Find the temperature at which there is 1% probability that a state with energy 2ev is occupied given that Fermi energy is 1.5eV
To find the temperature at which there is a 1% probability that a state with energy 2eV is occupied, we need to use the Fermi-Dirac distribution function.
The Fermi-Dirac distribution function is given by:
f(E) = 1 / (1 + exp((E - E_f) / kT))
where:
- f(E) is the probability that a state with energy E is occupied
- E_f is the Fermi energy
- k is the Boltzmann constant
- T is the temperature
In this case, we have E = 2eV and f(E) = 0.01 (1% probability). We are given that E_f = 1.5eV.
Plugging in the values, we get:
0.01 = 1 / (1 + exp((2eV - 1.5eV) / kT))
Simplifying further:
0.01(1 + exp(0.5eV / kT)) = 1
Expanding the expression:
0.01 + 0.01exp(0.5eV / kT) = 1
0.01exp(0.5eV / kT) = 1 - 0.01
0.01exp(0.5eV / kT) = 0.99
Taking the natural logarithm of both sides:
ln(0.01exp(0.5eV / kT)) = ln(0.99)
0.5eV / kT = ln(0.99)
Solving for T:
T = 0.5eV / (k * ln(0.99))
The value of k is approximately 8.617333262145e-5eV/K (Boltzmann constant).
Plugging in the values, we can find the temperature T.