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The work function for potassium is 2.24eV.
If potassium metal is illuminated with light of wavelength 310 nm,
find the maximum kinetic energy of the photoelectrons.

I have tried everything and still keep getting the wrong answer.

The speed of light is 3 × 108 and Planck’s constant is 6.62607 × 10−34
and the answer suppose to be in units of eV.

i used the equation MaxKE = h*c/ (wavelength x10^-9)

and then subtract (2.24 * 1.602x10^-19)

its not correct apparently, please help

  • Physics -

    i figured it out after all - thanks

  • Physics -

    Please include your units for physical quantities. Numbers are not enough.

  • Physics -

    kmax = hf - e
    kmax = h ( f becomes : c over lambda)- e

    e = ( Evev * 1.6 *10exp -19)

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