0
$\begingroup$

In this undergrad nuclear physics problem I am asked to find the kinetic energy threshold for an electron colliding with a still proton to create an electron-positron pair. So in short: $$e^- + p^+ \rightarrow e^- + e^+$$

First I am not sure whether this reaction can even happen in real life. Knowledge expected from me at this point is to apply 4-momentum conservation laws.

So I approached it with a standard procedure: I wrote the 4-momentum conservation law, and then squared it. $P^2$ is the same in any reference frame, so I chose the lab reference frame for the left side ($e^- + P^+$) and the center-of-mass reference frame for the $e^--e^+$ pair. And to get the threshold energy for incoming electron, I just chose the kinetic energy of the resulting pair in their center of mass frame to be zero.

The general result for kinetic energy threshold for reaction type $a+b \rightarrow c+d$ is: \begin{equation} \label{gen} T_a^{threshold} = \frac{(m_c+m_d)^2c^2-(m_a+m_b)^2c^2}{2m_b} \end{equation}

The result I got was in line with the previous one: $$T_e^{threshold} = \frac{3m_e^2-m_p^2-2m_em_p}{2m_p}$$

Note that the result $$T_e^{threshold} < 0$$

My questions are:

  • How do I interpret the result?
  • Is the general formula for $T_a^{threshold}$ valid for any value of masses $m_a,m_b,m_c,m_d$, ie. even if it yields $T_a^{threshold} < 0$?
  • Is this reaction even possible, or it is merely a dummy reaction with a sole purpose of illustrating the calculation?
$\endgroup$
5
  • 3
    $\begingroup$ It violates baryon and lepton number conservation. $\endgroup$
    – anna v
    Commented Jul 4, 2022 at 14:14
  • $\begingroup$ the negative energy threshold arises because you made a 938 GeV proton turn into a 511 keV positron. $\endgroup$
    – JEB
    Commented Jul 4, 2022 at 15:53
  • $\begingroup$ @JEB So calculation is sound, were it not for the fact that the reaction is altogether impossible, as anna v explained. But is there such a case where the threshold energy is negative for a real possible reaction? $\endgroup$
    – lojle
    Commented Jul 4, 2022 at 16:14
  • 1
    $\begingroup$ Surely your assignment is $\rm e^- + p^+ \to e^- + p^+ + e^-e^+$? $\endgroup$
    – rob
    Commented Jul 4, 2022 at 19:04
  • $\begingroup$ @rob Yeah, now I'm pretty sure it is. But it still might be possible that: $$T_a^{threshold} = \frac{(m_c+m_d)^2c^2-(m_a+m_b)^2c^2}{2m_b}<0$$ in a reaction? On the other hand there might be a law I'm currently unaware of that forbids this... $\endgroup$
    – lojle
    Commented Jul 4, 2022 at 19:14

1 Answer 1

3
$\begingroup$

As discussed in the comments, your reaction

$$ \rm e^- p^+ \not\rightarrow e^-e^+ $$

violates the conservation of baryon number and lepton number, and is therefore forbidden. Your homework assignment is almost certainly about

$$ \rm e^- p^+ \to e^-p^+e^-e^+ $$

which has a positive threshold energy.

However, you clarify that your real question is about whether it's possible to have a reaction whose threshold energy is negative, and how you would interpret such a system. The answer is yes, it's possible, and the meaning is that the reaction occurs at zero interaction energy.

For example, consider the capture of a free neutron on some other nucleus $^A Z$ with proton number $Z$ and mass number $A$. For nuclei which are not on the neutron drip line, the bound nucleus $^{A+1}Z$ has less total mass than the unbound system $^AZ+\rm n$. (Proof: if the unbound system were less massive, the nucleus could decay by neutron emission.) Consider a capture which results in a photon emission,

$$ \rm n + {}^1H \to {}^2H + \gamma + 2.2\,MeV $$

or in nucleon or cluster emission,

\begin{align} \rm n + {}^{14}N &\to \rm {}^{14}C + p \\ \rm n + {}^6Li &\to \rm{}^3H + {}^4He \end{align}

For each of these, you can verify that the total mass on the right-hand side is less than the total mass on the left-hand side. The observable effect is that these reactions can take place when the kinetic energy on the left-hand side is zero. All of these reactions take place with thermal neutrons, with milli-eV kinetic energies, even though the energy released in the reaction is mega-eV.

In these negative-threshold reactions, we talk about the threshold of the reverse reaction. For instance, some of the literature on the low-energy mass-two nuclear system is based on neutron capture on hydrogen, but other literature talks about "threshold photodissociation of deuterium," which means

$$ \rm \gamma + {}^2H \to {}^1H + n $$

where the photon energy is close to the 2.2 MeV minimum.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.