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 will use the formula for conductivity (\( \sigma \)) which is given by:

\[ \sigma = q \cdot (n_e \cdot \mu_e + n_h \cdot \mu_h) \]

Where:
- \( \sigma \) = conductivity (S/m)
- \( q \) = charge of an electron (\( 1.6 \times 10^{-19} \) C)
- \( n_e \) = electron concentration (\( 5 \times 10^{18} \, m^{-3} \))
- \( \mu_e \) = electron mobility (\( 2.0 \, m^2/V \cdot s \))
- \( n_h \) = hole concentration (\( 5 \times 10^{19} \, m^{-3} \))
- \( \mu_h \) = hole mobility (\( 0.01 \, m^2/V \cdot s \))

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

### Step 2: Substitute 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 first term \( n_e \cdot \mu_e \)
\[ n_e \cdot \mu_e = 5 \times 10^{18} \cdot 2.0 = 10 \times 10^{18} = 1 \times 10^{19} \]

### Step 4: Calculate the second term \( n_h \cdot \mu_h \)
\[ n_h \cdot \mu_h = 5 \times 10^{19} \cdot 0.01 = 0.5 \times 10^{19} = 5 \times 10^{18} \]

### Step 5: Sum the contributions from electrons and holes
\[ n_e \cdot \mu_e + n_h \cdot \mu_h = 1 \times 10^{19} + 5 \times 10^{18} = 1.5 \times 10^{19} \]

### Step 6: Substitute back into the conductivity equation
\[ \sigma = 1.6 \times 10^{-19} \cdot (1.5 \times 10^{19}) \]

### Step 7: Calculate the conductivity
\[ \sigma = 1.6 \cdot 1.5 = 2.4 \, S/m \]

### Step 8: Final result
The conductivity of the semiconductor sample is:
\[ \sigma = 2.4 \, S/m \]