When LED is forward biased, then

Options

A. electrons from the n - type side cross the p - n junction and recombine with holes in the p - type side
B. electrons and holes neutralise each other in depletion region
C. at junction electrons and holes remain at rest
D. none of these

Answer

To solve the question "When LED is forward biased, then," we need to analyze the behavior of an LED (Light Emitting Diode) when it is connected in a forward-biased configuration. Let's break down the steps:

### Step-by-Step Solution:

1. **Understanding LED Structure**:
- An LED is a type of diode formed by a p-n junction. The p-type semiconductor has holes as the majority carriers, while the n-type semiconductor has electrons as the majority carriers.

2. **Forward Biasing**:
- In a forward-biased condition, the p-type side is connected to the positive terminal of a power supply, and the n-type side is connected to the negative terminal. This configuration reduces the barrier potential at the junction.

3. **Movement of Charge Carriers**:
- When the LED is forward biased, holes from the p-type region move towards the n-type region, and electrons from the n-type region move towards the p-type region.

4. **Recombination of Charge Carriers**:
- As electrons from the n-type region cross the junction, they recombine with holes in the p-type region. This recombination process is crucial because it leads to the emission of energy.

5. **Energy Emission**:
- In the case of an LED, when electrons recombine with holes, the energy released is in the form of light rather than heat. This is what makes LEDs different from regular diodes.

6. **Evaluating the Options**:
- Option A states that "electrons from n-type side cross the p-n junction and recombine with the holes in the p-type side." This statement is true and accurately describes the process that occurs when an LED is forward biased.
- Option B is incorrect because electrons and holes do not neutralize each other in the depletion region; instead, they recombine across the junction.
- Option C is also incorrect because electrons and holes do not remain at rest; they are actively moving and recombining.
- Option D is not applicable since option A is correct.

### Conclusion:
The correct answer is **Option A**: "electrons from n-type side cross the p-n junction and recombine with the holes in the p-type side."

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