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EEJ MAIN Physics QUESTION #7065
Question 1
Consider an n-type semiconductor where the electron density is $10^{19} \text{ m}^{-3}$ and the electron mobility is $1.6 \text{ m}^2/(\text{V}\cdot\text{s})$. If we neglect the contribution of holes, what is the approximate resistivity of this semiconductor?
  • $2 \ \Omega\text{m}$
  • $4 \ \Omega\text{m}$
  • $0.4 \ \Omega\text{m}$✔️
  • $0.2 \ \Omega\text{m}$
Correct Answer Explanation

Resistivity ($\rho$) is the reciprocal of conductivity ($\sigma$).

$\sigma = ne\mu_e$

Given: $n = 10^{19} \text{ m}^{-3}$, $\mu_e = 1.6 \text{ m}^2/\text{Vs}$, and $e = 1.6 \times 10^{-19} \text{ C}$.

$\sigma = 10^{19} \times (1.6 \times 10^{-19}) \times 1.6 = 1.6 \times 1.6 = 2.56 \ \Omega^{-1}\text{m}^{-1}$.

$\rho = \frac{1}{\sigma} = \frac{1}{2.56} \approx 0.39 \ \Omega\text{m}$.

The closest value provided is $0.4 \ \Omega\text{m}$.