Calculating the Q Value for ^63_29Cu(p,n) ^63_30Zn

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SUMMARY

The discussion centers on calculating the Q value for the nuclear reaction ^63_29Cu(p,n) ^63_30Zn, followed by a beta-plus decay. The Q value for the entire process is determined to be 2.36 MeV, which is attributed to the energy of the emitted positron. The formula used for calculating the Q value is Q=[M(A,Z) - M(A,Z-1) - 2m_e]c^2, but adjustments must be made to account for the kinetic energy of the incoming proton. The total reaction energy is primarily contained in the positron, while the contributions from the neutrino and the copper nucleus are negligible.

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Brewer
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Homework Statement


The nuclear reaction ^63_29Cu(p,n) ^63_30Zn is immeadiately followed by /Beta^+. Write the equation for this /Beta^+ decay.

If the maximum energy of the positrons is 2.36 MeV, find the Q value of the original (p,n) reaction.


Homework Equations


Q=[M(A,Z) - M(A,Z-1) -2m_e]c^2


The Attempt at a Solution


Decay chain:
First: Copper + proton ---> Zinc + neutron
Second: Zinc ---> Copper + positron + neutrino

So Zinc can be substituted into the first equation to give that:
proton ---> neutron + positron + neutrino

Now I get confused. I pulled the above equation from my notes, and I know that its the energy of the system, but other than that I don't understand.

I've spoken to my lecturer about it, but he didn't make much sense with it so I'm hoping one of you could help me better with it.

As I understand it Q_{tot} = Q_1 + Q_2 where Q_2 is the second decay, given by the previous equation, but can be assumed to be the max KE of the positrons (as the neutrion is massless etc). But then I get stuck, as I can't see how I'd find the total energy of the system, much less the energy of the original (p,n) reaction.

Could someone point me in the right direction?

Thanks

Brewer
 
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Ok, first let's understand the energetics of what is going on. In your original reaction you are shooting protons at copper nuclei and this is causing a proton capture reaction. You could calculate the Q value for this reaction using masses only if the proton didn't carry kinetic energy. Since the proton must be carrying a fair amount of kinetic energy to overcome the Coulomb barrier (positive charges repel), simply using:

Q=[m_{Zn}+m_n-m_{Cu}-m_p]c^2

will not work.

Secondly, we know that the energy of the positron is 2.36 MeV. Since the big copper nucleus is unlikely to carry away much kinetic energy (because of its big mass), and the neutrino is unlikely to carry away much kinetic energy (because of its small mass) we can assume (as you already stated) that ALL of the total reaction energy is found in the positron. The final energy is the Q value for the whole process. Thus, the Q value for the total reaction is 2.36 MeV.

Next, you have a formula for the regular Q-value of the beta plus decay. Compute it. It should be smaller than 2.36 MeV by some amount. Where could this energy have come from?
 

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