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 \( e = 1.6 \times 10^{-19} \, C \))
Show Hint
- 1.65
- 1.25
- 0.85
- 1.95
The Correct Option is B
Solution and Explanation
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.
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