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 establish the relationship between the number densities of acceptor atoms (N_A), donor atoms (N_D), electrons (n_e), and holes (n_h) in a semiconductor.

### Step-by-Step Solution:

1. **Understanding the Role of Donor and Acceptor Atoms**:
- Donor atoms (N_D) introduce free electrons into the semiconductor. Each donor atom can contribute one free electron.
- Acceptor atoms (N_A) create holes by accepting electrons. Each acceptor atom can create one hole.

2. **Establishing Charge Neutrality**:
- For the semiconductor to be electrically neutral, the total number of electrons must equal the total number of holes. This can be expressed as:
\[ n_e + N_D = n_h + N_A \]
- Here, \(n_e\) is the number density of electrons contributed by donor atoms, and \(n_h\) is the number density of holes contributed by acceptor atoms.

3. **Rearranging the Equation**:
- We can rearrange the equation to isolate the terms:
\[ n_e - n_h = N_A - N_D \]
- This equation shows the balance between the number of electrons and holes in terms of the densities of donor and acceptor atoms.

4. **Final Relationship**:
- The final relationship that satisfies the condition for charge neutrality in the semiconductor is:
\[ n_e + N_D = n_h + N_A \]
- This equation indicates that the total number of electrons (from donors and intrinsic carriers) equals the total number of holes (from acceptors and intrinsic carriers).

### Conclusion:
The correct relationship that satisfies the condition for charge neutrality in a semiconductor with donor and acceptor atoms is:
\[ n_e + N_D = n_h + N_A \]