How Does Pressure Affect Volta Potential in Identical Metal Samples?

[Gregg]

1.

I need to find the volta potential between two samples of the same metal.

2.

The electron density is given, $\rho$, and each sample are subject to different pressures (1 atm and 100 atm), and it has bulk modulus $K=1 \times 10^{11} \text{N m}^{-2}$

Is the density given assumed to be the density at 1 atm?

3.

I know that the Fermi energy is

$\epsilon_F = \frac{\hbar^2}{2 m_e}\left(\frac{3\pi ^2N}{V}\right)^{\frac{2}{3}}$

$N/V = \rho$ .

$K = - V \frac{d P}{dV}$

So to find the difference in Fermi energies $\rho_1$ is given at 1 atm, and then you can find the other one somehow? Do you assume that the samples are identical and find a new electron density? How do I find the Volta potential once I have the difference in Fermi energies?

 

[fzero]

$N$ is fixed, so you don't really need to solve for the new density. But you have enough information to do so if you want (using the change in volume).

 

[Gregg]

Do you think that is the way you solve this?

 

[fzero]

Do you think that is the way you solve this?

I do. The potential energy difference for an electron is the difference between the Fermi energies. The electrostatic potential is related to that by dividing by the charge.

By saying that you don't have to solve for the density, I don't want to discourage you from doing it that way. There's a few ways to arrange the formulas, but they all boil down to computing the change in volume of the sample under pressure.

 

[Gregg]

I'm unsure how to compute the volume change.

$K dV = -V dP$

But $V=V(P)$ so I don't think it's that. It is probably quite simple. Could it be:

$\Delta V = -\frac{V \Delta P}{K}$?

If so I have

$\Delta V = -99 \frac{V_0}{10^{11}} \times (1.0133 \times 10^5 \text{ N m}^{-2}) = V_0 1.003\times 10^{-4}$So the other $\rho_2$ is $N/V_2 = 1.001 \rho_1$$\Delta \epsilon_F = \frac{\hbar^2}{2m_e} \left[ ( 3 \pi^2 \rho_1) ^{2/3}-( 3 \pi^2 \rho_2) ^{2/3} \right]$

Now I've found the change in Fermi energy, the potential difference for the electron is going to be this. To get the Volta potential divide by the charge?

 

[fzero]

I'm unsure how to compute the volume change.

$K dV = -V dP$

But $V=V(P)$ so I don't think it's that. It is probably quite simple.

I would say that since $dP/dV$ is specified, we should treat $V$ as the independent variable and $P=P(V)$ is the dependent variable.

Could it be:

$\Delta V = -\frac{V \Delta P}{K}$?

If so I have

$\Delta V = -99 \frac{V_0}{10^{11}} \times (1.0133 \times 10^5 \text{ N m}^{-2}) = V_0 1.003\times 10^{-4}$

The problem here is that $\Delta P/P$ is large, so we should really do an integral to find the result.

So the other $\rho_2$ is $N/V_2 = 1.001 \rho_1$


$\Delta \epsilon_F = \frac{\hbar^2}{2m_e} \left[ ( 3 \pi^2 \rho_1) ^{2/3}-( 3 \pi^2 \rho_2) ^{2/3} \right]$

Now I've found the change in Fermi energy, the potential difference for the electron is going to be this. To get the Volta potential divide by the charge?

Electric potential and electric potential energy are related by the charge of whatever charge carrier is involved.

 

[Gregg]

By doing an integral I end up with $-K \Delta V = V\Delta P$ as well don't I? I get

$U = -5.3570 \times 10^{-23}$I remember now that $U = qV = -eV$ So, I get $V = 0.3343 \text{ mV}$Is this reasonable?

 

[fzero]

By doing an integral I end up with $-K \Delta V = V\Delta P$ as well don't I?

No, the integral gives a log function.

I get

$U = -\text{5.356978328904951\grave{ }*{}^{\wedge}-23}$


I remember now that $U = qV = -eV$


So, I get $V = 1.035 \text{ mV}$


Is this reasonable?

I don't know what the answer is supposed to be. I'd guess it should be [STRIKE]within[/STRIKE] a few percent of the Fermi energy, but I haven't tried to work it out.

 

[Gregg]

Oh, how did I miss that... do you mean:

$\int_{V_1}^{V_2} \frac{dV}{V} = -\int_{P_1}^{P_2} \frac{dP}{K}$

$V_2 = V_1 \exp{-\Delta P \over K}$

So that,

$\rho_2 = \rho_1 \exp{\Delta P \over K}$

Using

$V = \frac{1}{-e}\frac{\hbar^2}{2 m_e}( (2 \pi^2)^{2/3} (\rho_1^{2/3}-\rho_2^{2/3}))$

And here I end up with $0.5015 \text{ mV}$

 

[fzero]

It looks ok to me.