Question

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

Short answer

For $I^{129}$atom , The total binding energy = $1029.38028MeV$

The binding energy per nucleon= $7.979MeV$

For $X{e}^{129}$atom, The total binding energy = $1029.56658MeV$

The binding energy per nucleon= $7.981MeV$

Step-by-step solution

Given information.

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.

Calculation for binding energy and binding energy per nucleon for I129.

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= $\frac{BE}{A}$

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

 Step3: Calculation for binding energy and binding energy per nucleon for Xe129.

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 = $\frac{BE}{A}MeV$

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

= $7.981MeV$