00:01
For this exercise, we have the system shown here on the figure, where a photon represented by this blue curvy line is incident on an electron.
00:12
And then the photon scatters off at a certain angle theta from the original path.
00:19
And we know the other exercise tells us that the incident photon had a wavelength of 0 .1 .1 .2.
00:32
148 nanometers, which is the same as 1 .48 times 10 to the minus 10 meters.
00:43
And the scattered photon has a wavelength of 1 .149 nanometers.
00:55
And this is 1 .49 times 10 to the minus 10 meters.
01:01
And the first thing we're asked is to calculate the angle theta.
01:07
So according to compton's, the compton's formula we have that the variation in the lambda in the wavelength is given by h over the mass of the electron times the speed of light, one minus the cosine of the scattering angle.
01:31
And we know the variation of the of the wavelength to be 10 to the minus 12 meters.
01:43
That's, uh, you just have to subtract the scattered angle from the incident angle.
01:48
That's going to be 0 .01 times 10 to the minus 10th.
01:53
And that's 10 to the minus 12 meters.
01:57
This is going to be equal to h over mec.
02:01
That constant is just 2 .43 times 10 to the minus 12.
02:08
It's useful to remind ourselves of this, to always have this constant at hand when working with components scattering, times one minus the cosine of theta.
02:25
The 10 to the minus 12th will cancel out here.
02:28
So we're going to have 1 over 2 .43 equals to 1 minus cosine of theta.
02:38
So we're going to have that the cosine of theta is 1 minus 1 over 2 .43.
02:51
And this is 0 .59...