What is the conductivity of a semiconductor (in `Omega^(-1) m^(-1)`) if electron density `= 5 xx 10^(12) cm^(-3)` and hole density `= 8 xx 10^(13) cm^(-3)`?
`(mu_(e) = 2.3 V^(-1) s^(-1)m^(2), mu_(h) = 0.01 m^(2) V^(-1) s^(-1))`

Options

A. a. `5.634`
B. b. `1.968`
C. c. `3.421`
D. d. `8.964`

Answer

To find the conductivity of the semiconductor, we will use the formula for conductivity (\(\sigma\)) given by:

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

Where:
- \(e\) is the charge of an electron (\(1.6 \times 10^{-19} \, \text{C}\))
- \(\mu_e\) is the mobility of electrons
- \(n_e\) is the electron density
- \(\mu_h\) is the mobility of holes
- \(n_h\) is the hole density

### Step 1: Convert densities from \(\text{cm}^{-3}\) to \(\text{m}^{-3}\)

Given:
- Electron density \(n_e = 5 \times 10^{12} \, \text{cm}^{-3}\)
- Hole density \(n_h = 8 \times 10^{13} \, \text{cm}^{-3}\)

To convert these values to \(\text{m}^{-3}\):
\[ 1 \, \text{cm}^{-3} = 10^6 \, \text{m}^{-3} \]

Thus,
\[ n_e = 5 \times 10^{12} \, \text{cm}^{-3} = 5 \times 10^{12} \times 10^6 \, \text{m}^{-3} = 5 \times 10^{18} \, \text{m}^{-3} \]
\[ n_h = 8 \times 10^{13} \, \text{cm}^{-3} = 8 \times 10^{13} \times 10^6 \, \text{m}^{-3} = 8 \times 10^{19} \, \text{m}^{-3} \]

### Step 2: Substitute the values into the conductivity formula

Given:
- \(\mu_e = 2.3 \, \text{m}^2 \, \text{V}^{-1} \, \text{s}^{-1}\)
- \(\mu_h = 0.01 \, \text{m}^2 \, \text{V}^{-1} \, \text{s}^{-1}\)

Now substituting the values into the formula:
\[ \sigma = e \cdot (\mu_e \cdot n_e + \mu_h \cdot n_h) \]
\[ \sigma = (1.6 \times 10^{-19}) \cdot \left( (2.3 \cdot 5 \times 10^{18}) + (0.01 \cdot 8 \times 10^{19}) \right) \]

### Step 3: Calculate the terms inside the parentheses

Calculating the first term:
\[ \mu_e \cdot n_e = 2.3 \cdot 5 \times 10^{18} = 11.5 \times 10^{18} \]

Calculating the second term:
\[ \mu_h \cdot n_h = 0.01 \cdot 8 \times 10^{19} = 0.08 \times 10^{19} = 8 \times 10^{17} \]

Adding these two results:
\[ 11.5 \times 10^{18} + 0.08 \times 10^{19} = 11.5 \times 10^{18} + 8 \times 10^{18} = 19.5 \times 10^{18} \]

### Step 4: Calculate the conductivity

Now substituting back into the conductivity formula:
\[ \sigma = (1.6 \times 10^{-19}) \cdot (19.5 \times 10^{18}) \]
\[ \sigma = 1.6 \cdot 19.5 \times 10^{-1} = 31.2 \times 10^{-1} = 3.12 \, \Omega^{-1} \, \text{m}^{-1} \]

### Step 5: Final answer

Thus, the conductivity of the semiconductor is:
\[ \sigma \approx 3.12 \, \Omega^{-1} \, \text{m}^{-1} \]