Question

The intrinsic carrier concentration of a semiconductor is \( 2.5 \times 10^{16} \, /m^3 \) at 300 K. If the electron and hole mobilities are \( 0.15 \, {m}^2/{Vs} \) and \( 0.05 \, {m}^2/{Vs} \), respectively, then the intrinsic resistivity of the semiconductor (in \( k\Omega \cdot m \)) at 300 K is _________. (Charge of an electron \( e = 1.6 \times 10^{-19} \, C \))

• A. 1.65
• B. 1.25
• C. 0.85
• D. 1.95

Hint:

The resistivity of intrinsic semiconductors is calculated using the mobilities of both electrons and holes, the intrinsic carrier concentration, and the charge of the electron. Remember to adjust units appropriately, especially when converting from ohms to kilo-ohms.

Answer 1

The Correct Option is B

Step 1: The intrinsic resistivity \( \rho_i \) of a semiconductor is given by the formula: \[ \rho_i = \frac{1}{q \cdot n_i \cdot (\mu_e + \mu_h)} \] where:
\( q \) is the charge of an electron \( (1.6 \times 10^{-19} \, C) \),
\( n_i \) is the intrinsic carrier concentration \( (2.5 \times 10^{16} / m^3) \),
\( \mu_e \) is the electron mobility \( (0.15 \, {m}^2/{Vs}) \),
\( \mu_h \) is the hole mobility \( (0.05 \, {m}^2/{Vs}) \). 
Step 2: Substituting the values into the formula: \[ \rho_i = \frac{1}{(1.6 \times 10^{-19}) \cdot (2.5 \times 10^{16}) \cdot (0.15 + 0.05)} = \frac{1}{(1.6 \times 10^{-19}) \cdot (2.5 \times 10^{16}) \cdot (0.2)} \] \[ \rho_i = \frac{1}{8 \times 10^{-3}} = 0.125 \, \Omega \cdot m \] To convert to \( k\Omega \cdot m \), we multiply by 1000: \[ \rho_i = 1.25 \, k\Omega \cdot m \] Thus, the correct answer is 1.25.