Question

Complete this nuclear reaction by selecting which particle would go in the

blank.
19
Ne →
10
+ OF
19
9

Answer 1

Missing particles:
`^(0)_(1)e^(+)` (a positron) and an electron neutrino
`\\u_(\rm e)`.

The nuclear equation would be:

`{\rm ^(19)_(10) Ne} \to ^(0)_(1)e^(+) + {\rm ^(19)_{\phantom{1}9}F }+ \\u_(e)`.

Step-by-step explanation:

The mass number of a particle is the number on the top-right corner of its symbol.

The atomic number of a particle is the number on the lower-right corner of its symbol.

The nuclear reaction here resembles a beta-plus decay. The mass numbers of the two nuclei are equal. However, the atomic number of the product nucleus is lower than that of the reactant nucleus by
`1`.

A beta decay may either be a beta-plus decay or a beta-minus decay. In a beta-plus decay, a positively-charged positron
`^(0)_(1)e^(+)` and an electron neutrino
`\\u_e` would be released. On the other hand, in a beta-minus decay, a negatively-charged electron
`\rm ^(0)_(1)e^(-)` and an electron antineutrino
`\overline{\\u}_e` would be released.

Electric charge needs to be conserved in nuclear reactions, including this one.

The atomic number of the
`\rm Ne` nucleus on the left-hand side is
`10`, meaning that the nucleus has a charge of
`+10`. On the other hand, the atomic number of the
`\rm F` nucleus on the right-hand side shows that this nucleus carries a charge of only
`+9`.

By the conservation of electric charge, the particles on the right-hand side must carry a positive charge of
`+1`. That rules out the possibility of the combination of one negatively-charged electron
`\rm ^(0)_(1)e^(-)` (with a charge of
`-1`) and an electron antineutrino
`\overline{\\u}_e` (with no electric charge at all.)

Hence, the only possibility is that the missing particle is a positron (and an electron neutrino
`\\u_e`, which carries no electric charge.)