Question

The Fermi energy for silver is $ E = 5.5eV $ . At $ T = {0^o}C $ , what are the probabilities that states with the following energies are occupied: (a) $ E = 4.4eV $ , (b) $ E = 5.4eV $ , (c) $ E = 5.5eV $ , (d) $ E = 5.6eV $ , and (e) $ 6.4eV $ ? (f) At what temperature is the probability $ 0.16 $ that a state with energy $ E = 5.6eV $ is occupied?

Answer 1

Hint: Fermi energy is a concept in quantum mechanics which usually refers to the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero temperature.

Complete answer:
The formula for finding the occupancy probability is
 $ P(E) = \dfrac{1}{{{e^{^{(E - {E_F})/kT}}} + 1}}.......(1) $
And we also know that at $ T = {0^o}C $ , $ kT = 0.02353eV $
Also, $ {E_F} = 5.5eV $
(a) In this part we need to find the probability that the state with $ E = 4.4eV $ is occupied,
 $ P(E) = \dfrac{1}{{{e^{^{(E - {E_F})/kT}}} + 1}} $
On putting the values, we get,
 $ P(4.4eV) = \dfrac{1}{{{e^{^{(4.4 - 5.5)/0.02353}}} + 1}} $
On further solving, we get,
 $ P(4.4eV) = 1 $
(b) ) In this part we need to find the probability that the state with $ E = 5.4eV $ is occupied,
 $ P(5.4eV) = \dfrac{1}{{{e^{^{(5.4 - 5.5)/0.02353}}} + 1}} $
On further solving, we get,
 $ P(5.4eV) = 0.9859 $
(c) In this part we need to find the probability that the state with $ E = 5.5eV $ is occupied,
 $ P(5.5eV) = \dfrac{1}{{{e^{^{(5.5 - 5.5)/0.02353}}} + 1}} $
On further solving, we get,
 $ P(5.5eV) = 0.5 $
(d) In this part we need to find the probability that the state with $ E = 5.6eV $ is occupied,
 $ P(5.6eV) = \dfrac{1}{{{e^{^{(5.6 - 5.5)/0.02353}}} + 1}} $
On further solving, we get,
 $ P(5.6eV) = 0.0141 $
(e) In this part we need to find the probability that the state with $ E = 6.6eV $ is occupied,
 $ P(6.6eV) = \dfrac{1}{{{e^{^{(6.6 - 5.5)/0.02353}}} + 1}} $
On further solving, we get,
 $ P(6.6eV) = 2.447 \times {10^{ - 17}} $
(f) To find $ T $ , we need to solve equation (1)
 $ P({e^{\Delta E/kT}} + 1) = 1 $
 $ {e^{\Delta E/kT}} = \dfrac{1}{P} - 1 $ $ T = \dfrac{{\Delta E}}{{k\ln \left( {\dfrac{1}{P} - 1} \right)}} $
 $ \dfrac{{\Delta E}}{{kT}} = \ln \left( {\dfrac{1}{P} - 1} \right) $
 $ T = \dfrac{{\Delta E}}{{k\ln \left( {\dfrac{1}{P} - 1} \right)}} $
so, here we need to find the temperature of whose probability is $ 0.16 $ , that is $ \Delta E = 0.1eV $ , substitute to
 $ T = \dfrac{{\Delta E}}{{k\ln \left( {\dfrac{1}{P} - 1} \right)}} $
On putting the required values, we get,
 $ T = \dfrac{{0.1eV}}{{(8.62 \times {{10}^{ - 5}}eV/K)\ln \left( {\dfrac{1}{{1.6}} - 1} \right)}} $
 $ T = 699.6K $
So, the final answers are:
(a) $ P(4.4eV) = 1 $
(b) $ P(5.4eV) = 0.9859 $
(c) $ P(5.5eV) = 0.5 $
(d) $ P(5.6eV) = 0.0141 $
(e) $ P(6.6eV) = 2.447 \times {10^{ - 17}} $
(f) $ T = 699.6K $

Note:
Fermi energy has many applications in the day to day life. It is used in semiconductors and insulators. It is also used to describe insulators, conductors and semiconductors. It is also used to understand the concept of white dwarfs.