Nuclear Reaction: 6.61 MeV Protons Incident on 27/13 Al

[jjson775]

Homework Statement

A beam of 6.61 MeV protons is incident on a target of 27/13 Al. Those that collide produce the reaction:

p + 27/13 Al —-> 27/14 Si + n

27/14 Si has mass 26.986 721 u. Neglecting any recoil of the product nucleus, determine the kinetic energy of the emerging neutrons.

Relevant Equations

Q = (Ma + Mx -My - Mb) c^2

 

[bobob]

Have you looked in the table of mass excesses for the nuclei in the reaction to determine the Q-value?

Edit: I just typed table of mass excesses into google, opened the pdf and within the first couple of pages, there is an example of exactly what you need to do.

 

[jjson775]

I haven’t had a problem working with atomic masses for binding/disintegration or reaction energy. I don’t know how to deal with the energetic proton In this particular problem.

 

[bobob]

OK, you have the expression for the Q value. Do you have the mass of $^{27}_{13}Al$? What is the meaning of the Q value? What does the incoming proton add to all of this besides its mass? If the $^{27}_{14}Si$ recoil energy is zero, where does the kinetic energy from the proton go?

 

[jjson775]

Using the mass only of the proton, Q is close to zero. So the kinetic energy of the neutron should be very close to the kinetic energy of the proton. Not the right answer.

 

[bobob]

You have 4 masses: $p,~n,~^{27}_{13}Al,~^{27}_{14}Si$. You have to add energy to the $Al$ nucleus to make the $Si$ nucleus, so some of the proton energy goes into making the reaction happen. What are you using for the masses of the two nuclei? How much of the proton energy goes into the binding energy of the $Si$ nucleus? Hint: If the proton energy is just at the threshold for the reaction to occur, how much energy will the neutron have?

 

[jjson775]

I think I understand the phenomena but can’t get the right answer and want to make sure I haven’t missed something important. The answer to your hint question is zero. Using a mass of 26.981538 for Al and the mass in the homework statement for Si, I get Q = -6.18 and neutron energy of 0.43 MeV. The answer is supposed to be 1 MeV. See picture attached. The energy of the neutron is the energy of the proton less what is required to make the reaction happen. Thanks for your patience.