What Is the Minimum Energy of a Photon in a Compton Collision with a Proton?

[vsage]

For some reason I don't feel like I was given all of the information on this question:

A proton at rest is struck by a photon in a Compton collision. If the recoil kinetic energy of the proton is 4.5 MeV, what is the minimum energy of the incident photon (in MeV)? (Take the mass of the proton to be 938 MeV/c2.)

I applied energy conservation but it didn't seem to work. (Yes the title is misleading, I confused it with something else)

 

[dextercioby]

It's an elastic collision.Both energy & momentum are conserved.Write the balance of energy:total energy before=total energy after.

The photon is massless.

Daniel.

 

[vsage]

Alright so then I have by conservation of energy: (subscript 0 = photon initial 1 = proton initial 2 = proton final 3 = photon final)

$$E_0 + E_1 = E_2 + E_3$$
$$p_0c + 938 = 938 + 4.5 + p_3c$$

and conservation of momentum

$$p_0 + p_1 = p_2 + p_3, p_1 = 0$$
$$p_0 = p_2 + p_3$$

$$p_2c$$ can be rewritten in terms of kinetic energy and the mass of a proton as $$p_2c = \sqrt{{KE}^2 + 2KEm_pc^2}$$

and, calculating it out, I get $$p_2c = 92 MeV$$
so now I have $$p_0c = 92 + p_3c$$

What can I do with this? substituting in 92 doesn't make any sense really. I think I did something wrong somewhere. I did assume a 1 dimensional collision though. I could probably work something in 2 dimensions but I don't see the relevance at this point.

 

[dextercioby]

Do you agree that
$$p_{0}c=4.5MeV+p_{3}c$$ ...?

Daniel.

P.S.What does the problem ask you...?

 

[vsage]

Right. The minimum energy I would have thought would have been where the photon has no momentum after collision (and therefore no energy), but $$p_0c = 4.5$$ doesn't appear to be the answer according to this computer script. This is probably such a silly question too.

 

[jtbell]

The minimum energy I would have thought would have been where the photon has no momentum after collision (and therefore no energy)

Uh, if a photon has no energy and momentum, it doesn't even exist!

You also need to use the equation that relates the scattered photon energy to the scattering angle:

$$\frac{1}{E'} - \frac{1}{E} = \frac {1}{m_p c^2}(1 - \cos \theta)$$

where E is the incoming photon energy and E' is the scattered photon energy. You've seen this equation or something like it (probably using wavelength instead of energy), right?

Your desired outcome corresponds to a particular scattering angle in the range of 0 to 180 degrees. Deduce that angle somehow, and together with the energy conservation equation, you have two equations in the two unknowns E and E'.

 

[vsage]

I had originally been using that formula jtbell, but i had been plugging in the proton's energies for some reason. It's good to hear confirmation though!. I'll see what I can do with it by combining that dextercioby said with that. Thanks guys!