Question

Calculate the binding energy per nucleon of technetium -$98$ .

Short answer

Binding energy per nucleons is $8.61\frac{\text{MeV}}{\text{nucleon}}$.

Step-by-step solution

Given data

Total number of nucleons, $A=98$.

Technetium-98 has $Z($ proton $)=43$ and $N($neutron$)=A-Z=98-43=55$.

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, $M_{\frac{q8}{43}}\text{Tc}=97.907115\text{u}$.

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:

$\begin{array}{l}\frac{B.E}{\text{nucleon}}=({\text{Zm}}_{\text{H}}+N{m}_n-M_{\frac{\text{?}}{43}\text{Tc}})c^2\\ \frac{B.E}{\text{nucleon}}=\frac{\left[43\right(1.007825u)+55(1.008665\text{u})-97.907115\text{u}]c^2}{98}\end{array}$

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

$\begin{array}{l}\frac{\text{B.E.}}{\text{nucleon}}=\frac{\left[43\right(1.007825\text{u})+55(1.008665\text{u})-97.907115\text{u}]\left(931.5\frac{\text{MeV}}{\text{u}}\right)}{98\text{nucleons}}\\ \frac{\text{B.E.}}{\text{nucleon}}=8.61\frac{\text{MeV}}{\text{nucleon}}\end{array}$

Therefore, binding energy per nucleons is $8.61\frac{\text{MeV}}{\text{nucleon}}$.