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Let us take a nuclear reaction as follows: $$\rm _7^{16}N +{}_2^4He \rightarrow {}_8^{19}O+{}_1^1H$$

Now the question is that we have to find the minimum kinetic energy of the Helium atom for the nuclear reaction to occur. Given are the masses of the atoms/nuclei.

Now I can easily calculate the $Q$ value of the reaction by the mass defect formula, which I had calculated. However, I am stuck here, as I can't seem to figure out what to do next after obtaining the $Q$ value, how can I use it to find kinetic energies?

One interesting case is that my calculation for $Q$ value yielded a negative value which suggests that it is an endothermic reaction, so I wonder if its of significance. Please help here, every help is massively appreciated.

EDIT: So I have understood the significance of negative value of $Q$ and subsequent requiremnet of kinetic energy of the bombarding nucleus, but I still cant figure out the actual calculation of the kinetic energies.

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  • $\begingroup$ Now, this was a great question. $\endgroup$
    – Cantor Dust Drachen
    Commented May 1, 2023 at 8:54

2 Answers 2

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A reaction where the $Q$-value is positive can happen even if there is zero kinetic energy in the initial state. For example, most neutron capture reactions can proceed with thermal neutrons, whose energies (milli-eV) are negligible compared to the energies (mega-eV) associated with the nucleon exchange.

A reaction whose $Q$-value is negative can only happen if there is enough energy in the initial state for energy to be conserved, with nonnegative kinetic energy, in the final state. Compare with the photoelectric effect, where high-energy photons can induce electron emission, but photons with energy below the “work function” cannot. The work function, as a cost of extracting an electron from a metal, is analogous to the magnitude of a negative $Q$-value in a kinematic reaction like the one you have described here.

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  • $\begingroup$ Ok, so I understand why kinetic energy is required but how would I calculate given that I only know masses of atoms and subsequent $Q$ value? $\endgroup$
    – Kshitij Kumar
    Commented Apr 30, 2023 at 17:16
  • $\begingroup$ Suppose the kinetic energy in the final state is zero. (This is sometimes called a “threshold reaction.”) Using the masses and conservation of energy, what is the kinetic energy in the initial state? $\endgroup$
    – rob ♦
    Commented Apr 30, 2023 at 18:26
  • $\begingroup$ Kinetic energy of helium nuclues? $\endgroup$
    – Kshitij Kumar
    Commented Apr 30, 2023 at 18:35
  • $\begingroup$ @KshitijKumar If the final kinetic energy is zero, we are in the center of momentum reference frame, and the initial kinetic energy is shared between the two nuclei. You can assign it all to the lighter nucleus if you are willing to approximate. Since the ratio of the momenta goes like the ratio of the masses, assigning all of the momentum to the helium is like taking the limit that $14\gg 4$, which is not great but also not terrible. $\endgroup$
    – rob ♦
    Commented May 1, 2023 at 1:48
  • $\begingroup$ Oh, I see we need to use centre of mass frame and we can use kinetic energy as $1/2\mu v_{rel}^2$ as initial kinetic energy, now what can I do? $\endgroup$
    – Kshitij Kumar
    Commented May 1, 2023 at 3:47
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Ok so on suggestions of @rob, I have tried to do as follows:

Let us take the Centre of mass frame, and let take $V_{cm}=0$, so: $$K_{cm}=\frac{1}{2}\mu v_{rel}^2$$

Now let us take oxygen nuclei and hydrogen nuclei to be at rest after collision for threshold energy,:

$$|Q|=K_{cm}$$ $$|Q|=\frac{M_{N}M_{\alpha}}{2(M_{N}+M_{\alpha})}v_{rel}^2$$ $$\left(1+\frac{M_{\alpha}}{M_N}\right)|Q|=\frac{1}{2}M_{\alpha}v_{rel}^2=K_{threshold}$$

I would like anyone to review my work, to see if its correct or not?

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