Question

Three photo diodes $D_1,D_2$ and $D_3$ are made of semiconductors having band gaps of $2.5eV,2eV$ and $3eV$, respectively. Which ones will be able to detect light of wavelength $6000\stackrel{0}{A}$ ?

$\text{Formula Used}:E=\frac{hc}{\lambda }$

Answer 1

Given :

Wavelength of light

$\lambda =6000\stackrel{o}{A}=6\times {10}^{-7}m$

As we know,

Energy of incident light photon,

$E=hv=\frac{hc}{\lambda }$

$1eV=1.6\times {10}^{-19}J$

Where,

$h=6.626\times {10}^{-34}Js$

$E=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{6\times {10}^{-7}\times 1.6\times {10}^{-19}}$

$E=2.06eV$

For the incident radiation to be detected by the photodiode, energy of incident radiation photon should be greater than the band gap.

This is true only for $D_2$. Therefore, only $D_2$ will detect this radiation.

$\text{Final Answer}:E=2.06eV$

$D_2$ will detect this radiation.