Concept:The built-in potential barrier \( V_0 \) creates an internal electric field directed from the positive donor ions (N-side) to the negative acceptor ions (P-side), preventing further diffusion of majority carriers.
Formula:$$V_0 = \frac{k T}{q} \ln\left(\frac{N_A N_D}{n_i^2}\right)$$
Solution:- The internal electric field exerts an electrostatic force that opposes the movement of free electrons from the N-region into the P-region.
- Simultaneously, it repels mobile holes attempting to cross from the P-region into the N-region.
- Therefore, the potential barrier prevents the diffusion of majority carriers from both sides.
Why other options are incorrect:- Option A: The built-in electric field assists the drift of minority carriers; it does not oppose them.
- Option C: Electrons in the P-region are minority carriers whose transit is aided by the barrier field.
- Option D: Holes in the N-region are minority carriers whose drift across the junction is facilitated by the internal field.
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