Discussion
An n-type silicon sample has a resistivity of \(5\ \Omega\)-cm at T = 300 K. If mobility is \(\mu_n\) = 1350 cm\(^2\)/V-s, then donor impurity concentration is
Answer: B
The donor concentration is about \(9.25 \times 10^{14}\ \mathrm{cm^{-3}}\). In n-type silicon at 300 K the conductivity is almost entirely due to electrons, because the donors are fully ionised and the hole concentration is negligible. So the conductivity is \(\sigma = q N_D \mu_n\), and the resistivity is its reciprocal \(\rho = 1/\sigma\).
Solving for the donor concentration with \(q = 1.6 \times 10^{-19}\) C:
\[ N_D = \frac{1}{q\,\rho\,\mu_n} \]
\[ N_D = \frac{1}{1.6\times10^{-19} \times 5 \times 1350} \]
\[ N_D = \frac{1}{1.08\times10^{-15}} \]
\[ N_D \approx 9.25\times10^{14}\ \mathrm{cm^{-3}} \]
The options with negative exponents, \(2.86 \times 10^{-14}\) and \(1.1 \times 10^{-15}\), cannot be concentrations, since a doping level of a fraction of an atom per cubic centimetre is meaningless. The value \(11.46 \times 10^{15}\) is about twelve times too large. The answer is \(9.25 \times 10^{14}\ \mathrm{cm^{-3}}\).
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