# Photoelectric Effect (Physics 420)

In a photoelectric effect electrons of kinetic energy 2 eV are ejected from a material having a work function of 3.2 eV.

a) What is the frequency of the photons?

a) What is the speed of the electrons that are ejected?

b) What is the kinetic energy of the electrons if the energy of the photons is increased by a factor of two?

$Photon energy = Work + Kinetic energy$

$h*F = 3.2 +2 =5.2 eV =5.2*1.6*10^{-19} =8.32*10^{-19} J$

$F = 1.256*10^{15} Hz$

$m*v^2/2 =Ec$

$V = sqrt(2Ec/m) =8.38*10^5 m/s$

$2h*F = W +Ec2$

$h*F = W+ Ec1$

$Ec2-Ec1 = h*F$

$Ec2 = h*F+Ec1 =5.2 +2 =7.2 eV =11.52*10^{-19} Joule$