Video Transcript
What is the critical frequency of a metallic surface that has a work function of 4.41
times 10 to the power negative 19 joules? Use ℎ equals 6.625 times 10 to the power negative 34 joule-seconds.
To begin, let’s recall that the critical frequency, also known as the threshold or
cutoff frequency, is a property of a metal. It gives the minimum frequency, represented by 𝑓 naught, of an incident photon that
can overcome that metal’s work function and induce the photoelectric effect. We should also recall the formula 𝑊 equals ℎ𝑓 minus 𝐸 max, where 𝑊 is the work
function of the metal surface, ℎ is the Planck constant, 𝑓 is the frequency of the
incident photon, and 𝐸 max is the maximum kinetic energy of a photoelectron.
Now, at the threshold frequency, 𝐸 max is equal to zero, since the energy of the
incident photon is barely enough to overcome the work function. Let’s modify the formula to reflect this. We can solve this for the threshold frequency by simply dividing both sides by ℎ so
that the expression reads 𝑓 naught equals 𝑊 over ℎ.
Here, we’ve been told that a metal surface has a work function 𝑊 equal to 4.41 times
10 to the power negative 19 joules. And we also know the value of the Planck constant ℎ. Since both of these values are expressed in their appropriate SI-derived units, we
know that this formula will give us a frequency value in proper units of hertz.
So we’re ready to substitute these terms into the formula to solve for the threshold
frequency. Doing this and grabbing a calculator, we get a result of 6.6566 and so on times 10 to
the power 14 hertz. Choosing to round this to two decimal places, we have our final answer. We’ve found that the threshold, or critical, frequency of this metallic surface is
6.66 times 10 to the power 14 hertz.
