Question

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=hν−hν0KE=hν−hν0

where KEKEKE is the kinetic energy of the emitted electron, h=6.63×10−34J⋅sh=6.63×10−34J⋅s is Planck's constant, ννnu is the frequency of the light, and ν0ν0nu_0 is the threshold frequency of the metal.

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

KE=E−ϕKE=E−ϕ

where EEE is the energy of the light and ϕϕphi is the binding energy of the electron in the metal.

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.60×1015Hz1.60×1015Hz ?

Express your answer in joules to three significant figures.

Answer 1