# Defining Expressions from the Equation of State

1. Dec 31, 2009

### TFM

1. The problem statement, all variables and given/known data

Starting from the expression for g(E) derived in the lectures, show that for a solid of volume V containing N free electrons which obey Fermi-Dirac Statistics, $$E_F, k_F, and v_F$$ can be expressed as:

$$E_F = \left( \frac{N}{V} \right)^{1/3}(3\pi^2)^{2/3}\left( \frac{(h/2\pi)^2}{2m} \right)$$

$$k_F = \left( \frac{N}{V} \right)^{1/3}(3\pi^2)^{1/3}$$

$$E_F = \left( \frac{N}{V} \right)^{1/3}(3\pi^2)^{1/3}\left( \frac{(h/2\pi)^2}{m} \right)$$

Calculate approximate values of these three parameters and of (N/V), $$T_F$$ and of $$\lambda_F$$ for Na metal. Compare the values of $$T_F, \lambda_F v_F$$ with Room Temperature, the interatomic spacing in Na and the velocity of light, respectively.

For Na: AW = 23, $$\rho \pprox$$ 1 gram.cm-3

Hint:

$$\int^?_?g(E)de = ?$$

2. Relevant equations

3. The attempt at a solution

Hi

Now I know that

$$g(E) = \left( \frac{V}{2\pi^2} \right)\left( \frac{2m}{\hbar^2} \right)^{3/2}E^{1/2}$$

Now I think that if I remember rightly,

$$\int^?_?g(E)de = ?$$

is the number of states?

2. Dec 31, 2009

### Mute

Many quantities you want to calculate in statistical mechanics are defined in terms of sums. For example, the total number of particles is

$$N = \sum_j \langle n_j\rangle;$$

in general the sum is hard to do, but our saving grace is that we often have enough states that we can approximate the sum as an integral. If we just straightforwardly converted the sum over indices j to an integral,

$$\sum_j \rightarrow \int_0^\infty dj$$
then we'd find that wasn't very useful since we'd have to express everything in the integrand as functions of j. So instead we want to do an integral over something like the energy, $\varepsilon$, but to do so we need this function $g(\varepsilon)$, called the density of states, to account for the fact that the E values are uniformly spaced out like the j values. So, what we get is then

$$\sum_j \rightarrow \int_0^\infty d\varepsilon~g(\varepsilon);$$
accordingly,

$$N = \sum_j \langle n_j \rangle \rightarrow \int_0^\infty d\varepsilon~g(\varepsilon) \langle n(\varepsilon) \rangle,$$

with things like the total energy, E, etc.

Note that one could also do this procedure for integrating over the momentum k:

$$\sum_j \rightarrow \int_0^\infty d^3k~f(\mathbf{k});$$

where $f(\mathbf{k}) = V/(2\pi)^3$.

Does this help?

3. Dec 31, 2009

### TFM

Yes that does make sense.

So for the first one, do I need to do the integral

$$\int^?_?g(E)de = ?$$

Between 0 and $$E_F$$,

Which should give me something like N = x

And then rearrange it for $$E_F$$?

4. Jan 1, 2010

### Mute

The total energy of the system is

$$E = \sum_j \varepsilon_j \langle n_j \rangle$$.

Convert that to an integral. You'll need to know $\langle n(\varepsilon)\rangle$ and how it behaves at low temperature.

5. Jan 2, 2010

### TFM

Well that integral would be:

$$E = \int_0^{\infty} d\varepsilon~g(\varepsilon) \epsilon \langle n(\varepsilon) \rangle,$$

$$\langle n(\varepsilon) \rangle$$

At low temperatures is thsu the part which goes to the graph like this:

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At low temperatures

6. Jan 2, 2010

### Mute

Sorry, I misread your original post - I thought you were looking for the total energy in terms of N and V, not just the fermi energy.

You are correct in that if you want the total number of particles N it is just

$$N = \sum_j \langle n_j \rangle \rightarrow \int_0^\infty d\varepsilon~g(\varepsilon)\langle n(\varepsilon) \rangle \simeq \int_0^{\varepsilon_F} d\varepsilon~g(\varepsilon)$$

This will give you N in terms of the fermi energy and V, so then just solve for the fermi energy.

You can then determine the fermi momentum in terms of the relation between energy and momentum.

Sorry for any confusion.

Last edited: Jan 2, 2010
7. Jan 2, 2010

### TFM

So I was right to assume to do the integral:

$$N = \int_0^{\varepsilon_F} d\varepsilon~g(\varepsilon)$$

And rearrange for $$E_F =$$

8. Jan 2, 2010

### Mute

Yes, but make sure that you understand the reason for that: You're approximating a sum as an integral, and you used the property of <n> at low temperatures to simplify the integral, cutting it off at the fermi energy. If T/T_F were not much less than one you would have to use the actual functional form of <n>, as it would not approximately be a step function.

9. Jan 3, 2010

### TFM

That makes sense, at low temperatures, the energy is approximately the step function where all the energy are below the Fermi energy, but as the temperature increases, it goes away from the step function, and more electrons have energy greater then the Fermi Energy.

Also, have I made a mistake somewhere in my algebra - I cannot see how you go from the 2pi^2 in the g(E) to 3pi^2:

$$N = \int^{E_F}_0 \frac{V}{2\pi^2}\frac{2m}{\hbar^2}^{3/2}E^{1/2}$$

$$N = \left( \frac{V}{2\pi^2}\frac{2m}{\hbar^2}^{3/2}E^{3/2} \right)^{E_F}_0$$

$$N = \left( \frac{V}{2\pi^2}\frac{2m}{\hbar^2}^{3/2}(E_F)^{3/2} - 0 \right)$$

rearrange for EF:

$$(E_F)^{3/2} = \frac{N}{ \left( \frac{V}{2\pi^2}\frac{2m}{\hbar^2}^{3/2}} \right)$$

Rearrange:

$$(E_F)^{3/2} = \frac{N}{ \left( \frac{V}{2\pi^2} \right)}\frac{2m}{\hbar^2}^{-3/2}$$

$$(E_F)^{3/2} = \frac{N*2\pi^2}{V}\left(\frac{\hbar^2}{2m}\right)^{3/2}}$$

$$(E_F)^{3/2} = \frac{N}{V}\left(2\pi^2\right)\left(\frac{\hbar^2}{2m}\right)^{3/2}}$$

10. Jan 3, 2010

### Mute

You forgot the factor of 2/3 when you integrated the E^(1/2).

$$\int dx~x^{1/2} = \frac{2}{3}x^{3/2} + C$$

11. Jan 4, 2010

### TFM

Yes. I realised that this morning

For the second expression,

$$k_F = \left( \frac{N}{V} \right)^{1/3}(3\pi^2)^{1/3}$$

I assume I need to use the expression for g(k)

$$g(k) = \frac{Vk^2}{2\pi^2}$$

into

$$N = \int^{k_F}_0 g(k)dk$$

I have done this, but it gives me :

$$k_F = \left( \frac{N}{V} \right)^{1/3}(6\pi^2)^{1/3}$$

Calculations:

$$N = \int^{k_F}_0 g(k) dk$$

$$N = \int^{k_F}_0 \frac{Vk^2}{2\pi^2} dk$$

$$N = \left[\frac{1}{3}\frac{Vk^3}{2\pi^2} \right]^{k_F}_0$$

$$N = \left[\frac{Vk^3}{6\pi^2} \right]^{k_F}_0$$

$$N = \left[\frac{Vk_F^3}{6\pi^2} - 0 \right]$$

$$N = \frac{Vk_F^3}{6\pi^2}$$

Then just rearranged to give:

$$k_F = \left( \frac{N}{V} \right)^{1/3}(6\pi^2)^{1/3}$$

Last edited: Jan 4, 2010
12. Jan 5, 2010

### Mute

You may need to include a factor of 2 in g(k) to account for spin degeneracy, unless the problem says to ignore this.

Also, note that you have free fermions - you can use the relation between energy and momentum to derive k_F from E_F.

13. Jan 5, 2010

### TFM

I have found the statement I need in my notes:

$$g(e)de = 2g(k) dk$$

For the third one, I now need to find:

$$v_F = \left( \frac{N}{V} \right)^{1/3}(3\pi^2)^{1/3}\left( \frac{(h/2\pi)^2}{m} \right)$$

So I am assuming I need to convert the Energy term into something with Velocity

Would I be correct in using the kinetic energy equation E = 1/2mv^2?

14. Jan 5, 2010

### Mute

You could do that, or you could use the simpler relation $p = \hbar k = mv$.

15. Jan 6, 2010

### TFM

Okay thanks.

I subsitituted it into the previous calculation here:

$$N = \frac{V}{3\pi^2}k_F^3$$

using:

$$k_F = \frac{mv_f}{\hbar}$$

$$k_F^3 = \frac{m^3v_f^3}{\hbar^3}$$

$$N = \frac{V}{3\pi^2}\frac{m^3v_f^3}{\hbar^3}$$

$$\frac{N}{V} 3\pi^2\frac{\hbar^3}{m^3} = v_f^3$$

then cube it gives me the right answer,

$$\left(\frac{N}{V}\right)^{1/3} \left(3\pi^2\right)^3\frac{\hbar}{m} = v_f$$

For the last part:

I am assuming we need to get the value for everything, but what is the best way to work out N/V, especially as we need that before we can work out E k and v.

As it says approximate, should I use the given Atomic Weight and density of Na to work out the amount of atoms in a metre cubed, as N is the atoms per volume, and V would be 1, so this would give me N/V?

Last edited: Jan 6, 2010