719. Problem 49.47 (RHK) X rays with a wavelength of 71.0 pm eject photoelectrons from a gold foil, the electrons originating from deep within the gold atoms. The ejected electrons move in circular paths of radius r in a region of uniform magnetic field. Experiment shows that . We have to find (a) the maximum kinetic energy of the photoelectrons and (b) the work done in removing the electrons from the gold atoms that make up the foil. |
Solution: Click For PDF Version (a) Energy of a photon in terms of its wavelength is given by the relation Therefore, the energy of a photon of wavelength 71.0 pm is Let the maximum speed of the photoelectrons ejected from the gold foil be As these electrons move in the magnetic field B in circular orbits of radius r, we have the condition It is experimentally measured that . Therefore, the maximum speed of the photoelectrons is
Therefore, the kinetic energy of electrons moving with speed of will be (b) And, the work done in removing the electrons from the gold metal is |