What is the conductivity of a semiconductor sample having electron concentration of `5 xx 10^(18) m^(-3)` hole concentration of `5 xx 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 xx 10^(-19)C)`

Options

A. `1.83 (Omega-m)^(-1)`
B. `1.68 (Omega-m)^(-1)`
C. `1.20 (Omega-m)^(-1)`
D. `0.59 (W-m)^(-1)`

Answer

To find the conductivity of the semiconductor sample, we can use the formula for conductivity (σ) given by:

\[ \sigma = q \cdot (n \cdot \mu_e + p \cdot \mu_h) \]

where:
- \( q \) is the charge of an electron,
- \( n \) is the electron concentration,
- \( p \) is the hole concentration,
- \( \mu_e \) is the electron mobility,
- \( \mu_h \) is the hole mobility.

### Step 1: Identify the given values
- Electron concentration (\( n \)) = \( 5 \times 10^{18} \, m^{-3} \)
- Hole concentration (\( p \)) = \( 5 \times 10^{19} \, m^{-3} \)
- Electron mobility (\( \mu_e \)) = \( 2.0 \, m^{2} V^{-1} s^{-1} \)
- Hole mobility (\( \mu_h \)) = \( 0.01 \, m^{2} V^{-1} s^{-1} \)
- Charge of an electron (\( q \)) = \( 1.6 \times 10^{-19} \, C \)

### Step 2: Substitute the values into the formula
Now, substituting the values into the conductivity formula:

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

### Step 3: Calculate the contributions from electrons and holes
1. Calculate the contribution from electrons:
\[ n \cdot \mu_e = (5 \times 10^{18}) \cdot (2.0) = 10 \times 10^{18} = 1.0 \times 10^{19} \]

2. Calculate the contribution from holes:
\[ p \cdot \mu_h = (5 \times 10^{19}) \cdot (0.01) = 0.05 \times 10^{19} \]

### Step 4: Combine the contributions
Now, combine the contributions:
\[ n \cdot \mu_e + p \cdot \mu_h = 1.0 \times 10^{19} + 0.05 \times 10^{19} = 1.05 \times 10^{19} \]

### Step 5: Calculate the conductivity
Now, substitute this back into the conductivity formula:
\[ \sigma = (1.6 \times 10^{-19}) \cdot (1.05 \times 10^{19}) \]

Calculating this gives:
\[ \sigma = 1.68 \, m^{-1} \]

### Final Answer
The conductivity of the semiconductor sample is:
\[ \sigma = 1.68 \, S/m \]

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