Question 1 of 7

Question 1 (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Approach. (a) mass of the unit cell over its volume; (b) Fermi–Dirac occupation with $E_F$ taken at mid-gap for pure (intrinsic) Si; (c) mass-action law $n_op_o=n_i^2$ with $p_o\approx N_a$; (d) donors add electrons, so $E_F$ must rise to keep $n_o$ larger.

1. Part (a) — density. A diamond-cubic cell of side $a$ holds 8 atoms of molar mass $M=28.1\,\text{g/mol}$: $$\rho=\frac{8M}{A_N\,a^3}=\frac{8(28.1)}{(6.02\times10^{23})(5.43\times10^{-8}\,\text{cm})^3}=\boxed{2.33\ \text{g/cm}^3}$$ (the textbook density of silicon, confirming the arithmetic).
2. Part (b) — occupation probability. Pure (intrinsic) Si has $E_F=E_i\approx$ mid-gap, so $E_c-E_F\approx E_g/2=0.56\,\text{eV}$. The state in question sits halfway between $E_F$ and $E_c$: $E-E_F=(E_c-E_F)/2=0.28\,\text{eV}$. With $kT=0.026\,\text{eV}$: $$f(E)=\frac{1}{1+e^{(E-E_F)/kT}}=\frac{1}{1+e^{0.28/0.026}}=\boxed{2.10\times10^{-5}}$$
3. Part (c) — equilibrium electron concentration. With $N_a=10^{17}\,\text{cm}^{-3}\gg n_i$, essentially every acceptor ionizes and $p_o\approx N_a$. From the mass-action law: $$n_o=\frac{n_i^2}{p_o}=\frac{(1.5\times10^{10})^2}{10^{17}}=\boxed{2.25\times10^3\ \text{cm}^{-3}}$$
4. Part (d) — effect of donor doping. Adding donor atoms raises the majority electron concentration $n_o$ far above $n_i$. Since $n_o=n_i e^{(E_F-E_i)/kT}$, a larger $n_o$ requires a larger $E_F-E_i$: the Fermi level moves up, away from mid-gap and toward the conduction-band edge $E_c$ (symmetrically, acceptor doping would push it down toward $E_v$).