723. Problem 39.29P (HRW) Suppose the fractional efficiency of a caesium surface (with work function 1.80 eV) is ; that is, on average one electron is emitted for every photons that fall on the surface. We have to find the current of electrons emitted from such a surface if it were illuminated with 600 nm light from a 2.00 mW laser and all the emitted electrons took part in the charge flow. |
Solution: Click For PDF Version Power of the laser beam which is illuminating the caesium surface is 2.00 mW and wavelength of its light is 600 nm. Energy of a 600 nm photon can be found from the relation The number of photons that strike the caesium surface can be obtained by dividing the power of the laser beam with the energy of each photon. We find As the fractional efficiency of the given caesium surface is , the number of photoelectrons emitted per second will be As charge of an electron is , current of the photoelectrons will be |