Q144 / 180 · KCET 2021 · Physics

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 $D_{3}$
  • Correct answer
    C
    $D_{2}$ only
  • D
    All of these

Solution

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 $\mathrm{nm}$.

Hence, the correct answer is option: 'C'.

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