Three photodiodes ${\mathrm{D}}_1,{\mathrm{D}}_2,{\mathrm{D}}_3$ are made of semiconductors having band gaps of 2.5ev, 2ev and 3ev which can detect light of wave length $6000{\mathrm{A}}^{\mathrm{o}}$ 

 

Given,

$D_1$ photo diode's band gap = 2.5eV

$D_2$photo diode's band gap = 2eV

$D_3$photo diode's band gap = 3eV

The incident light's wavelength is 6000A. Consequently, incident light's energy is

$E=\frac{hc}{\lambda }$

Substituting the values,

'$$\begin{gathered} E=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{6\times {10}^{-7}\times 1.6\times {10}^{-19}} \\ E=2.06eV \end{gathered}$$

The incident radiation must have a photon energy greater than the band gap in order for the photodiode to detect it. This only applies to $D_2$, though. Only $D_2$ will thus be able to detect this radiation.

Hence the correct answer is $D_2.$