Scattering of a Neutron and Proton

[physnut_05]

I have a problem here that has me a bit suck:

In scattering between incident neutrons and target protons at rest, show that the angle between the two scattered particles is always 90 degrees.

I have set up the conservation of energy and momentum equations and have:

Energy: T_n=T'_n+T_p
Momentum: p_n=p'_n*cos(X)+p_p*cos(Y) AND p'_n*sin(X)=p_p*sin(Y)

Here X and Y are the angles between the scattered neutron and proton with respect to the horizontal.

I've manipulated these around, found an expression for cos(X), but that doesn't seem to get me much of anywhere. A bump in the correct direction would be appreciated.

 

[bcrowell]

This is a classic cute fact about kinematics. You can verify it by making one coin collide with another on a tabletop. Note that it does not claim that the neutron is always scattered at 90 degrees relative to its initial direction of motion; it claims that the final velocity vector of the neutron is at 90 degrees with respect to the final velocity vector of the proton. For that reason, it's not particularly helpful to solve for the angle you're notating X, because what you're really trying to prove is that |X-Y|=90.

The whole thing comes out much more nicely if you do it without imposing a coordinate system at all. Just define notation for three vectors, and write both conservation laws purely in terms of those vectors, with no operations other than vector addition and vector dot products. Use the graphical interpretation of the vector addition as a triangle formed by the vectors.

The reason this problem can be stated in terms of neutron-proton scattering is that the masses are very nearly equal, and there are no internal degrees of freedom that can suck up any energy (assuming that the beam energy is fairly low). It would fail, for example, if you were scattering neutrons off of gold nuclei, since the gold nucleus could be left in an excited state.

 

[physnut_05]

Thanks for the help, it does make a lot more sense doing it that way.