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

[bayan]

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

 

[DrClaude]

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).