Suppose an atom decays via a $\beta^-$ decay, it's atomic number has increased by one, and an electron is emitted.
$$ _Z^AX \quad \to \quad_{Z+1}^{\quad A}X'\ + \ _{-1}^{\ \ 0}e^- \ + \ \bar v _e $$ here a neutron turns into a proton emitting an electron and an electron - anti neutrino. The electron has velocity that is a fraction of the speed of light, and travels away from the influence of the daughter nuclei.
Now the daughter nuclei has $Z+1$ protons and $Z$ electrons, making it a positive ion of a new element. Similar pondering will suggest a $\beta^+$ decay of an atom will make a negetive ion of some new element.
similarly for $\alpha$ decay we have a negetive ion with 2 more electrons than protons (in symbolic form, $_{Z-2}^{A-4}X'^{\ \ \mathbf {2-}}$).

Now if we have a sample of a pure radioisotope that decays via $\beta^-$ decay in vaccum, as time passes and atoms decay we should observe that the sample becomes richer in positively charged protons, and deficient in electrons, thus acquiring a net positive charge. Similarly other in types of decay (exept $\gamma$ decay) we should have samples acquiring net negetive charges.

Now to the Questions:

  1. Is the above reasoning correct? Do samples of radioactive materials produce ions and acquire net charges
  2. If 1) is not correct then what happens to samples radioactive atoms kept in a isolated system in vaccum?.
  3. Has anyone actually performed such an experiment that you know of?
  • 1
    $\begingroup$ To answer some of your questions in a very concrete way, I advise you to do a search for "self powered neutron detector" . $\endgroup$ Commented Jan 22, 2022 at 8:34
  • $\begingroup$ Related, related. $\endgroup$
    – rob
    Commented Jan 23, 2022 at 4:44

1 Answer 1


Note that you are also creating extra protons (Z+1), so the net charge in the system remains the same.

You are creating extra negative electrons, but in a closed system they will recombine with positively charged ions.


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