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.