What is the binding energy per nucleon of the americium isotope ${}_{\mathbf{95}}{}^{\mathbf{244}}\mathbf{Am}$? Here are some atomic masses and the neutron mass.

$$\begin{gathered} {}_{\mathbf{95}}{}^{\mathbf{244}}\mathbf{Am}\mathbf{}\mathbf{}\mathbf{}\mathbf{244}\mathbf{.}\mathbf{064279}\mathbf{}\mathbf{u}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{}^{\mathbf{1}}\mathbf{H}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{007825}\mathbf{}\mathbf{u} \\ \mathbf{n}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{008665}\mathbf{}\mathbf{u} \end{gathered}$$

The given isotope americium is${}_{95}{}^{244}Am$.

The atomic mass unit of the isotope americium,$M_{Am}=244.064279u$

The atomic mass of the hydrogen,$M_H=1.007825u$

The atomic mass unit of neutron,$M_n=1.008665u$

The binding energy of an element is defined as the amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. It can simply also be stated as the product of mass defect with the square of the speed of light. This relation will give the required binding energy of the isotope americium. Now, using this value and dividing it by the number of nucleons of the isotope, we can get the binding energy per nucleon of the americium isotope.

Formulae:

The binding energy of an atom,

$$\begin{gathered} ∆{\mathrm{E}}_{\mathrm{be}}=∆{\mathrm{mc}}^2 \\ =\left[{\mathrm{ZM}}_{\mathrm{H}}+\left(\mathrm{A}-\mathrm{Z}\right){\mathrm{M}}_{\mathrm{n}}-{\mathrm{M}}_{\mathrm{atom}}\right]{\mathrm{c}}^2 \end{gathered}$$ (1)

where,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. In principle, nuclear masses should be used, but the mass of the Z electrons included in $Z{M}_H$is canceled by the mass of the Z electrons included in $M_{atom}$, so the result is the same.

The binding energy per nucleon of an atom,

$∆E_{BEpernudeon}=\frac{∆E_{be}}{A}$ (2)

At first, we can calculate the mass excess or the mass defect for the americium isotope by Z = 95 using equation (1) as follows:

$$\begin{gathered} ∆m=\left[95\left(1.007825u\right)+\left(244-95\right)\left(1.008665u\right)-\left(244.064279u\right)\right] \\ =1.970181u \end{gathered}$$

Now, the binding energy can be calculated by converting the amu value of mass defect into MeV considering equation (1) as follows:

$$\begin{gathered} ∆{\mathrm{E}}_{\mathrm{be}}=(1.970181\mathrm{u}\left)\right(931.5\mathrm{Me}/\mathrm{u}) \\ =1835.212\mathrm{MeV} \end{gathered}$$

Thus, according to the concept the binding energy per nucleon of americium isotope with nucleon numbercan be calculated using equation (2) as follows:

$$\begin{gathered} ∆E_{BEpernucleon}=1835.212MeV/244 \\ =7.52MeV \end{gathered}$$

Hence, the required value of energy is 7.52 MeV.