Question

What is the conductivity of a semiconductor sample having electron concentration of $5 \times 10^{18} \, m^{-3}$, hole concentration of $5 \times 10^{19} \, m^{-3} $ , electron mobility of $2.0 \, m^2 \, V^{-1} \, s^{-1}$ and hole mobility of $0.01 \, m^2 \, V^{-1} \, s^{-1}$ ? (Take charge of electron as $1.6 \times 10^{-19} C$)

• A. $1.68 (\Omega - m)^{-1}$
• B. $1.83 (\Omega - m)^{-1}$
• C. $0.59 (\Omega - m)^{-1}$
• D. $1.20 (\Omega - m)^{-1}$

Answer 1

The Correct Option is A

To find the conductivity of the semiconductor, we need to use the formula for the conductivity of a semiconductor, which takes into account both electron and hole contributions:

\[\sigma = q \cdot (n \cdot \mu_n + p \cdot \mu_p)\]

Where:

• \(\sigma\) is the conductivity.
• \(q = 1.6 \times 10^{-19} \, C\) is the charge of an electron.
• \(n = 5 \times 10^{18} \, m^{-3}\) is the electron concentration.
• \(p = 5 \times 10^{19} \, m^{-3}\) is the hole concentration.
• \(\mu_n = 2.0 \, m^2 \, V^{-1} \, s^{-1}\) is the electron mobility.
• \(\mu_p = 0.01 \, m^2 \, V^{-1} \, s^{-1}\) is the hole mobility.

Substitute the given values into the conductivity formula:

\[ \sigma = 1.6 \times 10^{-19} \cdot ( 5 \times 10^{18} \cdot 2.0 + 5 \times 10^{19} \cdot 0.01 ) \]

Calculate the contributions from electrons and holes separately:

\[ \text{Electron contribution} = 5 \times 10^{18} \cdot 2.0 = 10 \times 10^{18} = 10^{19} \]

\[ \text{Hole contribution} = 5 \times 10^{19} \cdot 0.01 = 0.05 \times 10^{19} = 5 \times 10^{17} \]

Add the contributions:

\[ \text{Total} = 10^{19} + 5 \times 10^{17} = 10.5 \times 10^{18} \]

Now, calculate the conductivity:

\[ \sigma = 1.6 \times 10^{-19} \cdot 10.5 \times 10^{18} \]

Simplify the multiplication:

\[ \sigma = 1.68 \text{ S/m} \, (\Omega^{-1}m^{-1}) \]

Thus, the conductivity of the semiconductor is 1.68 (\Omega^{-1}m^{-1}), which corresponds to the correct answer choice:

• 1.68 (\Omega^{-1}m^{-1})