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 (Ge) given the mobility of electrons and conductivity, we can use the relationship between conductivity, electron concentration, charge, and mobility.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mobility of electrons, \( \mu = 5000 \, \text{cm}^2/\text{V} \cdot \text{s} \)
- Conductivity, \( \sigma = 5 \, \text{mho/cm} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

2. **Use the Formula for Conductivity:**
The conductivity \( \sigma \) of a semiconductor can be expressed as:
\[ \sigma = n \cdot e \cdot \mu \]
where:
- \( n \) is the electron concentration (impurity concentration),
- \( e \) is the charge of an electron,
- \( \mu \) is the mobility of electrons.

3. **Rearranging the Formula:**
To find the electron concentration \( n \), we rearrange the formula:
\[ n = \frac{\sigma}{e \cdot \mu} \]

4. **Substituting the Values:**
Now substitute the known values into the equation:
\[ n = \frac{5 \, \text{mho/cm}}{(1.6 \times 10^{-19} \, \text{C}) \cdot (5000 \, \text{cm}^2/\text{V} \cdot \text{s})} \]

5. **Calculating the Denominator:**
First, calculate the denominator:
\[ e \cdot \mu = (1.6 \times 10^{-19}) \cdot (5000) = 8.0 \times 10^{-16} \, \text{C} \cdot \text{cm}^2/\text{V} \cdot \text{s} \]

6. **Calculating the Electron Concentration:**
Now substitute back into the equation for \( n \):
\[ n = \frac{5}{8.0 \times 10^{-16}} = 6.25 \times 10^{15} \, \text{cm}^{-3} \]

7. **Final Result:**
The impurity concentration is:
\[ n = 6.25 \times 10^{15} \, \text{cm}^{-3} \]

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

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