For pure ''Ge'' semiconductor quantity of ''e'' and hole is `10^(19) e//m^(3)` if we doped donor impurity in it with density `10^(23) e//m^(3)` then quantity of hole `(e//m^(3))` in semiconductor is :-

Options

A. `10^(15)`
B. `10^(19)`
C. `10^(23)`
D. `10^(27)`

Explanation

To solve the problem, we need to determine the quantity of holes in a doped germanium (Ge) semiconductor after adding donor impurities.

### Step-by-Step Solution:

1. **Identify Given Values:**
- For pure germanium, the concentration of electrons (n) and holes (p) is given as:
\[ n = p = 10^{19} \, \text{m}^{-3} \]
- The concentration of donor impurities (N_d) is given as:
\[ N_d = 10^{23} \, \text{m}^{-3} \]

2. **Understand the Effect of Doping:**
- When donor impurities are added to a semiconductor, they contribute additional electrons to the conduction band. This increases the electron concentration (n) and decreases the hole concentration (p) due to the charge neutrality condition.

3. **Apply Charge Neutrality Condition:**
- In a doped semiconductor, the relationship between electrons (n), holes (p), and donor concentration (N_d) can be expressed as:
\[ n \cdot p = n_i^2 \]
- Where \( n_i \) is the intrinsic carrier concentration. For germanium, we can assume \( n_i \) is approximately \( 10^{19} \, \text{m}^{-3} \).

4. **Calculate New Electron Concentration:**
- After doping, the new electron concentration (n') can be approximated as:
\[ n' = n + N_d = 10^{19} + 10^{23} \approx 10^{23} \, \text{m}^{-3} \]
- (Since \( 10^{23} \) is much larger than \( 10^{19} \), we can ignore \( 10^{19} \) in this case.)

5. **Use the Charge Neutrality Condition to Find Holes:**
- Now, substituting the values into the equation:
\[ n' \cdot p' = n_i^2 \]
- Rearranging gives us:
\[ p' = \frac{n_i^2}{n'} \]
- Substituting \( n_i = 10^{19} \, \text{m}^{-3} \) and \( n' = 10^{23} \, \text{m}^{-3} \):
\[ p' = \frac{(10^{19})^2}{10^{23}} = \frac{10^{38}}{10^{23}} = 10^{15} \, \text{m}^{-3} \]

6. **Conclusion:**
- The quantity of holes in the doped germanium semiconductor is:
\[ p' = 10^{15} \, \text{m}^{-3} \]

### Final Answer:
The quantity of holes in the doped germanium semiconductor is \( 10^{15} \, \text{m}^{-3} \).