00:01
In this exercise, we have photons with an energy of 4 kiloelectron volts, scattering of electrons, according to the compton scattering.
00:10
And we want to know what is the biggest difference in the photon in energy.
00:16
That is, the energy of the initial photon minus the energy of the final photon, the scattered photon, divide as a fraction of the incident photon's energy.
00:29
So this is what we're going to calculate.
00:31
Notice that if we want delta e equals to e minus e prime to be maximum, then e prime has to be minimum.
00:43
We have to calculate what is the minimum energy of the scattered photon.
00:48
Now remember that the energy of a photon is given by ahc over lambda.
00:54
Here i'm going to use lambda prime for the wavelength of the scattered photon.
01:00
So that if we want the smallest energy e -prime, we're looking for the biggest wavelength lambda prime.
01:11
Now, according to compton's formula, we have that the wavelength of the scattered photon equals the wavelength of the incident photon, plus h over mcm is the mass of the electron, 1 minus cosine theta.
01:29
Notice that the angle for which the for which lambda prime is going to be the largest is 100 theta equals 180 degrees.
01:43
Because when that's true, when theta equals 108 degrees, the cosine of theta is minimum, which is minus 1.
01:52
And we're going to have lambda prime equals lambda plus 2h over mc.
02:01
The biggest wavelength possible...