If `n_(e)` and `n_(h)` are the number of electrons and holes in a semiconductor heavily doped with phosphorus, then

Options

A. `n_(e) gt gt h_(h)`
B. `n_(e) lt lt h_(h)`
C. `n_(e)leh_(h)`
D. `n_(e)=h_(h)`

Explanation

To solve the question regarding the number of electrons (`n_e`) and holes (`n_h`) in a semiconductor heavily doped with phosphorus, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Type of Doping**:
- Phosphorus is a pentavalent element, meaning it has five valence electrons. When added to a semiconductor (like silicon, which has four valence electrons), it donates an extra electron.

2. **Determine the Type of Semiconductor**:
- Doping a semiconductor with a pentavalent impurity like phosphorus creates an N-type semiconductor. This is because the extra electrons from the phosphorus atoms increase the electron concentration in the material.

3. **Understand Electron and Hole Concentrations**:
- In an N-type semiconductor, the majority carriers are electrons, while holes are the minority carriers. This means that the concentration of electrons (`n_e`) is significantly higher than the concentration of holes (`n_h`).

4. **Conclusion**:
- Since the semiconductor is heavily doped with phosphorus, we can conclude that the electron concentration (`n_e`) is much greater than the hole concentration (`n_h`). Therefore, we can express this relationship as:
\[ n_e \gg n_h \]

### Final Answer:
In a semiconductor heavily doped with phosphorus, the number of electrons (`n_e`) is much greater than the number of holes (`n_h`), i.e., \( n_e > n_h \).

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