Question

According to the semiempirical binding energy formula, what should be binding energy per nucleon of technetium -$98$ .

Short answer

The binding energy per nucleon of technetium - $98$is $-7.75716\text{MeV}/nucleon$.

Step-by-step solution

Given data

The given atom is technetium -$98$

Formula of Mass defect

Mass defect = Actual mass - Theoretical mass $\frac{\text{Bindingenergy}}{\text{Nucleon}}=\frac{\text{Massdefect}931.5\text{MeV}}{\text{Totalnumberofnucleons}}$ ……(1)

Calculation for the binding energy

The mass of technetium - $98$is$98.0\text{amu}$.

Technetium - $98$ contains $43$ number of protons and $55$ number of neutrons.

Theoretical mass of technetium$-98$.

Hydrogen atom mass$=43\times 1.007825\text{amu}=43.336475\text{amu}$.

Neutron mass$=55\times 1.008665\text{amu}=55.476575\text{amu}$ .

Hence, theoretical mass of technetium$-98=(43.336475+55.476575)\text{amu}$ .

Theoretical mass$=98.81305\text{amu}$.

Substitute the values of theoretical mass and actual mass of technetium - $98$.

Mass defect.$=(98.0-98.81305)\text{amu}=-0.81305\text{amu}$

Substitute the value of mass defect in equation (1).

$\begin{array}{l}\frac{\text{Bindingenergy}}{\text{Nucleon}}=\frac{-0.81305\times 931.5\text{MeV}}{98}\\ \frac{\text{Bindingenergy}}{\text{Nucleon}}=-7.75716\text{MeV}/nucleon\end{array}$