Answer:
Missing particles: [tex]^{0}_{1}e^{+}[/tex] (a positron) and an electron neutrino [tex]\nu_{\rm e}[/tex].
The nuclear equation would be:
[tex]{\rm ^{19}_{10} Ne} \to ^{0}_{1}e^{+} + {\rm ^{19}_{\phantom{1}9}F }+ \nu_{e}[/tex].
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 [tex]1[/tex].
A beta decay may either be a beta-plus decay or a beta-minus decay. In a beta-plus decay, a positively-charged positron [tex]^{0}_{1}e^{+}[/tex] and an electron neutrino [tex]\nu_e[/tex] would be released. On the other hand, in a beta-minus decay, a negatively-charged electron [tex]\rm ^{0}_{1}e^{-}[/tex] and an electron antineutrino [tex]\overline{\nu}_e[/tex] would be released.
Electric charge needs to be conserved in nuclear reactions, including this one.
The atomic number of the [tex]\rm Ne[/tex] nucleus on the left-hand side is [tex]10[/tex], meaning that the nucleus has a charge of [tex]+10[/tex]. On the other hand, the atomic number of the [tex]\rm F[/tex] nucleus on the right-hand side shows that this nucleus carries a charge of only [tex]+9[/tex].
By the conservation of electric charge, the particles on the right-hand side must carry a positive charge of [tex]+1[/tex]. That rules out the possibility of the combination of one negatively-charged electron [tex]\rm ^{0}_{1}e^{-}[/tex] (with a charge of [tex]-1[/tex]) and an electron antineutrino [tex]\overline{\nu}_e[/tex] (with no electric charge at all.)
Hence, the only possibility is that the missing particle is a positron (and an electron neutrino [tex]\nu_e[/tex], which carries no electric charge.)
