Asked by Physics Student
The work function energy of potassium is 2.2 eV.
The surface of potassium is illuminated with electromagnetic radiation of a specific wavelength.
Experiments show that the photoelectrons from potassium have kinetic energy in the range zero to 1.9 eV.
Determine the wavelength of the incident electromagnetic radiation.
The surface of potassium is illuminated with electromagnetic radiation of a specific wavelength.
Experiments show that the photoelectrons from potassium have kinetic energy in the range zero to 1.9 eV.
Determine the wavelength of the incident electromagnetic radiation.
Answers
Answered by
bobpursley
work function= 2.2-1.9 ev
remember E=hc/lambda
energy work funciton= .3 ev= .3*(1Joule / 6.24•1018eV)
so lambda= speedlight/energyWorkFuncionInJoules
remember E=hc/lambda
energy work funciton= .3 ev= .3*(1Joule / 6.24•1018eV)
so lambda= speedlight/energyWorkFuncionInJoules
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