Discussion
In a p-type semiconductor, the Fermi level lies 0.4 eV above the valence band. If the concentration of the acceptor atoms is trippled and kT = 0.03 eV, the new position of the Fermi level will be :
Answer: D
The Fermi level moves down to 0.367 eV above the valence band.
In a p-type semiconductor with full ionisation, \(p \approx N_{A}\) and \(E_{F} - E_{V} = kT\,\text{ln}(N_{V}/N_{A})\). Tripling \(N_{A}\) lowers this by \(kT\,\text{ln}\,3\):
\[ \begin{aligned} \Delta &= 0.03 \times \text{ln}\,3 = 0.03 \times 1.099 \\ &= 0.033\ \text{eV} \\ E_{F} - E_{V} &= 0.400 - 0.033 = 0.367\ \text{eV} \end{aligned} \]
More acceptors make the material more strongly p-type, so the Fermi level moves closer to the valence band, which rules out 0.525 eV. 0.250 eV and 0.100 eV are far too large a shift for a threefold change in doping.
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