Question

In a p–i–n photodiode, a pulse with $8\times10^{12}$ incident photons at $\lambda_0=1.55\,\mu$m yields $4\times10^{12}$ electrons collected. The quantum efficiency at this wavelength is _______%.

• A. 50
• B. 54.2
• C. 62.5
• D. 80

Hint:

QE counts quanta: one collected electron per incident photon $\Rightarrow$ $100%$. Halving that count $\Rightarrow$ $50%$.

Answer 1

The Correct Option is A

Quantum efficiency $\eta$ (at a given $\lambda$) is the ratio of collected carriers to incident photons: \[ \eta=\frac{N_e}{N_\gamma}\times 100%=\frac{4\times10^{12}}{8\times10^{12}}\times100%=50%. \] (The wavelength is irrelevant once $N_\gamma$ and $N_e$ are given.) Final Answer: $50%$ (Option A).