Mobility of electron in N-type Ge is `5000 cm^(2)//"volt"` sec and conductivity `5 "mho/cm"`. If effect of holes is negligible then impurity concentration will be

Options

A. `6.25xx10^(15)//cm^(3)`
B. `9.25xx10^(14)//cm^(3)`
C. `6xx10^(13)//cm^(3)`
D. `9xx10^(13)//cm^(3)`

Answer

To find the impurity concentration in N-type germanium given the mobility of electrons and conductivity, we can use the formula for conductivity:

\[ \sigma = n_e \cdot e \cdot \mu_e \]

Where:
- \(\sigma\) is the conductivity (in mho/cm),
- \(n_e\) is the electron concentration (in cm\(^{-3}\)),
- \(e\) is the charge of an electron (\(1.6 \times 10^{-19}\) coulombs),
- \(\mu_e\) is the mobility of electrons (in cm\(^2\)/(volt·sec)).

### Step-by-Step Solution:

1. **Identify the given values**:
- Mobility of electrons, \(\mu_e = 5000 \, \text{cm}^2/\text{volt sec}\)
- Conductivity, \(\sigma = 5 \, \text{mho/cm}\)
- Charge of an electron, \(e = 1.6 \times 10^{-19} \, \text{C}\)

2. **Rearrange the formula for conductivity to solve for electron concentration \(n_e\)**:
\[ n_e = \frac{\sigma}{e \cdot \mu_e} \]

3. **Substitute the known values into the equation**:
\[ n_e = \frac{5 \, \text{mho/cm}}{(1.6 \times 10^{-19} \, \text{C}) \cdot (5000 \, \text{cm}^2/\text{volt sec})} \]

4. **Calculate the denominator**:
\[ e \cdot \mu_e = 1.6 \times 10^{-19} \times 5000 = 8.0 \times 10^{-16} \, \text{C cm}^2/\text{volt sec} \]

5. **Now calculate \(n_e\)**:
\[ n_e = \frac{5}{8.0 \times 10^{-16}} = 6.25 \times 10^{15} \, \text{cm}^{-3} \]

6. **Conclusion**:
The impurity concentration \(n_e\) in N-type germanium is \(6.25 \times 10^{15} \, \text{cm}^{-3}\).

### Final Answer:
The impurity concentration is \(6.25 \times 10^{15} \, \text{cm}^{-3}\).

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