 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-2/a10/a/v/solution

Using the requested exponential law with a temperature-independent prefactor,

ln \frac{σ ( 420 )}{σ ( 400 )} = \frac{E _{g}}{2 k _{B}} (\frac{1}{400} - \frac{1}{420}) .

The conductivity ratio is two, so

E_{g} = \frac{2 k _{B} ln 2}{1/400 - 1/420} = \frac{2 ( 8.61733 \times 1 0 ^{- 5} ) l n 2}{1/400 - 1/420} eV,

E_{g} ≃ 1.0035 eV ≃ 1.00 eV .

The denominator is positive, as required for a positive gap and rising conductivity. If one instead retains the constant-DOS factor

T

while keeping mobility constant, the inferred gap becomes

E_{g} = \frac{2 k _{B} l n [ 2 ( 400/420 )]}{1/400 - 1/420} ≃ 0.933 eV .

That is a different prefactor model; $1.00 eV$ is the estimate from the relation specified for this part.

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