Question

Find the temperature at which there is 1%

probability that a state with energy 0.5 eV above Fermi energy can be occupied.​

Answer 1

Answer:

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Explanation:

Probability, f(E)= 1% = 1/100

E – EF = 0.5 eV

T = ?

Substituting the values, we get:

Taking ln on both sides, we get:

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Answer 2

To find the temperature

Given:

energy : 0.5 eV

Explanation:

Probability, f(E)= 1% = 1/100

E – EF = 0.5 eV

T = ?

F(E) = 1/ (1 + exp(E - E(F))/K(b)T

K(B) =1.381 × 10⁻²³J/K

       =1.381 × 10⁻²³ × 6.24 × 10¹⁸ eV/K

Substituting the value we get:

1/100 = 1/ 1 + exp[0.5/(861.744 × 10⁻5T)]

  100 = 1 + exp[5801.87/T]

  100 ≈ exp[ 5801.87/T]

Taking In on both sides, we get:

 Ln 100 = 5801.87/ T

          T = 5801.87 / 4.605 = 1259.98 K

Answer = 1259.98 K