Question

For a solid metal having a Fermi energy of 8.480 eV , what is the probability, at room temperature, that a state having an energy of 8.540 eV is occupied by an electron?

Answer 1

Solution:

fermi level , EF = 8.48 eV

energy of state, E = 8.540 eV

room temperature , T = 25 degree C

T = 298 K

for the probability of finding electron in energfy level ,

f(E) = 1/(1 + e^((E -EF)/(k*T)))

f(E) = 1/(1 + e^((8.540 - 8.480) (1.602 10^-19)/(1.381 10^-23 298)))

f(E) = 0.0882

the probability that a state having an energy of 8.540 eV is occupied by an electron is 0.0882

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answered by: anonymous