Concept:Responsivity \( \mathcal{R} \) relates optical power input to electrical current output based on photon energy: \( E_{\text{photon}} = \frac{h c}{\lambda} \).
Formula:$$\mathcal{R} = \frac{I_{\text{photo}}}{P_{\text{opt}}} = \frac{\eta q \left(\frac{P_{\text{opt}}}{h f}\right)}{P_{\text{opt}}} = \frac{\eta q}{h f} = \frac{\eta q \lambda}{h c}$$
Solution:- Each incident photon carries energy \( E = \frac{h c}{\lambda} \).
- The photon arrival rate is \( \Phi = \frac{P_{\text{opt}}}{h c / \lambda} = \frac{P_{\text{opt}} \lambda}{h c} \).
- Assuming \( 100\% \) quantum efficiency (\( \eta = 1 \)), the generated current is \( I = q \Phi = \frac{q P_{\text{opt}} \lambda}{h c} \).
- Dividing current by optical power gives \( \mathcal{R} = \frac{q \lambda}{h c} \).
Why other options are incorrect:- Option A: \( hf/q \) has units of Volts, not Amperes/Watt.
- Option B: \( \lambda \) belongs in the numerator because longer wavelengths contain more photons per watt.
- Option D: This inverts the photon energy ratio.
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