Question

Calculate the binding energy per nucleon for ${}_{\mathbf{1}}{}^{\mathbf{2}}\mathbf{H}$and ${}_{\mathbf{1}}{}^{\mathbf{3}}\mathbf{H}$. The atomic masses are${}_{\mathbf{1}}{}^{\mathbf{2}}\mathbf{H}\mathbf{,}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{01410}\mathbf{u}$, and${}_{\mathbf{1}}{}^{\mathbf{3}}\mathbf{H}\mathbf{,}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{01605}\mathbf{u}$.

Short answer

The binding energy per nucleon for is$7.78\times {10}^{-14}\mathrm{J}/\mathrm{nucleon}$.

The binding energy per nucleon for is $4.53\times {10}^{-13}\text{J/nucleon}$.

Step-by-step solution

Binding energy

The energy accountable for breaking a whole system (nucleus of an atom) into its constituent parts termed as nucleons is known as bonding energy. It is expressed in kJ/mole of nuclei.

Calculate the binding energy for H12.

Use the given method to calculate the mass defect for

$\begin{array}{c}\Delta m=2.01410-(1\times 1.00782+1\times 1.00866)\\ \Delta m=2.01410-1.00782-1.00866\Delta m=-2.38\times {10}^{-3}\frac{\mathrm{amu}}{\mathrm{nucleus}}\\ \Delta m=\left(-2.38\times {10}^{-3}\frac{\mathrm{amu}}{\mathrm{nucleus}}\right)\times \left(1.66\times {10}^{-27}\frac{\mathrm{kg}}{\mathrm{amu}}\right)\\ \Delta m=-3.9508\times {10}^{-30}\frac{\mathrm{kg}}{\mathrm{nucleus}}\end{array}$

The Einstein mass energy equation is represented by the following equation,

$\Delta E=\Delta m.c^2\dots \dots \dots \dots \left(1\right)$

Or,

$\begin{array}{c}\Delta E=-3.9508\times {10}^{-30}\times {\left(3\times {10}^8{\mathrm{ms}}^{-1}\right)}^2\\ \Delta E=-35.55\times {10}^{-14}\mathrm{J}/\mathrm{nucleon}\end{array}$

Use this formula to calculate the binding energy,

$\mathrm{Binding}\mathrm{energy}=\frac{\Delta E}{2nucleon}=\frac{35.55\times {10}^{-14}}{2}=17.78\times {10}^{-14}\mathrm{J}/\mathrm{nucleon}$

Calculate the binding energy for H13.

Use the given method to calculate the mass defect for

$\begin{array}{l}\Delta m=3.01605-(1\times 1.00782+2\times 1.00866)\\ \Delta m=3.01605-1.00782-2.01732\\ \Delta m=-0.00909\mathrm{amulnucleon}\\ \Delta m=(-0.00909\mathrm{amulnucleon})\times \left(1.66\times {10}^{-27}\mathrm{kg}/\mathrm{amu}\right)\\ \Delta m=0.0150894\times {10}^{-27}\mathrm{kg}/\mathrm{nucleus}\end{array}$

The Einstein mass energy can be calculated as:

$\begin{array}{l}\Delta E=\Delta m.c^2\\ \Delta E=\left(-0.0150894\times {10}^{-27}\right)\times {\left(3\times {10}^8\right)}^2\mathrm{J}/\mathrm{nucleus}\\ \Delta E=-0.1358046\times {10}^{-11}\mathrm{J}/\mathrm{nucleus}\end{array}$

Now, the binding energy per nucleon can be calculated as:

$\begin{array}{l}=\left(\frac{0.1358046\times {10}^{-11}}{3}\times \frac{J}{\mathrm{nucleon}}\right)\\ =4.53\times {10}^{-13}J/\mathrm{nucleon}\end{array}$