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A $p-n$ photodiode is fabricated from a semiconductor with a band gap of $2.5 \mathrm{eV}$. It can detect a signal of wavelength
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1889 Upvotes
Verified Answer
The correct answer is:
$4000 Å$
Key Idea Only signals having wavelength less than threshold wavelength will be detected.
Energy
$$
\begin{aligned}
& \mathrm{E}=\mathrm{hv}=\mathrm{h} \frac{\mathrm{c}}{\lambda} \\
& \lambda=\frac{\mathrm{hc}}{\mathrm{E}}
\end{aligned}
$$
$$
\Rightarrow \quad \lambda=\frac{\mathrm{hc}}{\mathrm{E}}
$$
Substituting the values of and in the above equation
$$
\lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2.5 \times 1.6 \times 10^{-19}}=5000 Å
$$
As $4000 Å < 5000 Å$
Signal of wavelength $4000 Å$ can be detected by the photodiode.
Energy
$$
\begin{aligned}
& \mathrm{E}=\mathrm{hv}=\mathrm{h} \frac{\mathrm{c}}{\lambda} \\
& \lambda=\frac{\mathrm{hc}}{\mathrm{E}}
\end{aligned}
$$
$$
\Rightarrow \quad \lambda=\frac{\mathrm{hc}}{\mathrm{E}}
$$
Substituting the values of and in the above equation
$$
\lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2.5 \times 1.6 \times 10^{-19}}=5000 Å
$$
As $4000 Å < 5000 Å$
Signal of wavelength $4000 Å$ can be detected by the photodiode.
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