### Video Transcript

A copper cathode in a vacuum
chamber is illuminated with light from a laser, causing electrons to be emitted from
the surface of the metal. The light has a frequency of 1.80
times 10 to the 15th hertz. The maximum kinetic energy of the
ejected electrons is 2.80 electron volts. What is the work function of
copper? Use a value of 4.14 times 10 to the
negative 15th electron volt seconds for the value of the Planck constant. Give your answer in electron volts
to three significant figures.

Okay, so in this exercise, we have
a copper cathode. And let’s say that this is that
cathode. It’s in a vacuum chamber being
illuminated, which causes some electrons to be ejected from the copper. We’re told that the frequency of
the incoming radiation, we’ll call it 𝐹, is 1.80 times 10 to the 15th hertz and
that the maximum kinetic energy of the electrons ejected from the copper, we’ll call
it 𝐾𝐸 sub 𝑚, is 2.80 electron volts. Recall that one electron volt is
equal to the amount of energy a charge of one electron gets when it’s moved across
the potential difference of one volt.

Knowing all this, we want to solve
for the work function of copper. To figure this out, we can recall a
relationship between frequency, 𝐹, maximum kinetic energy, 𝐾𝐸 sub 𝑚, and the
work function of a material, called 𝑊. That relationship states that
Planck’s constant ℎ multiplied by the frequency of incoming radiation minus the work
function of a material is equal to the maximum kinetic energy of ejected
electrons. Since we want to solve for the work
function 𝑊, we’ll rearrange this equation by subtracting 𝐾𝐸 sub 𝑚 from both
sides and adding the work function 𝑊 to both sides. When we do that, we find that 𝑊 is
equal to ℎ times 𝑓 minus 𝐾𝐸 sub 𝑚.

Note that a work function 𝑊 is
specific to a particular material. In our case, we’re solving for the
work function of our copper cathode. To help us solve for it, we’re
given the frequency 𝑓 of incoming radiation as well as the maximum kinetic energy
of ejected electrons. We’re also told to treat Planck’s
constant ℎ as 4.14 times 10 to the negative 15th electron volt seconds. So let’s substitute those three
values into this equation. With those values plugged in, let’s
take a look for a second at the units involved.

In our first term, we have electron
volt seconds multiplied by hertz. But hertz, the number of cycles
performed per second, can be replaced with one over seconds. And when we do this, we see that in
multiplying these two values together, that factor of seconds will cancel out. So now, we have one number in
electron volts being subtracted from another number in electron volts. Therefore, our final answer will be
in these desired units. And when we calculate the value of
this expression to three significant figures, we find a result of 4.65 electron
volts. This is the work function of
copper.