Question
(a) Calculate the number of photoelectrons per second ejected from a $1.00 \textrm{ mm}^2$ area of sodium metal by 500-nm EM radiation having an intensity of $1.30 \textrm{ kW/m}^2$ (the intensity of sunlight above the Earth’s atmosphere). (b) Given that the binding energy is 2.28 eV, what power is carried away by the electrons? (c) The electrons carry away less power than brought in by the photons. Where does the other power go? How can it be recovered?
1. $3.27 \times 10^{15} \textrm{ electrons/s}$
2. $0.105$
3. The remaining energy goes into the sodium metal in the form of mass equal to the binding energy. If the electron was ejected with less than the maximum kinetic energy some the energy would also be in the form of, probably, heat or an energetically excited state of sodium. The energy transformed into mass (which is the binding energy, in other words) will be released again as a photon or heat when the electron is recaptured by the sodium ion.
Solution Video