Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The question asks for the isotopes with a mass number of 17, known as $ A = 17 $ isobars.
From appendix c, the $ A = 17 $ isobars are:
$\bullet$ $ ^{17}\text{N} $ (Nitrogen).
$\bullet$ $ ^{17}\text{O} $ (Oxygen).
$\bullet$ $ ^{17}\text{F} $ (Fluorine).
$$\color{blue}{\bf [b]}$$
Of the three isobars, $\boxed{ ^{17}\text{O} }$ is stable. However, it is relatively rare in nature. Stability indicates that this isotope does not undergo radioactive decay under normal conditions.
$$\color{blue}{\bf [c]}$$
For the remaining two isobars, $ ^{17}\text{N} $ and $ ^{17}\text{F} $, we identify the decay processes and their daughter products:
$\bullet$ $ ^{17}\text{N} $ 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 process increases the atomic number by 1.
Thus, the daughter nucleus is $\boxed{ ^{17}\text{O} }$
$\bullet$ $ ^{17}\text{F} $ undergoes electron capture. During electron capture, a proton in the nucleus captures an orbital electron, converting into a neutron and emitting a neutrino ($ \nu $). This process decreases the atomic number by 1.
Thus, the daughter nucleus is $\boxed{ ^{17}\text{O} }$
