Question

The voltage across the diode indicates the energy given to charge carriers (electrons and holes, but more about that later in the course.) If the diode turns on at X volts, then the charge carriers are getting X electron-volts of energy. The charge carriers can give up this energy by emitting a photon. How does the energy the charge carriers get at the turn-on voltage compare to the energy of a red photon

Answer 1

Answer:

the charge carriers have an energy 2.8 10⁻¹⁹ J

Explanation:

The energy in a diode is conserved so the energy supplied must be equal to the energy emitted in the form of photons.

The energy of a photon is given by the Planck expression

             E = h f

the speed of light, wavelength and frequency are related

             c = λ f

we substitute

             E = $\frac{h \ c}{\lambda}$

a red photon has a wavelength of lam = 700 nm = 700 10⁻⁹ m

we calculate the energy

              E = 6.626 10⁻³⁴  3 10⁸/700 10⁻⁹

              E = 2.8397 10⁻¹⁹J

therefore the charge carriers have an energy 2.8 10⁻¹⁹ J,