Answer
$ ^{16}\text{O} $ is more tightly bound.
Work Step by Step
To calculate and compare the binding energy per nucleon for $ ^{14}\text{O} $ and $ ^{16}\text{O} $, we’ll apply the formula for nuclear binding energy:
$$
B =( Z m_H + N m_n - m_{\text{atom}})(931.49 \, \text{MeV/u})\tag 1
$$
where:
- $ Z $ is the number of protons,
- $ m_H $ is the mass of a proton,
- $ N $ is the number of neutrons,
- $ m_n $ is the mass of a neutron,
- $ m_{\text{atom}} $ is the mass of the nucleus.
The resulting binding energy $ B $ is then divided by the total number of nucleons to find the binding energy per nucleon.
For $ ^{14}\text{O} $: $Z=8$, $N=8$, and $ m_{\text{atom}}=\rm 14.008595 \;u$
Plug into (1);
$$
B_{ ^{14}\text{O} } =\left[ 8(1.007825 \, \text{u}) + 6(1.008665 \, \text{u}) - 14.008595 \, \text{u}\right](931.49 \, \text{MeV/u})
$$
$$
B_{ ^{14}\text{O} } =\color{blue}{\bf 98.7333}\, \text{MeV}
$$
Binding energy per nucleon for $ ^{14}\text{O} $:
$$
\dfrac{B_{ ^{14}\text{O} }}{{\rm nucleon }}=\frac{98.7333\, \text{MeV}}{14} =\color{red}{\bf 7.0524}\, \text{MeV}
$$
---
For $ ^{16}\text{O} $: $Z=8$, $N=8$, and $ m_{\text{atom}}=\rm 15.994915\; u$
Plug into (1);
$$
B_{^{16}\text{O} } =\left[ 8(1.007825 \, \text{u}) + 8(1.008665 \, \text{u}) - 15.994915\, \text{u}\right](931.49 \, \text{MeV/u})
$$
$$
B_{^{16}\text{O} } =\color{blue}{\bf 127.619}\, \text{MeV}
$$
Binding energy per nucleon for $ ^{16}\text{O} $:
$$
\dfrac{B_{ ^{16}\text{O} }}{{\rm nucleon }}\frac{127.619\, \text{MeV}}{16} = \color{red}{\bf 7.9762}\, \text{MeV}
$$
The binding energy per nucleon for $ ^{16}\text{O} \approx 7.98 \;\rm MeV$ is higher than the $ ^{14}\text{O} \approx 7.05 \;\rm MeV$.
Therefore, $ ^{16}\text{O} $ is more tightly bound.
