If a semiconductor photodiode can detect a photon with a maximum wavelength of 400nm, then its band gap energy is :
Planck’s constant $h = 6.63 \times 10^{- 34} J . s$ . Speed of light $c = 3 \times 10^{8} m / s$

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1.1eV

3.1eV

2.0eV

1.5eV

answer is B.

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Detailed Solution

Band gap = Energy of photon = $\frac{h c}{λ}$ =12400/ $λ$ in A^o
Band gap = $\frac{4.14 \times 10^{- 15} \times 3 \times 10^{8}}{4 \times 10^{- 7}} e v$ $= 3.1 e v$

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