Fermi energy question

1. Sep 28, 2015

orangeincup

1. The problem statement, all variables and given/known data
Part 1) Use the fermi dirac probability function for t=150k, t=300k, and t=600k to fill in the table below.

Part 2) Also show a sample calculation for (e-ef)=0.06eV and T=300k.

Part 3)(Same as part 2?) Calculate the probabilities of a state at E -EF =0.06 eV being empty for T =150 K , T = 300 K , and T = 600 K .

2. Relevant equations
F=1/((e^((E-Ef)/T)+1)
3. The attempt at a solution
I'm just learning about this topic now so bare with me.
So for E-Ef =(-0.15) in the first row, and T=150k..

F=1/((e^((-0.15+1)/150)+1)*100
=49.86% for (-.150) at 150K

1/((e^((-0.15+1)/300)+1) * 100
=49.93% for (-.150) at 300K

1/((e^(-0.15+1)/600)+1)*100
=49.96% for (-.150) at 600k

That is the three values for row 1, -.150 E-Ef

Repeat with the rest of the values using the same formula, switching the (-.150) for the appropriate value in the chart.

For part 2, I'm a bit lost. So for a sample calculation, would it be 1/((e^((-0.06+1)/300)+1)? That comes out to be 49.99%.

Part 3) The way I'm reading it, it's asking for basically the same as part 2, except it wants when the state is empty and not filled? So would it be 100%- the probability of an electron being inside? Here's my calculation for it using that logic:

1/((e^((-0.06+1)/150)+1) * 100 = 49.84%
100%-49.84%=50.2% the state is empty

Attached Files:

• ttable.png
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2. Sep 28, 2015

Staff: Mentor

Why do you add 1 in the exponents?
I think the Boltzmann constant is missing there.

For energies below the Fermi energy, the states should be more than 50% occupied.

Part 2: Just write down the formulas you used for part 1 I think. Pay attention to the sign of the energy difference.

Part 3:
That's the correct approach, but the number has to get fixed.

3. Sep 28, 2015

orangeincup

I redid all my calculations using the Boltzmann constant

1/((e^((-0.15)/(150*(8.62*10^-5)))+1)
=99.99%

1/((e^((-0.15)/(150*(8.62*10^-5)))+1)
=99.7%
1/((e^((-0.15)/(600*(8.62*10^-5)))+1)
=94.79%

Part 2...
1/((e^((0.06)/(300*(8.62*10^-5)))+1) = 0.089
=8.90%?

Part 3...

1/((e^((0.06)/(150*(8.62*10^-5)))+1) = 0.00956
=0.956%? 100%-0.956%=99.04%

1/((e^((0.06)/(300*(8.62*10^-5)))+1) = .0895
100%-8.95% = 91.05%

1/((e^((0.06)/(600*(8.62*10^-5)))+1)=.2387
100%-23.87%=76.13%

Last edited: Sep 28, 2015
4. Sep 28, 2015

Staff: Mentor

Those numbers look much more realistic.

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