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Three photo diodes $\mathrm{D}_1, \mathrm{D}_2$ and $\mathrm{D}_3$ are made of semiconductors having band gaps of $2.5 \mathrm{eV}, 2 \mathrm{eV}$ and $3 \mathrm{eV}$, respectively. Which ones will be able to detect light of wavelength $6000 Å$ ?
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Verified Answer
As given that, wavelength of light
$$
\lambda=6000 Å=6000 \times 10^{-10} \text { meter }
$$
So, Energy of the light photon
$$
\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}=2.06 \mathrm{eV}
$$
The incident radiation is detected by the photodiode $\mathrm{D}_2$ having energy should be greater than the band-gap or knee voltage $2 \mathrm{eV}$. So, it is only valid for diode $\mathrm{D}_2$. Diode $\mathrm{D}_2$ will detect less than incident radiation of $2.06 \mathrm{eV}$.
$$
\lambda=6000 Å=6000 \times 10^{-10} \text { meter }
$$
So, Energy of the light photon
$$
\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}=2.06 \mathrm{eV}
$$
The incident radiation is detected by the photodiode $\mathrm{D}_2$ having energy should be greater than the band-gap or knee voltage $2 \mathrm{eV}$. So, it is only valid for diode $\mathrm{D}_2$. Diode $\mathrm{D}_2$ will detect less than incident radiation of $2.06 \mathrm{eV}$.
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