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Fermi Energy for Silver is 5.5 eV. Find out the energy for which the probability of occupancy at 300K is 0.9.
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Given data:

  • T=300K
  • k=1.38×1023J/K
  • EF=5.5eV
  • f(E)=0.9

     

To find: E = ?

 

Solution:

kT=1.38×1023×3001.6×1019

kT=0.0259

f(E)=11+e(EEf)kT

0.9=11+e(E5.5)0.0259

1+e(E5.5)0.0259=10.9

1+e(E5.5)0.0259=1.1111

e(E5.5)0.0259=1.11111

e(E5.5)0.0259=0.1111

Taking log on both sides,

$\log {e^{(E-5.5)\over …

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