Question

The electron and hole concentrations in a semiconductor are $5 \times 10^{18}\text{ m}^{-3}$ and $8 \times 10^{19}\text{ m}^{-3}$ respectively. If the mobilities of the electron and hole are $0.24\text{ m}^2\text{V}^{-1}\text{s}^{-1}$ and $0.01\text{ m}^2\text{V}^{-1}\text{s}^{-1}$ respectively, then the conductivity of the semiconductor is:

• A. $0.48\text{ Sm}^{-1}$
• B. $0.16\text{ Sm}^{-1}$
• C. $0.32\text{ Sm}^{-1}$
• D. $0.64\text{ Sm}^{-1}$

Hint:

For semiconductors, both electrons and holes contribute to conductivity. Always add $n_e\mu_e$ and $n_h\mu_h$ before multiplying by the electronic charge $e$.

Answer 1

The Correct Option is C

Concept: The conductivity of a semiconductor is given by \[ \sigma=e(n_e\mu_e+n_h\mu_h) \] where \[ e=1.6\times10^{-19}\text{ C} \] \[ n_e,n_h=\text{electron and hole concentrations} \] \[ \mu_e,\mu_h=\text{electron and hole mobilities} \]

Step 1: Calculate the electron contribution \[ n_e\mu_e=(5\times10^{18})(0.24) \] \[ n_e\mu_e=1.2\times10^{18} \]

Step 2: Calculate the hole contribution \[ n_h\mu_h=(8\times10^{19})(0.01) \] \[ n_h\mu_h=8\times10^{17} \] \[ n_h\mu_h=0.8\times10^{18} \]

Step 3: Add both contributions \[ n_e\mu_e+n_h\mu_h =1.2\times10^{18}+0.8\times10^{18} \] \[ =2.0\times10^{18} \]

Step 4: Calculate conductivity \[ \sigma=(1.6\times10^{-19})(2.0\times10^{18}) \] \[ \sigma=3.2\times10^{-1} \] \[ \sigma=0.32\text{ Sm}^{-1} \] Therefore, \[ \boxed{\sigma=0.32\text{ Sm}^{-1}} \]