Cobalt-$\mathbf{\text{59}}$is the only stable isotope of this transition metal. One${}^{\mathbf{\text{59}}}\mathbf{\text{Co}}$atom has a mass of$\mathbf{\text{58.933198}}$amu. Calculate the binding energy

(a) per nucleon in MeV;

(b) per atom in MeV;

(c) per mole in kJ.

-$\text{1amu = 931.5MeV}$

- The atomic mass of${}^{\text{59}}\text{Cois58.933198amu}$

- Atomic number of Co is $27$, so it has$27$ protons

- The number of nucleons is $59$, therefore,

the number of neutrons is ($59$-$27$) $=32$.

- The mass of neutron is${\text{m}}_{\text{n}}\text{= 1.008665amu}$

- The mass of proton is${\text{m}}_{\text{p}}\text{= 1.007825amu}$

Now, let us calculate the change in mass

$$\begin{gathered} ∆\text{m =}\text{58.933198amu -}\left({\text{27×m}}_{\text{p}}{\text{+ 32×m}}_{\text{n}}\right) \\ =\text{58.933198amu -}\left(\text{27×1.007825amu+ 32×1.008665amu}\right) \\ =\text{- 0.555357amu} \end{gathered}$$

(a)

The binding energy per nucleon, in MeV is

$\text{BE per nucleon =}\frac{{\displaystyle \text{0.555357amu×}\frac{\text{931.5MeV}}{\text{1amu}}}}{{\displaystyle \text{59nucleons}}}\text{=}\text{8.768MeV/ nucleon}$

(b)

The binding energy per atom is

$$\begin{gathered} \text{BE per atom =}\text{8.768MeV/ nucleon×59nucleons/atom} \\ \text{=}\text{517.312MeV/ atom} \end{gathered}$$

(c)

The binding energy per mole in kJ

$$\begin{gathered} \text{BE per mol in k.J =}{\text{BE per atom×6.022×10}}^{\text{23}}\text{atoms /mol} \\ =\text{517.312MeV/ atom}{\times 6.022\times 10}^{\text{23}}\text{atoms /mol} \\ ={\text{3.115×10}}^{\text{26}}\text{MeV/mol×}\frac{{\displaystyle {\text{1.602×10}}^{\text{- 13}}\text{J}}}{{\displaystyle \text{1MeV}}} \\ ={\text{4.99×10}}^{\text{13}}\text{J/mol×}\frac{\text{1kJ}}{\text{1000J}} \\ ={\text{4.99×10}}^{\text{10}}\mathrm{kJ}/\mathrm{mol} \end{gathered}$$