A photodiode is designed to detect incident light of wavelength 663 nm . For the photodiode to successfully detect this light, what should be the maximum bandgap energy of the semiconductor material used? (Take h = 6.63 10⁻³⁴ J s , c = 3 10^8 m/s , 1 eV = 1.6 10⁻¹⁹ J )

Step-by-step solution

For a photodiode to detect light, the energy of the incident photons must be greater than or equal to the bandgap energy of the semiconductor ( E E_g ). Thus, the maximum allowed bandgap energy equals the photon energy. The energy of the incident photon in Joules is given by: E = hc Substituting the given values: E = 6.63 10⁻³⁴ 3 10^8 663 10⁻⁹ E = 19.89 10⁻²⁶ 663 10⁻⁹ = 0.03 10⁻¹⁷ J = 3.0 10⁻¹⁹ J To convert this energy into electron-volts (eV), divide by 1.6 10⁻¹⁹ J/eV : E_g = 3.0 10⁻¹⁹ 1.6 10⁻¹⁹ = 3.0 1.6 = 1.875