The first ionization potential of potassium is 419 kJ/mol. If UV-?radiation with wavelength ? = 250 nm is used to eject electrons from potassium, the energy of each photon in excess of the ionization potential provides kinetic energy to escaping electrons. The mass of an electron is 9.109 x 10-?31 kg. What will be the velocity of the escaping electrons under these conditions? What is the de Broglie wavelength of the escaping electrons?© BrainMass Inc. brainmass.com October 10, 2019, 5:49 am ad1c9bdddf
E = hc/wavelength = (6.625 x 10^-34 Js x 3 x 10^8 m/s ) / (250 x 10^-9 m) = 7.95 x 10^-19 J per photon
Per mole of photons E = 7.95 x 10^-19 J x 6.023 x 10^23 = ...
The expert examines escaping electrons for first ionization potential of potassium.