Answer: Equating \(n\) and \(p\) gives \(E_F=\dfrac{E_C+E_V}{2}+\dfrac{3kT}{4}\ln\dfrac{m_h^*}{m_e^*}\approx\dfrac{E_C+E_V}{2}\), i.e. the middle of the band gap.
Electron concentration in the conduction band. Using the Fermi–Dirac function with \(E_C-E_F\gg kT\):
\[n=N_C\,e^{-(E_C-E_F)/kT},\qquad N_C=2\left(\frac{2\pi m_e^*kT}{h^2}\right)^{3/2}\qquad\ldots(1)\]
Hole concentration in the valence band. The probability that a state is empty is \(1-f(E)\approx e^{-(E_F-E)/kT}\), so
\[p=N_V\,e^{-(E_F-E_V)/kT},\qquad N_V=2\left(\frac{2\pi m_h^*kT}{h^2}\right)^{3/2}\qquad\ldots(2)\]
Here \(N_C\), \(N_V\) are the effective densities of states, and \(m_e^*\), \(m_h^*\) the effective masses of electron and hole.
Intrinsic condition. In a pure semiconductor every electron that jumps to the conduction band leaves one hole, so \(n=p\):
\[N_C\,e^{-(E_C-E_F)/kT}=N_V\,e^{-(E_F-E_V)/kT}\]
\[\Rightarrow\ e^{(2E_F-E_C-E_V)/kT}=\frac{N_V}{N_C}=\left(\frac{m_h^*}{m_e^*}\right)^{3/2}\]
Taking logarithms:
\[2E_F-(E_C+E_V)=\frac32kT\ln\frac{m_h^*}{m_e^*}\ \Rightarrow\ E_F=\frac{E_C+E_V}{2}+\frac{3kT}{4}\ln\frac{m_h^*}{m_e^*}\]
Conclusion. If \(m_h^*=m_e^*\), the log term is zero (\(\ln1=0\)); even when they differ, \(\frac34kT\) is only about 0.02 eV at room temperature, which is small compared with the band gap (about 1.1 eV for Si). Hence
\[E_F\approx\frac{E_C+E_V}{2}\]
Taking the top of the valence band as zero (\(E_V=0\), \(E_C=E_g\)), \(E_F=E_g/2\): the Fermi level of an intrinsic semiconductor lies in the middle of the forbidden gap.
