How Is the Depth of the Potential Well Calculated for Free Electrons in Gold?

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Homework Help Overview

The discussion centers around calculating the depth of a potential well for free electrons in gold, specifically using the work function and Fermi energy. The original poster is attempting to find the depth of the well given the work function of 4.8 eV.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster assumes that the depth of the well is the sum of the Fermi energy and the work function but encounters difficulties with the calculations. They raise questions about the correct interpretation of variables and units, particularly regarding the calculation of electron density.

Discussion Status

Participants are actively engaging in checking the calculations and assumptions made by the original poster. Some have pointed out potential errors in the calculation of electron density and have requested to see the work shown for clarity. There is a collaborative effort to identify mistakes and clarify the approach without reaching a definitive conclusion.

Contextual Notes

There is a mention of the need to use appropriate units for mass in the calculations, indicating that unit conversion may be a point of confusion. The original poster's expectation of the answer being on the order of 1 eV is also noted, highlighting a specific target for their calculations.

Dawei
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Homework Statement



For free electrons in a metal, the depth of a potential well can be determined by observing that the work function is the energy required to remove an electron at the top of the occupied states from the metal; an electron in this state has the Fermi energy.

Assuming each atom provides one free electron to the gas, find the depth of the well for a free electron in gold (work function = 4.8 eV)

Homework Equations



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The Attempt at a Solution



I'm assuming that the answer will be just the fermi energy plus the work function, but I can't seem to get the right answer when plugging everything in.

For N/V, I used (Avogadro # x Density of gold) / (molar mass of gold) , and got 3.89 x 10^9 electrons / cubic meter

The answer should be on the order of 1 ev, but I can't seem to get this. The m here refers to the mass of a single electron, correct? But using that, I just don't get the right orders of magnitude. (I get about 5 x 10^6).
 
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That isn't nearly enough electrons in a cubic meter. You made a mistake in calculating N/V. Please show all of your work so we can see what you did wrong.
 
Yeah you're right. 10^9 was what I got when I took number of electrons / m^3 and raised it to the 1/3 power, to get N/L. I think I'm supposed to use the 1 dimensional equation.

edit: yeah I was just being dumb, I had it right, was just using the wrong units for mass. Thanks though.
 
Last edited:
You have the right equation. Just show your work on solving N/V so we can check it. Remember to show all units.
 

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