0:00
Hi there.
00:01
So for this problem, we have an x -ray photon that is scattered from a free electron address at an angle theta that is 110 degrees relative to the incident direction.
00:17
Now, if the scatter photon has a wavelength that is given, we call this wavelength prime, is equal to 0 .340.
00:31
Nanometers.
00:34
So the question is what is the wavelength of the incident button? so for part let's call this part a we need to find that wavelength for the incident photon.
00:44
So to determine this we use the equation for the compton shift that it states that the difference between the scotter wavelength minus the initial wavelength is equal to plans this divided by the mass of the electron times the speed of light.
01:10
This is a constant that we know it has a value of 2 .43 picometers and this times 1 minus the cosine of the angle given.
01:25
So what we need to do is to simply solve for the wavelength sub -zero of the incident button.
01:33
So we will have that that is equal to the wavelength prime minus 2 .43 picometers times one minus the cosine of the angle theta.
01:45
And now we just simply substitute those values.
01:49
So the wavelength that we are given is 0 .340 nanometers.
01:56
These minus 2 .43 picomeres.
01:57
This minus 2 .43 picomers.
02:01
Meters and this times one minus the cosine of the angle that we are given that is 110 degrees.
02:13
Now remember that nano means tens to the minus 9 and pico means tens to the minus 12.
02:24
So with that said, the wavelength that we obtain in here for the incident photon is equal to 0 .3.
02:45
137 well let's just add more decimal so we will have 6 69 and this in nanometers so that's a solution for part the first part of this problem now for par p we are asked about to determine the energy of the incident now, we know that this energy is, we can obtain it by just the product between plum's constant and the speed of light, and this divided by the wavelength of the incident photon.
03:34
So the product between plumst constant and the speed of light, that is a constant that we know, that is 1 ,240 electron balls times nanometers, and this divided by, 0 .33 669 nanometers...