# Inelastic Scattering of alpha particles on a target.

1. Dec 1, 2012

### jboyd2107

1. The problem statement, all variables and given/known data
It is desired to study the low-lying excited states of 35Cl (1.219, 1.763, 2.646, 2.694, 3.003, 3.163 MeV) through the 32S (alpha,proton) reaction. a) With incident alpha particles of 5.000 MeV, which of these excited states can be reached? b) Again with 5.000 MeV incident alphas, find the proton energies observed at 0, 45, and 90 degrees.

2. Relevant equations
Q = Tb*(1+(mb/mB) - Ta*(1-(ma/mB)) - 2((ma/mB)*(mb/mB)*Tb*Ta)^(1/2)*cos(theta)

Q=energy of reaction
Tb=kinetic energy of proton
Ta=kinetic energy of alpha
mb=mass of proton
ma=mass of alpha
mB=mass of 35Cl
theta=angle of proton scattering

Ex = Q(gs) - Q
Q=energy of reaction
Q(gs)=energy of ground state of residual nucleus
Ex=excited state energy (kinetic energy of residual nucleus?)

3. The attempt at a solution

Where to begin.... I'm stumped on this problem. Given the above equations, I feel I need to know the kinetic energy of the proton in order to obtain a certain value for the excited energy of the residual nucleus 35Cl. This leads me to believe that the first part of the question is conceptual and probably simply based on the kinetic energy of the alpha striking the 32S. I have also attempting solving the first part by considering the above reaction as elastic, even though it is not. Some guidance would be helpful.

2. Dec 2, 2012

### Staff: Mentor

You get most energy for the nuclear reaction if the proton moves with the same velocity as the 35Cl afterwards. This should correspond to cos(theta)=1 (your energy Q is negative?)
You can derive Q to find the value of Tb where Q gets minimal/maximal, or use relativistic kinematics to calculate it yourself.

In (b), you know Q (based on the reachable excited states) and have to solve for Tb.

Rough (partially non-relativistic) approximation:
p=5GeV for 36 nucleons gives 140 MeV/c momentum per nucleon or v ≈ 0.15. This corresponds to a kinetic energy of ~380 MeV, so ~4600 MeV are available for the reaction. I would expect that the real value is somewhere between 4000 and 4800, so all excitations are possible.

3. Dec 2, 2012

### jboyd2107

Part B is now very clear to me; however, I did not fully follow the line of thought listed in the bottom of the above comment. The given kinetic energy of the alpha particle beam is 5 MeV, not 5000 MeV. I could be wrong, but I believe part A may have only one answer?

4. Dec 2, 2012

### jboyd2107

does this question belong in the advanced physics forum?

5. Dec 3, 2012

### Staff: Mentor

Oh sorry, I got confused by "." and "," (where I live, they are used the other way round - you can write 1 million as "1.000.000,00" for example).
GeV would be weird as excitation energy anyway.

E=5MeV gives ~0.1c or p≈93 MeV or ~2.6 per nucleon after the reaction, which corresponds to 0.128 MeV kinetic energy. Same result, all excitations are possible.

What do you mean with "one answer"?