Be is often used as the neutron gun to supply neutrons to initiate nuclear reactions, particularly nuclear chain reactions. I always presumed the nucleus of lighter elements to be more stable than the heavier ones. Be with an atomic number of 4, meaning 4 protons, it is twice that of a helium nucleus. The helium nucleus is the same as an alpha particle, and is regarded as the most stable light nucleus.

Being that we can think of Be as two helium nuclei fused together, one would presume that the Be nucleus would be very hesitant to change its structure and give up neutrons. But this is what happens when it is struck with an alpha particle. Why is this so? I read that it has a neutron spin of 3/2. I don't yet understand the neutron spin property, but maybe this could explain it.

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    $\begingroup$ $^{8}$Be is not stable - it will spontaneously become two alpha particles. Note that the one isotope of Be found in nature is $^{9}$Be. $\endgroup$
    – Jon Custer
    Commented Sep 26, 2017 at 19:55
  • $\begingroup$ Are you saying Be is radioactive? $\endgroup$ Commented Sep 26, 2017 at 19:58
  • $\begingroup$ No, as I said, $^{9}$Be is stable. But $^{8}$Be is not, which one could think of as two alpha particles fused together. Except it is energetically more favorable to be two alpha particles. So, $^{9}$Be plus an alpha makes a $^{13}$C in an excited state, and this can couple to the state of $^{12}$C plus a neutron with a reasonable cross section. $\endgroup$
    – Jon Custer
    Commented Sep 26, 2017 at 20:09
  • $\begingroup$ Does neutron spin have anything to do with it? I think you answered the question. If you rewrite as answer instead of comment, I will up-vote and mark answered. $\endgroup$ Commented Sep 26, 2017 at 20:14
  • $\begingroup$ We could say that ${}^8\mathrm{Be}$ is radioactive, but you can't find it in the nature, because its half life $\approx 10^{-17} s$. In think, it would be an interesting (phylosophical) question, could we talk about the existence of an isotope which decays faster as the light would leave its nucleus. $\endgroup$
    – peterh
    Commented Sep 26, 2017 at 21:02

1 Answer 1


As noted, the (sole) stable isotope of Be is $^{9}$Be, with 4 protons and 5 neutrons. $^{8}$Be spontaneously dissociates into two $\alpha$ particles since it is more favorable energetically (this was the basis for the first human-controlled nuclear reaction, $^{7}$Li(p,$\alpha$)$\alpha$ by Cockroft and Walton in 1932 where they varied the proton energy).

The $^{9}$Be($\alpha$,n)$^{12}$C reaction, with Q=5.7MeV, is detailed in Richard G. Miller and R.W. Kavanagh, Nuclear Physics 88 492-500 (1966). The fundamental take away is that there are a number of resonances between the excited $^{13}$C nucleus and available $^{12}$C+n states, for various neutron energies. Furthermore there is the possibility of a 3$\alpha$+n output state mentioned in the paper, although that does not appear on nuclear energy level diagrams available at, e.g., tunl.duke.edu.

I wouldn't say there is anything particularly special about the overall reaction, just lots of opportunities to end up with a neutron out.


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