Calculate (in $MeV$) the binding energy per nucleon for $O^{14}$and $O^{16}$. Which is more tightly bound?

We have been given that,

$O^{14}$, $O^{16}$

We need to find the binding energy per nucleon for both atoms.

The binding energy is given by,$O^{14},Z=8,N=6,A=14$

$BE=\left(Z{m}_P+N{m}_N-m_A\right)MeV$

$$\begin{gathered} BE=\left(8\times 1.0087u+6\times 1.0078u-14.008596u\right)Mev/u \\ =(14.1164-14.0085)931.49MeV \\ =100.5077MeV \end{gathered}$$

Now,

$BE=100.5077MeV,A=14$

The binding energy per nucleon is = $\frac{BE}{A}$

$$\begin{gathered} =\frac{100.5077}{14}MeV \\ =7.179MeV \end{gathered}$$

Binding energy is given by,$O^{16},Z=8,N=8,A=16$

$BE=\left(Z{m}_P+N{m}_N-m_A\right)MeV$

$$\begin{gathered} BE=\left(8\times 1.0087u+8\times 1.0078u-15.9949u\right)931.49MeV/u \\ =\left(16.132-15.9947\right)931.49MeV \\ =127.8935MeV \end{gathered}$$

Now, $BE=127.8935MeV,A=16$

Binding energy per nucleon is = $\frac{BE}{A}MeV$

= $\frac{127.8935}{16}MeV$

= $7.9933MeV$

Since the binding energy per nucleon : $O^{16}>O^{14}$

Therefore, $O^{16}$ is more tightly bounded.