Compton Scattering: Determining the Energy of a Scattered Gamma Ray

AI Thread Summary
To determine the energy of a scattered gamma ray in Compton scattering, the equation E(gamma prime) = E(gamma)/(1 + (E(gamma)/(mc^2)(1 - cos(theta))) is appropriate. The energy of the gamma ray, 0.662 MeV, should be kept in MeV rather than converted to joules for simplicity. The mass of the electron must also be expressed in MeV/c^2 to maintain consistency in units. Using MeV avoids the complexity of converting to SI units. This approach streamlines the calculation process for the energy of the scattered gamma ray.
ChrisWM
Messages
13
Reaction score
1
Homework Statement
A 0.662 MeV gamma ray Compton scatters from an electron at an angle of 60°. What is the energy of the scattered gamma ray? (Gamma rays are photons and are treated identically to x-rays in the analysis of Compton scattering.)
Relevant Equations
E(gamma prime)= E(gamma)/(1+(E(gamma)/(mc^2)(1-cos(theta))) ?
I'm unsure of how to proceed here. Would I use the equation

E(gamma prime)= E(gamma)/(1+(E(gamma)/(mc^2)(1-cos(theta))) ?

Also, do I keep the .662 Mev as is or do I convert to joules?
 
Physics news on Phys.org
ChrisWM said:
Homework Statement:: A 0.662 MeV gamma ray Compton scatters from an electron at an angle of 60°. What is the energy of the scattered gamma ray? (Gamma rays are photons and are treated identically to x-rays in the analysis of Compton scattering.)
Relevant Equations:: E(gamma prime)= E(gamma)/(1+(E(gamma)/(mc^2)(1-cos(theta))) ?

I'm unsure of how to proceed here. Would I use the equation

E(gamma prime)= E(gamma)/(1+(E(gamma)/(mc^2)(1-cos(theta))) ?

Also, do I keep the .662 Mev as is or do I convert to joules?
If you already have an equation for the change in energy, then why not use it?

Definitely use ##MeV##. Note that ##\frac{E}{mc^2}## is dimensionless. If you have ##E## in ##MeV##, then you need ##m## in ##MeV/c^2##. Which is much simpler than converting everything to SI units in this case.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top