Question:
Attempt:
I need to find the theoretical peak for an 88KeV photon being back-scattered. I don't think I am understanding this question correctly.If the detector is placed 180 degrees behind photon scattered on the leg, then we are considering E' where $\theta = \pi$ from this equation:
I believe you can assume the relativistic mass m_o to be m_e the mass of an electron at rest. Even if I don't assume the change in the relativistic mass of the electron given the velocity of the electron to be relatively small, and suppose that all of the 88KeV was transferred to the electron.
$\frac{1}{2} mv^2 = 88KeV = 1.4099 \times 10^{-14} $ Joules
$v=\sqrt{\frac{1.4099 \times 10^{-14}\times2}{9.11\times10^{-31}}}$
$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$
Which resulted in mass increasing by a factor of 1.37, and E' given this change is not effected by much less than 1%, so I am not sure how the spectrum has such a high backscattering energy of 66KeV when my backscattering is pretty much 0. However, I do think the angle of the detector is placed differently for the spectrum, which I have to find later in the question. But I still don't think my theoretical Compton backscattering peak should be 0. Someone please tell me what I am doing so horribly wrong for finding the theoretical peak here. There is also another equation you can use to see the energy of the electron for the maximum electron recoil energy, and then $ E'+ E_{recoil} = E $, to acquire the E' for the photons backscattered energy.
h:plancks constant
v:frequency
m:electron mass, 9.11E-31 kg.
You can use the relation E=hv and replace the hv as 88KeV.