KCET · Physics · Semiconductors
Three photodiodes \(D_{1}, D_{2}\) and \(D_{3}\) are made of semiconductors having band gaps of \(2.5 \mathrm{eV}, 2 \mathrm{eV}\) and \(3 \mathrm{eV}\), respectively. Which one will be able to detect light of wavelength \(600 \mathrm{~nm}\) ?
- A \(D_{1}\) only
- B Both \(D_{1}\) and \(D_{3}\)
- C \(D_{2}\) only
- D All of these
Answer & Solution
Correct Answer
(C) \(D_{2}\) only
Step-by-step Solution
Detailed explanation
Given, \(E_{1}=2.5 \mathrm{eV}, E_{2}=2 \mathrm{eV}, E_{3}=3 \mathrm{eV}\) and \(\lambda=600 \mathrm{~nm}\)
As we know, \(E=\frac{1242}{\lambda}=\frac{1242}{600}\)
\(=2.07 \mathrm{eV}\)
Since, \(E_{1}\) and \(E_{3}\) is greater than \(E\). So, only \(D_{2}\) will able to detect the given light of wavelength 600 \(\mathrm{nm}\).
As we know, \(E=\frac{1242}{\lambda}=\frac{1242}{600}\)
\(=2.07 \mathrm{eV}\)
Since, \(E_{1}\) and \(E_{3}\) is greater than \(E\). So, only \(D_{2}\) will able to detect the given light of wavelength 600 \(\mathrm{nm}\).
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