Probability of a state above the Fermi energy in a metal atom at 700K

Introduction:
In a metal, the Fermi energy represents the highest energy level occupied by electrons at absolute zero temperature. At finite temperatures, there is a probability that some states above the Fermi energy can be occupied by electrons. The probability of a state being occupied above the Fermi energy can be calculated using statistical mechanics.

Given data:
Energy above the Fermi energy = 0.2 eV
Temperature = 700 K

Explanation:

Step 1: Calculate the probability using the Fermi-Dirac distribution function:

The probability of a state being occupied by an electron at a given energy level is given by the Fermi-Dirac distribution function:

P(E) = 1 / (1 + exp((E - EF) / kT))

where P(E) is the probability, E is the energy level, EF is the Fermi energy, k is the Boltzmann constant, and T is the temperature.

Step 2: Convert energy from eV to Joules:

To use the Boltzmann constant and temperature in SI units, we need to convert the energy from electron volts (eV) to joules (J):

1 eV = 1.602 × 10^-19 J

So, the energy above the Fermi energy becomes:

E = 0.2 eV * 1.602 × 10^-19 J/eV = 3.204 × 10^-20 J

Step 3: Calculate the probability:

Plugging in the values into the Fermi-Dirac distribution function:

P(E) = 1 / (1 + exp((3.204 × 10^-20 J - EF) / (k * 700 K)))

Simplifying further, we get:

P(E) = 1 / (1 + exp((3.204 × 10^-20 J - EF) / (1.38 × 10^-23 J/K * 700 K)))

P(E) = 1 / (1 + exp((3.204 × 10^-20 J - EF) / 9.66 × 10^-21 J))

Step 4: Calculate the probability for the given energy:

Given that the probability P(E) = 0.035, we can rearrange the equation to solve for EF:

0.035 = 1 / (1 + exp((3.204 × 10^-20 J - EF) / 9.66 × 10^-21 J))

Simplifying further:

0.035(1 + exp((3.204 × 10^-20 J - EF) / 9.66 × 10^-21 J)) = 1

1 + exp((3.204 × 10^-20 J - EF) / 9.66 × 10^-21 J) = 1 / 0.035

exp((3.204 × 10^-20 J - EF) / 9.66 × 10^-21 J) = 1 / 0.035 - 1

exp((3.204 × 10^-20 J - EF) / 9.66 × 10^-21 J) = 26.