What is the probability of an electron state being filled if it is located at the fremi level?
The correct answer and explanation is:
Correct Answer: 1/2
The probability of an electron state being filled at the Fermi level is one-half or 0.5.
Explanation:
In quantum mechanics and solid-state physics, the Fermi-Dirac distribution function describes the probability that an electron state at a specific energy level is occupied by an electron. The equation for this distribution is: f(E)=1e(E−EF)/kT+1f(E) = \frac{1}{e^{(E – E_F)/kT} + 1}
Where:
- f(E)f(E) is the probability that a state at energy EE is occupied.
- EFE_F is the Fermi energy (also called the Fermi level).
- kk is the Boltzmann constant.
- TT is the absolute temperature in Kelvin.
At absolute zero temperature (T = 0 K):
- All energy states below the Fermi level are completely filled, f(E)=1f(E) = 1.
- All states above the Fermi level are completely empty, f(E)=0f(E) = 0.
- At the Fermi level itself, the function is undefined exactly at T = 0, but it is conventionally taken as a step function, changing from 1 to 0 at E=EFE = E_F.
At any temperature above absolute zero (T > 0 K):
- Thermal energy allows some electrons to occupy higher energy states.
- At the Fermi level (E=EFE = E_F), the equation simplifies to:
f(EF)=1e(EF−EF)/kT+1=1e0+1=12f(E_F) = \frac{1}{e^{(E_F – E_F)/kT} + 1} = \frac{1}{e^{0} + 1} = \frac{1}{2}
This means there is a 50% chance that the state at the Fermi level is occupied. This property is very important in semiconductor physics and explains why electrons can be thermally excited from the valence to the conduction band. The 50% probability reflects the fact that the state is at the boundary between filled and empty states in the electron energy distribution.