Concept:Carrier freeze-out occurs when thermal energy \( k T \) falls below the dopant ionization energy \( \Delta E_D \) (\( \approx 0.045\text{ eV} \) in Si), causing free carriers to become trapped in localized donor/acceptor states and reducing conductivity.
Formula:$$n(T) \approx \sqrt{\frac{N_c N_D}{2}} \exp\left(-\frac{\Delta E_D}{2 k T}\right) \longrightarrow 0 \quad \text{as } T \to 0\text{ K}$$
Solution:- At very low temperatures, \( k T \ll \Delta E_D \).
- Electrons lack the thermal energy needed to escape donor atoms into the conduction band, becoming trapped ('frozen out') at neutral donor sites.
- This reduces free carrier density and causes semiconductor resistivity to rise dramatically toward insulator levels.
Why other options are incorrect:- Option B: Low temperatures freeze and stabilize crystal lattices rather than melting them.
- Option C: Dopant ionization is an electronic transition, not nuclear transmutation.
- Option D: Standard semiconductors become electrical insulators at cryogenic freeze-out, not superconductors.
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