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

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 =

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,

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.

im sorry i dont know