The probability that a state which is 0.2eV above the fermi energy in a metal atom at 700K is?Correct answer is '0.036'. Can you explain this answer?

Question

Answer

The Probability of a State 0.2eV Above the Fermi Energy in a Metal Atom at 700K

When considering the probability of a state being occupied in a metal atom, there are several factors to take into account. In this case, we are interested in the probability of a state that is 0.2eV above the Fermi energy at a temperature of 700K.

Fermi Energy

The Fermi energy is a concept in solid-state physics that represents the highest energy level that is occupied by an electron at absolute zero temperature. It acts as a reference point for determining the energy levels of electrons in a metal.

Boltzmann Distribution

The probability of a state being occupied can be determined using the Boltzmann distribution. According to this distribution, the probability of a state with energy E being occupied at a temperature T is given by the equation:

P(E) = exp(-E/kT)

Where P(E) is the probability, E is the energy of the state, k is the Boltzmann constant (8.617333262145 × 10^-5 eV/K), and T is the temperature in Kelvin.

Calculating the Probability

In this case, we want to calculate the probability of a state that is 0.2eV above the Fermi energy at a temperature of 700K. Let's plug in the values into the Boltzmann distribution equation:

P(E) = exp(-0.2eV/(8.617333262145 × 10^-5 eV/K * 700K))

P(E) = exp(-0.2eV/6.0321332855015e-2 eV)

P(E) = exp(-3.317)

P(E) ≈ 0.036

Conclusion

The probability of a state that is 0.2eV above the Fermi energy in a metal atom at 700K is approximately 0.036. This probability is calculated using the Boltzmann distribution, which takes into account the energy of the state and the temperature.

Answer

E - E_f= 0.2 eV
k_B = 8.62 x 10^-5eVK^-1
T = 700K

P = 0.036
The correct answer is: 0.036

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