Three photodiodes D_{1}, D_{2} and D_{3} are | vedclass

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(C) The energy of a photon corresponding to a wavelength $\lambda$ is given by $E = \frac{1240}{\lambda (nm)} eV$.For $\lambda = 600 nm$,the energy of

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$D_{2}$ only

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(C) The energy of a photon corresponding to a wavelength $\lambda$ is given by $E = \frac{1240}{\lambda (nm)} eV$.For $\lambda = 600 nm$,the energy of the incident photon is $E = \frac{1240}{600} \approx 2.07 eV$.$A$ photodiode can detect light only if the energy of the incident photon is greater than or equal to the band gap energy $(E_{g})$ of the semiconductor material $(E \ge E_{g})$.Comparing the photon energy $(2.07 eV)$ with the band gaps:For $D_{1}$: $E_{g} = 2.5 eV$. Since $2.07 eV For $D_{2}$: $E_{g} = 2 eV$. Since $2.07 eV > 2 eV$,$D_{2}$ can detect this light.For $D_{3}$: $E_{g} = 3 eV$. Since $2.07 eV Therefore,only $D_{2}$ will be able to detect the light of wavelength $600 nm$.

Three photodiodes $D_{1}, D_{2}$ and $D_{3}$ are made of semiconductors having band gaps of $2.5 eV, 2 eV$ and $3 eV$,respectively. Which one will be able to detect light of wavelength $600 nm$?

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