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jboyd2107
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Homework Statement
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.
Homework 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?)
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.