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ID: past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2016/paper-1/2/a/solution

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Past exam of the materials science course of the University of Cambridge / 2016 / Paper 1 / 2 / a / Solution by

Codex  0 2026-10-20

Assume one temperature-independent activation energy over this interval and an unchanged dopant-controlled carrier population. Write the Arrhenius equation as

D = D_{0} exp [- Q / (k_{B} T)]

. The two measurements give

\frac{Q}{k _{B}} = \frac{l n 10}{1/1000 - 1/1180} = 1.5095 \times 1 0^{4} K, Q = 1.301 eV = 125.5 kJ mo l^{- 1} .

Using

D (T) = 1 0^{- 13} exp [(Q / k_{B}) (1/1000 - 1/ T)]

yields

D (1250) = 2.047 \times 1 0^{- 12}, D (1110) = 4.463 \times 1 0^{- 13} m^{2} s^{- 1} .

The Nernst-Einstein relation includes the factor

1/ T

:

σ = n q^{2} D / (k_{B} T)

. For fixed

n

and a fixed correlation correction,

σ (1110) = 0.08 \frac{D ( 1110 )}{D ( 1250 )} \frac{1250}{1110} = 1.964 \times 1 0^{- 2} S m^{- 1} .

Using conductivity proportional to

D

alone would miss the temperature factor. Carrier association, a changing vacancy population or a changing diffusion correlation factor would invalidate this simple extrapolation.

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