Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
From appendix c, the $ A = 17 $ isobars are:
$\bullet$ $ ^{19}\text{O} $ (Oxygen)
$\bullet$ $ ^{19}\text{F} $ (Fluorine)
$\bullet$ $ ^{19}\text{Ne} $ (Neon)
$$\color{blue}{\bf [b]}$$
From appendix c, we can see that only $ \boxed{\;^{19}\text{F}\;} $ is stable, meaning it does not undergo radioactive decay under normal conditions.
$$\color{blue}{\bf [c]}$$
$\bullet$ $ ^{19}\text{O} $ undergoes beta-minus decay. In beta-minus decay, a neutron in the nucleus converts into a proton, emitting an electron ($ e^- $) and an antineutrino ($ \bar{\nu} $). This results in an increase in atomic number by 1.
Thus, the daughter nucleus is $\boxed{\; ^{19}\text{F}\;} $.
$\bullet$ $ ^{19}\text{Ne} $ undergoes beta-plus decay (or positron emission). In beta-plus decay, a proton in the nucleus transforms into a neutron, emitting a positron ($ e^+ $) and a neutrino ($ \nu $). This process decreases the atomic number by 1.
Thus, the daughter nucleus is $\boxed{\; ^{19}\text{F}\;} $.
