If `N_(A)` is number density of acceptor atoms added and `N_(D)` is number density of donor atoms added to a semiconductor, `n_(e) and n_(h)` are the number density of electrons and holes in it, then

Options

A. `n_(e) = N_(D), n_(h) = N_(A)`
B. `n_(e) = N_(A), n_(h) = N_(D)`
C. `n_(e) = N_(D) = n_(h) + N_(A)`
D. `n_(e) + N_(A) = n_(h) + N_(D)`

Answer

To solve the problem, we need to analyze the relationships between the number densities of acceptor atoms, donor atoms, electrons, and holes in a semiconductor.

### Step-by-Step Solution:

1. **Understand the Definitions**:
- \( N_A \): Number density of acceptor atoms.
- \( N_D \): Number density of donor atoms.
- \( n_e \): Number density of electrons.
- \( n_h \): Number density of holes.

2. **Charge Neutrality Condition**:
- In a semiconductor, the overall charge must be neutral. This means that the total positive charge (from holes and acceptor atoms) must equal the total negative charge (from electrons and donor atoms).
- Mathematically, this can be expressed as:
\[ N_A + n_h = N_D + n_e \]

3. **Analyzing the Options**:
- **Option A**: \( n_e = N_D \) and \( n_h = N_A \)
- **Option B**: \( n_e = N_A \) and \( n_h = N_D \)
- **Option C**: \( n_e = N_D \) and \( n_h = N_A \)

We need to determine which of these options satisfies the charge neutrality condition.

4. **Substituting into the Charge Neutrality Equation**:
- For **Option A**:
\[ N_A + N_A = N_D + N_D \quad \Rightarrow \quad 2N_A = 2N_D \quad \Rightarrow \quad N_A = N_D \quad \text{(not generally true)} \]
- For **Option B**:
\[ N_A + N_D = N_D + N_A \quad \Rightarrow \quad \text{(always true)} \]
- For **Option C**:
\[ N_A + N_A = N_D + N_D \quad \Rightarrow \quad 2N_A = 2N_D \quad \Rightarrow \quad N_A = N_D \quad \text{(not generally true)} \]

5. **Conclusion**:
- The only option that holds true under the charge neutrality condition is **Option B**: \( n_e = N_A \) and \( n_h = N_D \).

### Final Answer:
The correct answer is **Option B**: \( n_e = N_A \) and \( n_h = N_D \).