Question

A photodiode is made of a semiconductor with a bandgap of 1.42 eV. Given that Planck’s constant is \(6.626 \times 10^{-34}\) J·s, the speed of light in vacuum is \(3 \times 10^8\) m/s, and \(1 \, \text{eV} = 1.6 \times 10^{-19}\) J, the cut-off wavelength (in nanometers) of the photodiode is _________ (round off to one decimal place).

Answer 1

Correct Answer: 870

The energy of a photon \(E\) is related to the wavelength \(\lambda\) by the equation: \[ E = \frac{h c}{\lambda} \] Where:
- \(E\) is the energy of the photon,
- \(h = 6.626 \times 10^{-34} \, \text{J·s}\) is Planck’s constant,
- \(c = 3 \times 10^8 \, \text{m/s}\) is the speed of light,
- \(\lambda\) is the wavelength of the photon.
Substitute the bandgap energy \(E = 1.42 \, \text{eV} = 1.42 \times 1.6 \times 10^{-19} \, \text{J}\):
\[ \lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.42 \times 1.6 \times 10^{-19}} \] After calculating, the cut-off wavelength is approximately \( \boxed{870.0} \, \text{nm} \).