material at T = 300 K is 5.50 eV. The electrons in this material follow the Fermi-Dirac distribution function. a) Find the probability of an energy level at 5.50 eV being occupied by an electron. b) Repeat part (a) if the temperature is increased to T = 600 K. (Assume that EF is a constant.). c) Calculate the energy level where probability of finding an electron at room temperature is 70%. d) Calculate

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Question 2: The Fermi energy level for a particular material at T = 300 K is 5.50 eV. The electrons in this material follow the Fermi-Dirac distribution function. a) Find the probability of an energy level at 5.50 eV being occupied by an electron. b) Repeat part (a) if the temperature is increased to T = 600 K. (Assume that EF is a constant.). c) Calculate the energy level where probability of finding an electron at room temperature is 70%. d) Calculate the temperature at which there is a 7 percent probability that a state 0.4 eV below the Fermi level will be empty of an electron. Please use formula below
Physical Constants
h = 1.055 x 10-34 J.s
mo = 9.109 x 10-³1 Kg
K = 1.388 x 10-23 J/K
q = 1.602 x 10-1⁹ C
Density of States
9c(E)=
Fermi Function
Formula Sheet
mn√2m (EEc)
π² x ħ³
mn
mp
Ks
f(E)=
Semiconductor Parameters
Ge
0.56mo
0.4mo
16
&o=8.854 x 10-1² F/m
KT|T=300K = 0.026 eV
Es = Ks X Eo
1
9v (E)=
E-EE
1+eKT
GaAs
0.067mo
0.5mo
12.9
mp√ √2mp(Ev - E)
π² x ħ³
Si
1.18mo
0.8mo
11.9
Transcribed Image Text:Physical Constants h = 1.055 x 10-34 J.s mo = 9.109 x 10-³1 Kg K = 1.388 x 10-23 J/K q = 1.602 x 10-1⁹ C Density of States 9c(E)= Fermi Function Formula Sheet mn√2m (EEc) π² x ħ³ mn mp Ks f(E)= Semiconductor Parameters Ge 0.56mo 0.4mo 16 &o=8.854 x 10-1² F/m KT|T=300K = 0.026 eV Es = Ks X Eo 1 9v (E)= E-EE 1+eKT GaAs 0.067mo 0.5mo 12.9 mp√ √2mp(Ev - E) π² x ħ³ Si 1.18mo 0.8mo 11.9
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