Pure germanium has a band gap of 0.67 EV. The Fermi energy is in the middle of the gap. For temperatures of 250K, 300K and 350K, calculate the probability f(E) that a state at the bottom of the conduction band is occupied. b) For each temperature in part a), calculate the probability that a state at the top of the valence band is empty.

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To calculate the probability that a state at the top of the valence band is empty, we can use the Fermi-Dirac distribution function and the fact that the probability that a state is empty is just 1 minus the probability that it is occupied.

This means that:

At 250K:

f(E) = 1 / (e^((E - 0.335 eV)/(8.6*10^-5 eV/K * 250K)) + 1)

probability that a state at the top of the valence band is empty = 1 - f(E)

At 300K:

f(E) = 1 / (e^((E - 0.335 eV)/(8.6*10^-5 eV/K * 300K)) + 1)

probability that a state at the top of the valence band is empty = 1 - f(E)

At 350K:

f(E) = 1 / (e^((E - 0.335 eV)/(8.6*10^-5 eV/K * 350K)) + 1)

probability that a state at the top of the valence band is empty = 1 - f(E)

Thank you for answering my question Professor Smith.

I have one more question for you if you don't mind.

b) For each temperature in part (a) calculate the probability that a state at the top of the valence band is empty.

At temperatures of 250K, 300K, and 350K, the probability f(E) that a state at the bottom of the conduction band is occupied can be calculated using the Fermi-Dirac distribution function:

f(E) = 1 / (e^((E - Ef)/(kT)) + 1)

where E is the energy of the state, Ef is the Fermi energy, k is the Boltzmann constant, and T is the temperature in Kelvin. Since the Fermi energy is in the middle of the band gap, Ef = 0.67 eV / 2 = 0.335 eV.

At 250K:

f(E) = 1 / (e^((E - 0.335 eV)/(8.6*10^-5 eV/K * 250K)) + 1)

At 300K:

f(E) = 1 / (e^((E - 0.335 eV)/(8.6*10^-5 eV/K * 300K)) + 1)

At 350K:

f(E) = 1 / (e^((E - 0.335 eV)/(8.6*10^-5 eV/K * 350K)) + 1)