In an intrinsic semiconductor at room temperature, the number of electrons $n_e$ and holes $n_h$ are equal. If the temperature $T$ increases, how does the Fermi level $E_f$ shift?

Solution

1. The Fermi level $E_f$ for an intrinsic semiconductor is defined by the expression $E_f=\frac{E_c+E_v}{2}+\frac{3}{4}{k}_{B}T\ln \left(\frac{m_h^{\ast }}{m_e^{\ast }}\right)$.
2. In this expression, $E_c$ and $E_v$ represent the conduction and valence band edges, while $m_h^{\ast }$ and $m_e^{\ast }$ are the effective masses of holes and electrons.
3. Since the ratio of the effective masses $\frac{m_h^{\ast }}{m_e^{\ast }}$ is approximately unity, the logarithmic term becomes negligible.
4. Consequently, the Fermi level $E_f$ remains fixed at the midpoint of the band gap, $\frac{E_c+E_v}{2}$, regardless of temperature $T$.
5. Hence the answer is (C).