Question

If a semiconductor photodiode can detect a photon with a maximum wavelength of $400\text{nm}$, then its band gap energy is: $\text{Planck's constant,}h=6.63\times {10}^{-34}\text{Js}$
$\text{Speed of light,}c=3\times {10}^8{\text{ms}}^{-1}$

• A. $1.1\text{eV}$
• B. $2.0\text{eV}$
• C. $1.5\text{eV}$
• D. $3.1\text{eV}$

Answer 1

The correct option is D $3.1\text{eV}$

Given,
$\text{Wavelength of photon,}\lambda =400\text{nm}$

A photodiode can detect a wavelength corresponding to the energy of band gap.

$\therefore \text{Band gap,}{E}_g=\frac{hc}{\lambda }$

$\Rightarrow E_g=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}}$

$=4.96\times {10}^{-19}\text{V}=\frac{4.96\times {10}^{-19}}{1.6\times {10}^{-19}}\text{eV}$

$=3.1\text{eV}$

Hence, option $\left(D\right)$ is correct.