00:01
In this exercise, we have photons shining on a metal, and we have the information that for the wavelength of 400 nanometers, the photon can remove electrons from the metal, but with a wavelength of 700 nanometers, the photons cannot remove any electrons.
00:20
So for question a, we are asked if this metal could be cesium that has a work function of, 1 .8 electron volts.
00:34
So the first thing we need to notice is that since 400 nanometers photons remove electrons and 700 nanometers do not, the cutoff wavelength for this matter is between 400 nanometers.
00:54
Lambda c is the cutoff wavelength.
00:57
It's between 400 nanometers and 700 nanometers.
01:01
And the cutoff wavelength is the maximum wavelength that the photons can have so that they expel electrons from the metal.
01:10
We also know that the minimum energy that the photons photons will have to have in order to remove electrons from a metal is equal to the work function and the energy of a photon is given by hc over lambda.
01:26
So from this we can calculate what is the cutoff frequency, okay, the maximum frequency, isolating lambda c in this formula.
01:39
So you're going i have hc over phi.
01:43
So let's calculate the cutoff frequency for the cesium and see if it lies in this range that we've established here.
01:52
Hc is 1 ,140 that's in electron volts nanometers.
01:59
Phi is 1 .8 electron volts so that the cutoff frequency will be 689 nanometer, cutoff, length, i'm sorry.
02:14
It's going to be 689 nanometers...