Question

In an unbiased $p-n$ junction, holes diffuse from the $p$-region to $n$-region because of

• A. The attraction of free electrons of $n$-region.
• B. The higher hole concentration in $p$-region than that in $n$-region.
• C. The higher concentration of electrons in the $n$-region than that in the $p$-region.
• D. The potential difference across the p-n junction.

Answer 1

The Correct Option is B

The question asks why holes diffuse from the p-region to the n-region in an unbiased p-n junction.

To understand this, let's consider the characteristics of a p-n junction:

• The p-region is characterized by a high concentration of holes (positive charge carriers), while the n-region is characterized by a high concentration of electrons (negative charge carriers).
• In an unbiased junction, where no external voltage is applied, a natural concentration gradient exists across the junction due to the difference in carrier concentrations.

The diffusion process can be explained by the principle of concentration gradient: Particles (like holes) tend to move from an area of higher concentration to an area of lower concentration. In this case, holes from the p-region diffuse to the n-region to balance the concentration.

Let's examine the options:

1. The attraction of free electrons of n-region. This is incorrect; diffusion is primarily driven by concentration differences, not attraction.
2. The higher hole concentration in p-region than that in n-region. This is correct. The holes diffuse because there are more of them in the p-region compared to the n-region.
3. The higher concentration of electrons in the n-region than that in the p-region. While true, electrons move from high concentration regions, but the question specifically asks about hole movement.
4. The potential difference across the p-n junction. This is incorrect; the potential difference is zero in an unbiased junction.

Conclusion: Thus, the correct answer is that holes diffuse due to the higher hole concentration in the p-region than in the n-region, which is option 2.