / 4}^{\epsilon_f /2} \frac{\epsilon^{1/2} d\epsilon}{\exp[(\epsilon -\epsilon_f)/kT] +1}$$
correct? where C is a constant arising from the DoS. Don't know how to evaluate that integral though
Then that's just
$$\epsilon_f' = (\frac{3}{8 \pi})^{2/3} \frac{h^2}{2m} (\frac{N'}{V})^{2/3}$$Would that be the released energy, then? i.e. e_f-e_f' ? Or, rather, the difference in energy is obtained by integrating E all the way up to the initial fermi energy, minus E integrated up to the new fermi energy. COrrect me if I'm wrong.

The occupation fraction $\langle n \rangle = 1$ at $T=0$. Recall the heavyside function. Do it the usual way you would to find the fermi energy in terms of N, but now you have to change the limits of integration.

Yes, the energy difference is emitted to the surroundings, which occurs when you pull electrons out of their energy levels. The bound electrons now become free, releasing energy.

 

[erogard]

The occupation fraction $\langle n \rangle = 1$ at $T=0$. Recall the heavyside function. Do it the usual way you would to find the fermi energy in terms of N, but now you have to change the limits of integration.

A bit confused by this statement. Are you saying that what I wrote for obtaining the remaining number of particles, N', is inccorect?

And why would the given expresssion for the new Fermi energy be any different? Aren't we still using
$$N' = \int_0^{\epsilon_f'} D(\epsilon) d\epsilon$$
and solving for e_f '?

 

[unscientific]

A bit confused by this statement. Are you saying that what I wrote for obtaining the remaining number of particles, N', is inccorect?

And why would the given expresssion for the new Fermi energy be any different? Aren't we still using
$$N' = \int_0^{\epsilon_f'} D(\epsilon) d\epsilon$$
and solving for e_f '?

The expression for the fermi energy is still the same, but the value is changed, because N -> N' .$N' = \int_0^{\epsilon_f'} D(\epsilon) d\epsilon$ will solve for the new fermi energy in terms of N'. Simply replace the N by N' in your old expression for $\epsilon_F$.

Now we have to find N'.

Your expression is nearly correct:

$$<n_\epsilon> = \frac{1}{\exp[(\epsilon-\mu)kT]+1}$$

Consider these two cases. At $T = 0$, what is the value of $\langle n_\epsilon \rangle$ when $\epsilon > \epsilon_F = \mu$ and when $\epsilon < \epsilon_F = \mu$ ?

 

[erogard]

Consider these two cases. At $T = 0$, what is the value of $\langle n_\epsilon \rangle$ when $\epsilon > \epsilon_F = \mu$ and when $\epsilon < \epsilon_F = \mu$ ?

Well, as you pointed out, for very small T we are pretty much dealing with a step function, so it should be identically 1 for energies below the fermi energy, and 0 above it since the higher states would now be unoccupied. So are you saying I should just use n_e = 1 when I integrate the DoS from ef/2 to ef/2?

 

[unscientific]

Well, as you pointed out, for very small T we are pretty much dealing with a step function, so it should be identically 1 for energies below the fermi energy, and 0 above it since the higher states would now be unoccupied. So are you saying I should just use n_e = 1 when I integrate the DoS from ef/2 to ef/2?

That is right. Recall the definition of the fermi level: "Highest occupied energy level at T = 0. This means any energy less than fermi energy is definitely occupied, any energy higher than fermi energy is not occupied.