Q: A p-n Photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength
(a) 6000 Å
(b) 4000 nm
(c) 6000 nm
(d) 4000 Å
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Ans: (d)
Sol: $\large \lambda_{max} = \frac{1242}{E(eV)} $
$\large \lambda_{max} = \frac{1242}{2.5} $
= 496.8 nm = 4968 A°