Electrons are emitted from the surface of a metal when it's exposed to light. This is called the photoelectric effect. Each metal has a certain threshold frequency of light, below which nothing happens. Right at this threshold frequency, an electron is emitted. Above this frequency, the electron is emitted and the extra energy is transferred to the electron. The equation for this phenomenon is KE=hv-hv0 where KE is the kinetic energy of the emitted electron, h=6.63x10^-34 J. s is Planck's constant, v is the frequency of the light, and v0 is the threshold frequency of the metal.
Also, since E=hv, the equation can also be written as KE=E-E0
where E is the energy of the light and E0 is the threshold energy of the metal.
Here are some data collected on a sample of cesium exposed to various energies of light.
Light energy Electron emitted? Electron KE
(eV ) ( eV)
3.87 no __
3.88 no __
3.89 yes 0
3.90 yes 0.01
3.91 yes 0.02
What is the threshold frequency v0 of cesium?
Note that 1 eV (electron volt)=1.60x10^-19 J.
Express answer in hertz