Archived CMP. Probability of finding electron with E > fermi energy.

AI Thread Summary
The discussion centers on calculating the probability of finding an electron with energy between 5 eV and 5.5 eV at a temperature of 300 K, given a Fermi energy of 4.2 eV. The probability formula used incorporates the Fermi-Dirac distribution, and the calculated probability is approximately 7.07 x 10^-13%. A subsequent integration of the probability distribution yields an even smaller value of about 3.67 x 10^-16. Despite the low probabilities, the large number of conduction electrons in a metal can still result in a significant quantity of electrons at these energy levels. The calculations highlight the rarity of finding "loose" electrons at room temperature.
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Homework Statement


what is the probability of finding an electron with energy between 5eV and 5.5eV at T=300k, given that the fermi energy of the metal is 4.2 eV.


Homework Equations



##P(E,T)dE= \frac{3}{2} E_F^\frac {-3}{2} \frac{E^\frac{1}{2}}{e^\frac{E-E_F}{K_B T}+1}dE##


The Attempt at a Solution



I have got an answer and just wanted check and see if it is correct. I have used the eV value for ##K_B##

##P(5,300)0.5= \frac{3}{2}4.2^\frac {-3}{2} \frac{5^\frac{1}{2}}{e^\frac{5-4.2}{K_B 300}+1}0.5##


##P(5,300)0.5≈ 7.07*10^-13 % ##

Have I used the right method for solving this? the answer kind of makes sense as it suggests there is a very low probability of having **loose** electrons at 300K
 
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Probability distributions are made to be integrated:
$$
\frac{3}{2} E_F^{-3/2} \int_{5\ \mathrm{eV}}^{5.5\ \mathrm{eV}} \frac{E^{1/2}}{\exp[(E -E_F)/k_B T]} dE \approx 3.67 \times 10^{-16}
$$
for ##E_F = 4.2\ \mathrm{eV}## and ##T = 300\ \mathrm{K}##.

While this number may be small, the total number of conduction electrons can be huge (~Avogadro number).
 
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