Concept:Electrical conductivity is proportional to the concentration of mobile charge carriers (\( \sigma = e(n \mu_n + p \mu_p) \)). Without free carriers, resistivity becomes very high.
Formula:$$\rho = \frac{1}{\sigma} = \frac{1}{e(n \mu_n + p \mu_p)} \to \infty \quad (\text{as } n, p \to 0)$$
Solution:- Recombination near the interface removes mobile electrons and holes from the depletion layer.
- The remaining charges are immobile donor and acceptor ions bound in the crystal lattice.
- Without mobile charge carriers to transport current, the depletion layer exhibits high resistivity.
Why other options are incorrect:- Option A: High mobile carrier density characterizes the low-resistivity neutral N-region, not the depletion layer.
- Option B: The crystal lattice retains its crystalline structure across the junction.
- Option C: Recombination releases small amounts of thermal energy or photons; it does not drop the temperature to absolute zero.
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