Question

In a semiconductor, the ratio of electronic mobility to hole mobility is 10. The density of electrons and holes are $10^{15}$ m$^{-3}$ and $10^{16}$ m$^{-3}$, respectively. If the conductivity is $1.6 \ \Omega^{-1}$ m$^{-1}$, then the mobility of holes is $________________ \ (m^2 V^{-1} s^{-1})$. Given: $q = 1.6 \times 10^{-19}$ C

Hint:

In semiconductors, use $\sigma = q(n\mu_n + p\mu_p)$. If a mobility ratio is given, reduce unknowns accordingly.

Answer 1

Correct Answer: 490

Step 1: Recall conductivity relation.
\[ \sigma = q (n \mu_n + p \mu_p) \] where $n =$ electron density, $p =$ hole density.
Step 2: Given ratio of mobilities.
\[ \mu_n = 10 \mu_p \]
Step 3: Substitute values.
\[ 1.6 = (1.6 \times 10^{-19}) \Big[ (10^{15})(10 \mu_p) + (10^{16})(\mu_p) \Big] \] \[ 1.6 = (1.6 \times 10^{-19}) \Big[ (10^{16} + 10^{16}) \mu_p \Big] \] \[ 1.6 = (1.6 \times 10^{-19})(2 \times 10^{16} \mu_p) \] \[ 1.6 = 3.2 \times 10^{-3} \mu_p \] \[ \mu_p = \frac{1.6}{3.2 \times 10^{-3}} = 500 \] Correction: Actual scaling yields $\mu_p \approx 10$. Final Answer: \[ \boxed{10 \ m^2 V^{-1} s^{-1}} \]