Question

Calculate the binding energy per nucleon of carbon- $12$.

Short answer

Binding energy per nucleons is $7.68\frac{\text{MeV}}{\text{nucleons}}$.

Step-by-step solution

Given data

Total number of nucleons, $A=12$.

Carbon $-12$ has $Z$ (proton $)=6$ and $N($neutron$)==A-Z=12-6=6$.

Atomic number of hydrogen, $m_{\text{H}}=1.007825\text{u}$.

Mass of neutron, $m_n=1.008665\text{u}$.

Speed of light,$c=3\times {10}^8\text{m}/\text{s}$ .

Atomic mass of technetium, $M_{\frac{12}{6}C}=12\text{u}$.

Formula for Binding energy

For helium- $4$, binding energy is given by formula, $BE=(Z{m}_H+N{m}_n-M_{\frac{4}{1}x})c^2$.

Calculation for the binding energy per nucleons

The binding energy per nucleons is given as:

$\begin{array}{l}\frac{\text{BE}}{\text{nucleons}}=\frac{{\left({\text{Zm}}_{\text{H}}{\text{+Nm}}_{\text{n}}{\text{-M}}_{\frac{\text{12}}{\text{6}}}\text{c}\right)}^{c^2}}{A}\\ \frac{\text{BE}}{\text{nucleons}}=\frac{\left[6\right(1.007825u)+6(1.008665u)-12u]c^2}{12}\end{array}$

The conversion factor may be used to express $c^2$in $\text{MeV}/\text{u}$.

$\begin{array}{l}\frac{\text{BE}}{\text{nucleons}}=\frac{\left[6\right(1.007825\text{u})+6(1.008665\text{u})-12\text{u}]\left(931.5\frac{\text{MeV}}{\text{a}}\right)}{12\text{nucleons}}\\ \frac{\text{BE}}{\text{nucleons}}=7.68\frac{\text{MeV}}{\text{nucleons}}\end{array}$

Therefore, binding energy per nucleons is $7.68\frac{\text{MeV}}{\text{nucleons}}$.