 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2015/paper-2/section-a/4/solution

Count two electron spin states in each reciprocal-space cell of volume

(2 π)^{3} / V

n = \frac{2}{( 2 π ) ^{3}} \frac{4 π k _{F}^{3}}{3}, k_{F} = (3 π^{2} n)^{1/3} = 1.360 \times 1 0^{10} m^{- 1} .

E_{F} = \frac{ℏ ^{2}}{2 m _{e}} (3 π^{2} n)^{2/3} = 1.129 \times 1 0^{- 18} J ≃ 7.05 eV .

The energy zero is the bottom of the model's conduction band. The supplied carrier density already incorporates the one-electron-per-atom assumption; it must not be multiplied by another atomic count.

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