Question

Three photo diodes D1, D2, D3 are made up of semiconductors having band gap of 2.5 eV, 2 eV and 3eV respectively. Ehich one will be able to detect light of wavelength 6000 A?

Answer 1

max wavelength 1 = hc/E
= 6.6 x 10⁻³⁴ x 3 x 10⁸/ 2 x 1.6 x 10⁻¹⁹
= 6.18 x 10⁻⁷ m

max wavelength 2 = hc/E
= 6.6 x 10⁻³⁴ x 3 x 10⁸/ 2.5 x 1.6 x 10⁻¹⁹
= 3.96 x 10⁻⁷m

max wavelength 3 = hc/E
= 6.6 x 10⁻³⁴ x 3 x 10⁸/ 3 x 1.6 x 10⁻¹⁹
= 4.125 x 10⁻⁷m

as given A= 6 x 10⁻⁷m
The wave length on incident energy radiation must be greater for optical signals. which is true in case of D2

Answer 2

Answer:

The photodiode $D_{2}$ can detect the incident light.

Explanation:

Given band gap energies of photodiodes,

                   $D_{1}=2.5 eV, \ D_{2}=2 eV\ \&\ D_{3}=3eV$

         wavelength,  $\lambda = 6000 A=6\times 10^{-7} m$

Energy of a light photon, $E=\frac{hc}{\lambda}$

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

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

                                              $=\frac{6.6 \times 3} {6\times 1.6 }=2.062eV$

To detect the incident light, the incident photon energy should be greater than the bandgap energy of the photodiode.

Comparing the values of the diodes, the value of $D_{2}$ is less than the energy of photon, therefore only  $D_{2}$ can detect the incident light.