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

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In summary, the nuclear reaction ^63_29Cu(p,n) ^63_30Zn is immediately followed by beta plus decay, with a maximum energy of 2.36 MeV for the positrons. The Q value for the original (p,n) reaction can be found by subtracting the Q value for the beta plus decay from 2.36 MeV. This additional energy can be accounted for by the kinetic energy of the proton and the small mass of the neutrino.
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Brewer
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Homework Statement


The nuclear reaction [tex]^63_29[/tex]Cu(p,n) [tex]^63_30[/tex]Zn is immeadiately followed by [tex]/Beta^+[/tex]. Write the equation for this [tex]/Beta^+[/tex] decay.

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


Homework Equations


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


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 [tex]Q_{tot} = Q_1 + Q_2[/tex] 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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  • #2
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:

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

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?
 

What is the Q value for ^63_29Cu(p,n) ^63_30Zn?

The Q value for this reaction can be calculated using the mass difference between the reactants and products, according to the formula Q = (m_initial - m_final) * c^2, where m is the mass and c is the speed of light.

How do I determine the initial and final masses for this reaction?

The initial mass is the mass of the proton, which is approximately 1.0073 atomic mass units (amu). The final mass is the combined mass of the ^63_29Cu nucleus and the emitted neutron, which can be calculated by adding the mass of the ^63_29Cu nucleus (62.9296 amu) to the mass of the neutron (1.0087 amu).

What is the speed of light used in the Q value calculation?

The speed of light used in this calculation is the standard value of 2.998 x 10^8 meters per second (m/s).

Can the Q value be negative?

No, the Q value cannot be negative as it represents the energy released in the reaction, which is always a positive value. If the calculated Q value is negative, it indicates that the reaction is not energetically favorable.

How accurate is the Q value calculation?

The Q value calculation is typically accurate to within a few decimal places, depending on the precision of the mass values used. However, it is important to note that this is a theoretical calculation and the actual Q value may differ slightly in experimental measurements due to other factors such as nuclear binding energies and reaction energies.

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