00:01
Hi there, so for this problem we are told that photons of energy of 0 .2 mega electron balls, so the initial energy for these photons is 0 .2 mega electron balls are being scattered by three electrons that are initially at rest.
00:21
Now, for part a of this problem, we are asked about at what angle must the scattering occurs, so that the photon lose 1 and 10 % of its energy, of its initial energy.
00:38
So for this we need to use quantum scattering formula that states that the difference between the wavelength of the scattered photon minus the wavelength of the initial incident photon is equal to plums constant divided by the mass of an electron times the speed of light.
00:57
And this times 1 minus the cosine of the angle theta.
01:04
Now, in this case, what we need to determine is this angle of scattering theta.
01:13
Now, with that said, we are going to first write the wavelengths in terms of the energy.
01:21
Now, recall that the energy is related to the wavelength by just simply plants constant divided times the speed of light divided by the wavelength.
01:30
That is the energy of a photon.
01:32
So the initial energy is just plams constant times the speed of light, this divided by the initial wavelength.
01:41
So solving for the wavelength, we will obtain plams constant, the speed of light, and this divided by the initial wavelength.
01:48
So with that said, we will have that the difference in the wavelength can be written as just plums constant, the speed of light, we can take out that as a common term.
02:02
This 1 divided by the energy of the scattered phantom that we're gonna call you simply as e, and then this minus the initial wavelength, that is 1 divided by this.
02:17
And now what we need to do is also to write the energy of the scattered photon because we are told that the photon lose 10 % of its energy.
02:28
So that means that it's cutter photon is it has the energy, the initial energy, minus that, a 10 % of that, so that will be 0 .1 times 0 .2 mega -electron bolts.
02:43
So, of course, this will give us a value of 0 .18 megalletron volts.
02:58
Now, well, with that said, we know that this is equal by the content scattering formula, plam constant divided by the mass of the electron times the speed of light, this times one minus the cosine of theta.
03:14
So now, solving for the cosine of theta, we will obtain that that is equal to one minus the mass of the electron times the speed of light square, and this times one divided by the energy minus one divided by the initial energy.
03:32
So first, we're going to transform the energies that we are given from mega electron balls to joules.
03:43
So for the first one, for the initial energy, we'll have 0 .2.
03:48
Mega electron balls.
03:50
So mega means 10 to the 6.
03:53
And we know that one electron balls is equal to 1 .6 times 10 to the minus 19 joules.
04:02
So from this, we obtain a value of 3 .2 times 10 to the minus 14 joules.
04:18
And now we do the same for the energy of the scattered photon.
04:21
That is 0 .18 times 10 to the 6 electron balls.
04:25
That times 1 .6 times 10 to the minus 19 joules per one electron bowl...