Calculate (in $MeV$) the total binding energy and the binding energy per nucleon for $I^{129}$and for $X{e}^{129}$.

We have been given that,

$I^{129}$, $X{e}^{129}$

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

Binding energy is given by, $I^{129},Z=53,N=76,A=129$

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

$Z=$atomic number, $m_P=$mass of proton,$N=$number of neutron ,

$m_N=$mass of neutron,$m_A=$mass of atom

$$\begin{gathered} BE=\left(53\times 1.00783u+76\times 1.00783u-128.90498u\right)931.49MeV/u \\ =(53.41499+76.59508-128.90498)931.49MeV/u \\ =1.10509\times 931.49MeV \\ =1029.38028MeV \end{gathered}$$

Now ,

$BE=1029.38028MeV,A=129$

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

$$\begin{gathered} =\frac{1029.38028}{129}MeV \\ =7.979MeV \end{gathered}$$

Binding energy is given by ,

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

For $X{e}^{129}$,

$Z=54,N=75,A=129$

$$\begin{gathered} BE=\left(54\times 1.00783u+75\times 1.00783u-128.90478u\right)931.49MeV/u \\ =\left(54.42282+75.58725-128.90478\right)931.49MeV \\ =1.10529\times 931.49MeV \\ =1029.56658MeV \end{gathered}$$

Now,

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

= $\frac{1029.56658}{129}MeV$ $\left(A=129\right)$

= $7.981MeV$