Question

In beta decay, an electron, $e^{-}$ (or a positron, $e^{+}$) is emitted by a nucleus. What happens to the atom, among the following options?

• A. The atom gets positively charged if is emitted, or negatively charged if is emitted
• B. The atom gets negatively charged if is emitted, or positively charged if is emitted
• C. The atom stays neutral, since a proton is also created to balance the charge
• D. Any of the three can take place, depending on other factors.

Answer 1

The correct option is A The atom gets positively charged if is emitted, or negatively charged if is emitted

Writing down an example of ${\beta }^{-}$ decay,
${}_6^{14}C{\to }_7^{14}N+e^{-}+\overline{v},$
we'll see, if you work the numbers,
$mass{(}_6^{14}C)$ > $mass{(}_7^{14}N+e^{-}+\overline{v},)$
where $mass{(}_6^{14}C)$ refers to the mass of the nucleus and not the atom. This is true for every $\beta$ decay process. There is a $\Delta m$ that predominantly becomes the kinetic energy $\left(\Delta m{c}^2\right)$
of the outgoing electron (or positron).
The atomic energy levels are usually of the order of few tens of eV but the kinetic energy of the outgoing electron (or positron) is usually of the order of MeV. The electron (or positron) is too fast/energetic to be recaptured by the new atom of higher atomic number.
Thus in ${\beta }^{-}$ decay, where the atom gets a new proton in the nucleus but is short of an electron to balance the extra charge, it becomes a positive ion; similarly, a ${\beta }^{+}$ decay leaves an atom negatively charged, since a proton turns into a neutron but there is one excess electron in the atom already.
(In the example I took, the nitrogen will be a positive ion (cation). The representation ${}_7^{14}N$ refers only to the nucleus and not the atom itself, so we don't write ${}_7^{14}{N}^{+}.$)