A metal surface with a work function of 2.0 eV is illuminated by a beam of photons, each having a momentum of 1.0 10⁻²⁷ kg m/s . If the intensity of the incident beam is then increased by a factor of 5 , what will be the maximum kinetic energy of the emitted photoelectrons? (Given: c = 3 10^8 m/s , 1 eV = 1.6 10⁻¹⁹ J )

Step-by-step solution

The energy E of a single photon can be calculated from its momentum p using the relation E = pc . Given p = 1.0 10⁻²⁷ kg m/s and c = 3 10^8 m/s : E = (1.0 10⁻²⁷) (3 10^8) = 3.0 10⁻¹⁹ J Converting this energy into electron-volts (eV): E = 3.0 10⁻¹⁹ 1.6 10⁻¹⁹ eV = 1.875 eV The work function of the metal is = 2.0 eV . Since the incident photon energy ( 1.875 eV ) is less than the work function ( 2.0 eV ), no photoelectric emission will take place. Increasing the intensity of the incident beam by a factor of 5 only inc