Calculate the binding energy per nucleon for ${}_{10}{}^{20}Ne, {}_{26}{}^{56}Fe$ and ${}_{92}{}^{238}U$. Given that mass of neutron is $1.008665 amu$, mass of proton is $1.007825 amu,$ mass of ${}_{10}{}^{20}Ne$ is $19.992440 amu,$mass of ${}_{26}{}^{56}Fe$ is $55.93492 amu$ and mass of ${}_{92}{}^{238}U$ is $238.050783 amu$.

Binding energy of nucleus ${}_{Z}X^A$ is given by the equation 

$$\begin{gathered} {\left(BE\right)}_{Ne}=\left[\left(A-Z\right)m_n+z{m}_p-M\left({}_Z{}^{A}X\right)\right]c^2 \\ {\left(BE\right)}_{Ne}=\left(∆m\right)c^2=\left[\left(10m_n+10m_p\right)-M\left({}_{10}{}^{20}Ne\right)\right]c^2 \end{gathered}$$

where 

$$\begin{gathered} ∆m=\left(10\left(1.008665\right)+10\left(1.007825\right)\right)-19.992440 \\ ∆m=0.17246 amu \\ {\left(BE\right)}_{Ne}=\left(0.17246\right)\left(931.5\right)=160.64 MeV \end{gathered}$$

So the binding energy per nucleon is 

$\frac{{\left(BE\right)}_{Ne}}{A}=8.03 MeV/nucleon$

Similarly for ${}_{26}{}^{56}Fe$, we have 

${\left(BE\right)}_{Fe}=\left(∆m\right)c^2=\left[\left(30m_n+26m_p\right)-M\left({}_{26}{}^{56}Fe\right)\right]c^2$

where,

$∆m=0.52848 amu$

${\left(BE\right)}_{Fe}=\left(0.52848\right)\left(931.5\right)=492 MeV$

Hence binding energy per nucleon is 

$\frac{{\left(BE\right)}_{Fe}}{A}=8.79 MeV/nucleon$

Binding energy for ${}_{92}{}^{238}U$ is 

${\left(BE\right)}_U=\left[\left(146m_n+92m_p\right)-M\left({}_{92}{}^{238}U\right)\right]c^2$

where,

$$\begin{gathered} ∆m=1.934 amu \\ {\left(BE\right)}_U=\left(1.934\right)\left(931.5\right)=1802 MeV \end{gathered}$$

Binding energy per nucleon is 

$\frac{{\left(BE\right)}_U}{A}=\frac{1802}{238}=7.57 MeV$