Question

If a semiconductor photodiode can detect a photon with a maximum wavelength of$400nm$, then its band gap energy is? Planck’s constant $h=6.63\times {10}^{-34}J.s$. Speed of light $c=3\times {10}^{8}m/s$.

• A. $1.5eV$
• B. $2.0eV$
• C. $3.1eV$
• D. $1.1eV$

Answer 1

The correct option is C

$3.1eV$

Step 1: Given data:

The maximum wavelength $\lambda =400nm$

Planck’s constant $h=6.63\times {10}^{-34}J.s$

The Speed of light $c=3\times {10}^{8}m/s$

Step 2: Formula used:

Know that the formula of band gap energy $E=\left(\frac{hc}{\lambda }\right)$

Here, $h$is Planck’s constant, $\lambda$is a wavelength and the speed of light is $c$.

$$\begin{gathered} E=\left(\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}}\right) \\ E=\left(\frac{1240}{400}\right)eV \\ E=3.1eV \end{gathered}$$

Therefore the gap energy is $3.1eV$.

Hence, the correct option is (C).