Verify the binding energy per nucleon given in Table 42-1 for the plutonium isotope ${}^{\mathbf{239}}\mathit{P}\mathit{u}$.The mass of the neutral atom is 239.05216u.

1. The given plutonium isotope is ${}^{239}Pu$.
2. The mass of the neutral atom,$m_{pu}=239.05216u$
3. The atomic mass of the hydrogen, $m_H=1.007825u$
4. The atomic mass unit of neutron,$m_n=1.008665u$

The binding energy of an atom is as follows:

$∆E_{be}=\left[Z{M}_H+\left(A-Z\right)M_n-M_{atom}\right]c^{2}or∆m{c}^2$ …… (i)

Here,Zis the atomic number (number of protons),Ais the mass number (number of nucleons),$M_H$is the mass of a hydrogen atom,$M_n$is the mass of a neutron, and$M_{atom}$is the mass of an atom.

The binding energy per nucleon of an atom is as follows:

$∆E_{be/nucleon}=∆E_{be}/A$ …… (ii)

Determine the mass excess or the mass defect for the americium isotope with Z = 94 using equation (i) as follows:

$$\begin{gathered} ∆m=\left[94\left(1.007825u\right)+\left(239-94\right)\left(1.008665u\right)-\left(239.05216u\right)\right] \\ =1.94101u \end{gathered}$$

Now, the binding energy is calculated by converting the amu value of mass defect into as MeV considering equation (i) follows:

$$\begin{gathered} ∆{\mathrm{E}}_{\mathrm{be}}=\left(1.94101\mathrm{u}\right)\left(931.5\mathrm{M}\frac{\mathrm{eV}}{\mathrm{u}}\right) \\ =1808\mathrm{MeV} \end{gathered}$$

Thus, according to the concept the binding energy per nucleon of americium isotope with nucleon number A = 244 as can be calculated using equation (ii) as follows:

$$\begin{gathered} ∆{\mathrm{E}}_{\mathrm{be}/\mathrm{nucleon}}=\frac{1808}{239}\mathrm{MeV} \\ =7.56\mathrm{MeV} \end{gathered}$$

Hence, the required value of energy is 7.56 MeV.