The work function for potassium is 2.24 eV. If potassium metalis illuminated wit

Question

The work function for potassium is 2.24 eV. If potassium metalis illuminated with light of wavelength 280 nm
a) find the maximum kinetic energy of the photaelectrons. The speedof light is 3 e8 m/s and Planck's constant is 6.62607e-34 J*s.Answer in units of eV. b) find the cutoff wavelength. Answer in units of nm. The work function for potassium is 2.24 eV. If potassium metalis illuminated with light of wavelength 280 nm
a) find the maximum kinetic energy of the photaelectrons. The speedof light is 3 e8 m/s and Planck's constant is 6.62607e-34 J*s.Answer in units of eV. b) find the cutoff wavelength. Answer in units of nm.

Explanation / Answer

From Einstein's photo electric equation we have :               h f    = W   + 1/2 mv2 Where   W = Work function = 2.24 eV                                                = 2.24 x 1.602 x 10-19   J                                              =   3.588   x 10-19   J                              hf    = h c /                   = (6.62 x 10-34 * 3 x 108 ) / 280 x 10-9                     =   0.0709 x 10-17                    = 7.09 x 10-19   Hz (a)               Kinetic energy is :                        K.E    =    h f   - W                                 = 7.09   x 10-19     - 3.588   x 10-19                                   =   3.502   x 10-19   J (b)          Cutoff wavelength is :                       o     = h c / W                                = (6.62 x 10-34 * 3 x 108 ) / ( 2.24  x 1.602 x 10-19   J )                                =   ---------   Solve it. Hope this helps u!

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