Finding the Volta Potential: A Step-by-Step Guide

  • Thread starter Gregg
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In summary, the difference in Volta potentials between two samples of the same metal can be found by first computing the change in Fermi energies using the given equations and then dividing by the charge of the charge carrier involved. This can be done by finding the change in volume of the samples under pressure using the bulk modulus and then using this to find the change in electron density. Finally, the change in potential energy can be found and divided by the charge to find the Volta potential difference between the two samples.
  • #1
Gregg
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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?
 
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  • #2
[itex]N[/itex] 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).
 
  • #3
Do you think that is the way you solve this?
 
  • #4
Gregg said:
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.
 
  • #5
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?
 
  • #6
Gregg said:
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 [itex]dP/dV[/itex] is specified, we should treat [itex]V[/itex] as the independent variable and [itex]P=P(V)[/itex] 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 [itex]\Delta P/P[/itex] 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.
 
  • #7
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?
 
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  • #8
Gregg said:
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.
 
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  • #9
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} ##
 
  • #10
It looks ok to me.
 

What is the Volta Potential?

The Volta Potential is the difference in electric potential between two points in a material or system. It is often used to measure the work function of a material, which is a measure of the energy needed to remove an electron from the material's surface.

Why is it important to find the Volta Potential?

Finding the Volta Potential can provide valuable information about a material's properties, such as its conductivity and work function. It can also help in understanding the behavior of electronic devices and their interfaces.

What are the steps to finding the Volta Potential?

The steps to finding the Volta Potential include: 1) Preparing the sample and measuring its surface topography, 2) Measuring the sample's current-voltage (IV) characteristics, 3) Measuring the IV characteristics of a reference electrode, 4) Calculating the difference in potential between the sample and reference electrode, and 5) Adjusting for any offset in potential due to the reference electrode.

What equipment is needed to find the Volta Potential?

To find the Volta Potential, you will need a scanning probe microscope (SPM) for surface topography measurements, an electrometer for measuring the IV characteristics, a reference electrode, and a computer for data analysis.

Are there any limitations to finding the Volta Potential?

Yes, there are some limitations to finding the Volta Potential. For example, the accuracy of the measurement can be affected by environmental factors such as temperature and humidity. Additionally, the method may not work for all types of materials, and calibration of the equipment is crucial for accurate results.

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