Concept:Photodetection requires the incident photon energy to satisfy \( E_{\text{photon}} = \frac{h c}{\lambda} \ge E_g \), defining a cutoff wavelength of \( \lambda_{\text{max}} = \frac{h c}{E_g} \).
Formula:$$\lambda_{\text{max}} = \frac{h c}{E_g} = \frac{1240\text{ eV}\cdot\text{nm}}{1.10\text{ eV}} \approx 1127.27\text{ nm} \approx 1127\text{ nm}$$
Solution:- Given: \( E_g = 1.10\text{ eV} \) and \( h c \approx 1240\text{ eV}\cdot\text{nm} \).
- Cutoff wavelength: \( \lambda_{\text{max}} = \frac{1240}{1.10} \approx 1127\text{ nm} \) (in the near-infrared spectrum).
- Silicon is transparent to wavelengths longer than \( 1127\text{ nm} \).
Why other options are incorrect:- Option A: 620 nm corresponds to a bandgap of \( 2.0\text{ eV} \).
- Option B: 850 nm corresponds to a bandgap of \( 1.46\text{ eV} \) (GaAs).
- Option D: 2066 nm corresponds to a bandgap of \( 0.60\text{ eV} \).
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