Question

Background:


Electrons are emitted from the surface of a metal when it's exposedto light. This is called the photoelectric effect. Each metal has acertain threshold frequency of light, below which nothing happens.Right at this threshold frequency, an electron is emitted. Abovethis frequency, the electron is emitted and the extra energy istransferred to the electron.

The equation for this phenomenon is

KE = hv - hv_0

where KE is the kinetic energy of the emitted electron, h= 6.63times 10^{-34}J * s is Planck's constant, v is the frequency of thelight, and v_0 is the threshold frequency of the metal.

Also, since E = hv, the equation can also be written as

KE = E - E_0

where E is the energy of the light and E_0 is the threshold energyof the metal.

Question:

Here are some data collected on a sample of cesium exposed tovarious energies of light.

Light energy
eV Electron emitted? Electron KE
eV
3.87 no —
3.88 no —
3.89 yes 0
3.90 yes 0.01
3.91 yes 0.02

Part A
What is the threshold frequency v_0 of cesium?

Note that 1eV (electron volt) = 1.60\times 10^{-19}~J.
Express your answer in hertz. v_0 = ___________Hz