This should be easy, but I think I have a mind-block...

For $\beta^-$-decay, what is the maximum possible momentum for the electron? The two equations I can use are conservation of energy and conservation of momentum, but I have three unknowns: Momentum of electron, nucleus and anti-neutrino, so what am I missing?

Could I just set the kinetic energy of the nucleus to zero and work from there?

  • $\begingroup$ Unless you specify a reference frame, the momentum can be anything... it's conventional to analyze decays in the CM frame of the original particle. $\endgroup$ – David Z Aug 13 '11 at 23:12
  • $\begingroup$ I'd be interested in the lab frame... $\endgroup$ – Lagerbaer Aug 13 '11 at 23:41
  • $\begingroup$ In that case you have an underdetermined problem, because "lab frame" doesn't mean anything until you define it (e.g. by specifying the momentum of one of the particles in the lab frame). $\endgroup$ – David Z Aug 13 '11 at 23:49
  • $\begingroup$ And if the entire nucleus starts out with zero momentum? $\endgroup$ – Lagerbaer Aug 14 '11 at 0:38
  • 1
    $\begingroup$ That would be the CM frame of the original particle. But you do have to use the fact that $p_i = 0$ in the math. $\endgroup$ – David Z Aug 14 '11 at 2:06

Answer of Zassounotsukushi is correct, I'd like to expand it

consider center of mass frame, a resting neutron for simplicity. four-momentum is $\underline{p}_n=\binom{m_n}{\vec{0}}$. neutron decays into proton, electron, antineutrino: $n \rightarrow p^+ e^- \bar{\nu}$ $$ \binom{m_n}{\vec{0}}=\binom{E_p}{\vec{p}_p}+\binom{E_e}{\vec{p}_e}+\binom{E_\nu}{\vec{p}_\nu} $$ neutrino mass $(m_\nu < 0.2eV)$ is negligible $\Rightarrow E_\nu=|\vec{p}_\nu|$, to maximize electron energy neutrino has to be left without any $\Rightarrow E_\nu=0$ $$ \Rightarrow\binom{m_n}{\vec{0}}=\binom{E_p}{\vec{p}_p}+\binom{E_e}{\vec{p}_e} $$ for momentum conservation to be satisfied , $\vec{p}_p+\vec{p}_e=\vec0 $, proton momentum has to be $\vec{p}_p=-\vec{p}_e$ $$ \Rightarrow\binom{m_n}{\vec{0}}=\binom{E_p}{-\vec{p}_e}+\binom{E_e}{\vec{p}_e} $$ Energy-momentum relation can be used: $E_p=\sqrt{p_p^2+m_p^2}=\sqrt{p_e^2+m_p^2}$, $E_e=\sqrt{p_e^2+m_e^2}$

energy conservation: $$ \Rightarrow m_n = \sqrt{p_e^2 + m_p^2} + E_e $$ $$ \Rightarrow m_n = \sqrt{E_e^2 - m_e^2 + m_p^2} + E_e $$ $$ \Rightarrow (m_n - E_e)^2 = E_e^2 - m_e^2 + m_p^2 $$ $$ \Rightarrow E_{e,max} = \frac{m_n^2 - m_p^2 + m_e^2}{2m_n} $$ $$ \Rightarrow p_{e,max} = \sqrt{\left(\frac{m_n^2 - m_p^2 + m_e^2}{2m_n}\right)^2 - m_e^2} $$

the "mass defect" in transition $n\rightarrow p$ is in the term $m_n^2 - m_p^2$ included

I assumed $c=\hbar=1$

Edit after the question by AAAAAA:

if the proton is at rest, similar consideration about the momentum conservation leads to: $$ \Rightarrow\binom{m_n}{\vec{0}}=\binom{m_p}{0}+\binom{E_\nu}{\vec{p}_\nu}+\binom{E_e}{\vec{p}_e} $$ Now the $E_\nu$ has to be substituted approximately by $E_\nu\approx|\vec{p}_\nu|$ and thus $E_\nu\approx|\vec{p}_\nu| = |-\vec{p}_e| = \sqrt{E_e^2 - m_e^2}$

Energy conservation gives:

$$ \Rightarrow m_n = m_p + \sqrt{E_e^2 - m_e^2} + E_e $$ $$ \Rightarrow (m_n - m_p) - E_e = + \sqrt{E_e^2 - m_e^2} $$ with $\Delta m:=(m_n - m_p)$ $$ \Rightarrow (\Delta m - E_e)^2 = E_e^2 - m_e^2 $$ $$ \Rightarrow -2\Delta m E_e + \Delta m^2 = - m_e^2 $$ $$ \Rightarrow E_e = \frac{m_e^2 + \Delta m^2}{2\Delta m} $$

This results in a smaller electron energy (and momentum): $E_{e, \textrm{proton at rest}} = 748 keV$ while $E_{e, \textrm{neutrino at rest}} = 1.293 MeV$ above.

Maybe it's in limits valid to imagine that it's easier for the electron to "bounce off" a massive proton that off a light neutrino.

  • $\begingroup$ Still processing this, but I'm having trouble. Why do you say the neutrino mass is negligible? Isn't $m_n$ about equal to $m_p$? Yes, then $m_p\gg m_e$, but this doesn't seem consistent with the set of assumptions you employed. That is, of course, if I'm reading this correctly. $\endgroup$ – Alan Rominger Aug 14 '11 at 18:46
  • $\begingroup$ neutrino's mass is some $eV$-s, $m_e\approx511keV$ - about $10^5$ times larger. neutron mass is about $1.3 MeV$ larger than that of a proton, so the difference is large in comparison to electron mass. one can improve the above calculation by including the neutrino rest mass, but there is no data on it $\endgroup$ – IljaBek Aug 14 '11 at 19:00
  • 1
    $\begingroup$ Last line of the solution is wrong, either $p_{e,max}$ should be squared or right side should be rooted. $\endgroup$ – Rafal Nov 18 '15 at 8:14
  • 1
    $\begingroup$ Good question. Assuming the proton at rest would mean that the neutrino and the electron are sharing the same momentum in opposite directions. The calculation leads to the same E_max - I'll include it in the answer. $\endgroup$ – IljaBek May 1 '17 at 15:33
  • 1
    $\begingroup$ actually the calculation gave another result than the first glance! have a look. $\endgroup$ – IljaBek May 1 '17 at 16:06

The answer to your question is that the beta endpoint energy is equal to the total Q value of the reaction. Here is a graph of the spectrum from a beta emission, which is why we use internal conversion instead when looking for a monoenergetic source of beta particles. This graph is used in the German Wikipedia, the y-axis is counts per unit energy and the x-axis is energy (of the electron).

Beta decay spectrum

You asked for momentum instead, so use $E^2=(m_e c^2)^2 + (pc)^2$. Do not use the classical analog because it is extremely common that beta decay gives relativistic electrons.

  • $\begingroup$ Thanks. But doesn't that ignore the neutrino energy? $\endgroup$ – Lagerbaer Aug 14 '11 at 1:57
  • 2
    $\begingroup$ Why would I consider the neutrino energy? No really, I'm being completely serious. The question is about the maximum kinetic energy of the electron. The kinetic energy of the neutrino contains whatever energy from the reaction that is not in the kinetic energy of the electron. If I said the kinetic energy of the neutrino was $0$ then the spectrum would be one narrow peak at $E_{max}$. So aside from what I have said, what needs to be considered? $\endgroup$ – Alan Rominger Aug 14 '11 at 2:16
  • $\begingroup$ @Zass It is absolutely essential to consider the neutrino's momentum, and that implies that it carries off energy. So yes, you must worry about the neutrino. $\endgroup$ – dmckee --- ex-moderator kitten Aug 14 '11 at 2:54
  • 1
    $\begingroup$ @dmckee The neutrino's energy is the reason for the picture of the spectrum I posted. There would not be a spectrum if not for the energy of the neutrino, so yes, I considered the energy of the neutrino, I shouldn't have said I didn't. Another thing that I didn't do is ignore the energy of the neutrino. The entire discussion is about the energy of the neutrino so that we can establish that the energy of the neutrino for the beta endpoint energy is zero, so that the energy of the electron is equal to the reaction $Q$ value. $\endgroup$ – Alan Rominger Aug 14 '11 at 3:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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