Question

Oxygen-$\mathbf{16}$is one of the most stable nuclides. The mass of a${}^{\mathbf{\text{16}}}\mathbf{\text{O}}$atom is$\mathbf{\text{15.994915}}$amu. Calculate the binding energy

(a) per nucleon in MeV;

(b) per atom in MeV;

(c) per mole in kJ.

Short answer

The binding energies are

$$\begin{gathered} \text{a)7.976MeV/nucleon} \\ \text{b)127.616MeV/atom} \\ {\text{c)1.231×10}}^{\text{10}}\text{kJ/mol} \end{gathered}$$

Step-by-step solution

To calculate the binding energy(a)

-The atomic mass of ${}^{\text{16}}\text{O is 15.994915 amu}$

- Atomic number of O is$8$, so it has $8$protons

- The number nucleons is $16$, hence

the number of neutrons is $\left(16-8\right)8$.

- 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

$\begin{array}{rcl}& & ∆\text{m = 15.994915amu -}\left(8\times {\text{m}}_{\text{p}}{\text{+ 8×m}}_{\text{n}}\right)\\ & & \text{=15.994915amu -}\left(\text{8×1.007825amu + 8×1.008665amu}\right)\\ & & \text{=- 0.137005amu}\end{array}$

(a)

The binding energy per nucleon, in $\text{MeV}$is

$\text{BE per nucleon =}\frac{{\displaystyle \text{0.137005amu×}\frac{\text{931.5MeV}}{\text{1amu}}}}{{\displaystyle \text{16nucleons}}}=\text{7.976MeV/ nucleon}$

To calculate the binding energy(b)

(b)

The binding energy per atom is

$$\begin{gathered} \text{BE per atom = 7.976MeV/ nucleon×16 nucleons/atom} \\ =\text{127.616MeV/ atom} \end{gathered}$$

To calculate the binding energy(c)

(c)

The binding energy per mole in$\text{kJ}$

$$\begin{gathered} \text{BE per mol in kJ}={\text{BE per atom×6.022×10}}^{\text{23}}\text{atoms /moln} \\ =\text{127.616MeV/ atom×}{\mathrm{6.022\times 10}}^{\text{23}}\text{atoms /moln} \\ ={\text{7.685×10}}^{\text{25}}\text{MeV/mol×}\frac{{\displaystyle {\text{1.602×10}}^{\text{- 13}}\text{J}}}{{\displaystyle \text{1MeV}}} \\ {\text{=1.231×10}}^{\text{13}}\text{J/mol×}\frac{1{\displaystyle \text{kJ}}}{1000{\displaystyle \text{J}}} \\ \text{=}{\mathrm{1.231\times 10}}^{\text{10}}\text{kJ}/\mathrm{mol} \end{gathered}$$