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Fundamentals Of Physics
Found in: Page 1275

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Short Answer

Show that P(E), the occupancy probability in Eq. 41-6, is symmetrical about the value of the Fermi energy; that is, show that P(EF+ΔE)+P(EF-ΔE)=1.

It is shown that P(EF+ΔE)+P(EF-ΔE)=1 that is the occupancy probability is symmetrical about the value of the Fermi energy.

See the step by step solution

Step by Step Solution

Step 1: Understanding the concept of occupancy probability

The probability of a state to be occupied by and electron is referred to as the occupancy probability. At 0 K temperature, the states below the Fermi level have occupancy probability equal to and for the states above the Fermi level, its value is .


The occupancy probability of the state with energy E is-

PE=1eE-EF/KT+1 ( i )

Here EF is the Fermi energy, k=8.62×10-5 eV/K and T is the absolute temperature.

Step 2: Calculation of the given symmetrical condition of probability

Upon expansion in view of equation (i), the LHS value of the given equation can be solved as follows:

LHS=PEF+E+PEF-E =1eEF+E-EF/KT+1+1eEF-E-EF/KT+1 =1eE/KT+1+1e-E//KT+1 =eE/KT+1+eE/KT+1eE/KT+1e-E/KT+1

On further solving,

L.H.S=eE/KT+1+eE/KT+2eE/KT+eE/KT+2 =1 =R.H.S

Hence, the given condition is proved and this implies the symmetrical condition for occupancy probability.

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