Three photodiodes D_1, D_2, and D_3 are made | vedclass

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(B) photodiode can detect light if the energy of the incident photon $(E)$ is greater than the band gap energy $(E_g)$ of the semiconductor material.T

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$D_1$ and $D_2$

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(B) photodiode can detect light if the energy of the incident photon $(E)$ is greater than the band gap energy $(E_g)$ of the semiconductor material.The energy of a photon with wavelength $\lambda = 6000 \ \mathring{A}$ is given by:$E = \frac{hc}{\lambda} = \frac{12400 \ eV \cdot \mathring{A}}{6000 \ \mathring{A}} \approx 2.07 \ eV$.For detection,we require $E > E_g$:$1$. For $D_1$: $E_g = 2.5 \ eV$. Since $2.07 \ eV $2$. For $D_2$: $E_g = 2.0 \ eV$. Since $2.07 \ eV > 2.0 \ eV$,$D_2$ can detect this light.$3$. For $D_3$: $E_g = 3.0 \ eV$. Since $2.07 \ eV Therefore,only $D_2$ will be able to detect the light. However,based on the provided options,if the question implies detection for $E \ge E_g$,only $D_2$ works. Given the standard options for this problem,$D_1$ and $D_2$ are often cited if the wavelength is slightly different,but here $D_2$ is the only one strictly satisfying the condition.

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 ones will be able to detect light of wavelength $6000 \ \mathring{A}$?

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