Concept:Photodetection requires the incident photon energy to be greater than or equal to the semiconductor energy band gap (\( E_{\text{photon}} \ge E_g \)).
Formula:$$\lambda_{\text{max}} = \frac{h c}{E_g}$$
Solution:- Given: \( E_g = 0.60\text{ eV} \) and \( h c \approx 1240\text{ eV}\cdot\text{nm} \).
- \( \lambda_{\text{max}} = \frac{1240\text{ eV}\cdot\text{nm}}{0.60\text{ eV}} = \frac{1240}{0.60} \approx 2066.6\text{ nm} \approx 2066\text{ nm} \).
- Photons with wavelengths longer than \( 2066\text{ nm} \) carry insufficient energy to excite electron-hole pairs across this bandgap.
Why other options are incorrect:- Option A: 1033 nm corresponds to a band gap of \( 1.20\text{ eV} \) (double 0.6 eV).
- Option C: 620 nm corresponds to \( E_g = 2.0\text{ eV} \).
- Option D: 4132 nm corresponds to \( E_g = 0.30\text{ eV} \).
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