Question

The easiest fusion reaction to initiate is

${}_{\mathbf{1}}{}^{\mathbf{2}}{\mathbf{H}\mathbf{+}}_{\mathbf{1}}^{\mathbf{3}}\mathbf{H}{\to }_{\mathbf{2}}^{\mathbf{4}}{\mathbf{He}\mathbf{+}}_{\mathbf{0}}^{\mathbf{1}}\mathbf{n}$

Calculate the energy released per nucleus of and per mole of are produced. The atomic mass are as follows:

, 2.01410 u; , 3.01605 u; and , 4.00260 u. The masses of the electron and neutron are and 1.00866 u, respectively.

Short answer

The energy released per nucleus of ${}_2{}^4\mathrm{He}=-2.83\times {10}^{-12}\mathrm{J}/$nucleus .

The energy released per mole of ${}_2{}^4\mathrm{He}=-1.70\times {10}^{12}\mathrm{J}/\mathrm{mo}\mathrm{l}$.

Step-by-step solution

Fusion reaction.

Nuclear fusion refers to the process of combining two or multiple nuclei and further producing one or more new nuclei. This process requires a high amount of energy.

Calculate the mass defect of He24

The comparison of the sum of masses of neutrons and protons with helium nuclei resulted in the energy released.

For ${}_2{}^{4}H$nucleus:

• Number of protons will be equal to the number which is equal to 2.
• $\begin{array}{c}\mathrm{Number}\mathrm{of}\mathrm{neutrons}=\mathrm{mass}\mathrm{number}-\mathrm{atomic}\mathrm{number}\\ =2-1\\ =1\end{array}$
  
• Number of neutrons$=4-2=2$

• And, number of electrons = atomic number

The following chemical equation shows the nuclear reaction:

${}_1{}^{2}H+\frac{3}{1}H\to \frac{4}{2}He+\frac{1}{0}n$

The mass defect can be calculated as:

$\begin{array}{c}\Delta m=\left[{\mathrm{Massof}}_2^{4}He\mathrm{nucleus}+\mathrm{Massofneutron}\right]\\ \Delta m=(4.00260)+1.00866-[2.01410+3.01605]\\ \Delta m=-0.01889amu\end{array}$

Calculate the energy released per nucleus of He24.

The Einstein mass energy will be used to calculate the mass defect:

$\Delta E=\Delta m{c}^2.....\left(i\right)$

Where,

• $\Delta E=$Change in energy
• $\Delta m=$Change in mass
• role="math" localid="1663880247647" $c=$Velocity of light

The unit of mass is obtained in kg/nucleus.

$\begin{array}{l}\Delta m=\left(-0.01889amu\times \frac{1}{\mathrm{amu}}\right)\times \left(1.66\times {10}^{-27}\frac{kg}{nucleus}\right)\\ \Delta m=-3.14\times {10}^{-29}kg/nucleus\end{array}$

Now, put the calculated value in equation (1) to calculate the change in energy,

$\begin{array}{c}\mathrm{\Delta E}=\left(-3.14\times {10}^{-29}\frac{\mathrm{kg}}{\mathrm{nucleus}}\right)\times {\left(3.00\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2\\ =-2.822\times {10}^{-12}\mathrm{J}/\mathrm{nucleus}\end{array}$

The negative sign represents that the energy is released during the process.

Calculate the energy released per mole of He24

Now, the energy released during the formation of 1 mole of ${}_2{}^4\mathrm{He}$nuclei is given as:

$\begin{array}{c}\mathrm{Energy}=\left(-2.83\times {10}^{-12}\mathrm{J}/\mathrm{nucleus}\right)\times \left(6.022\times {10}^{23}\mathrm{nuclei}/\mathrm{mol}\right)\\ \mathrm{Energy}=-1.70\times {10}^{12}\mathrm{J}/\mathrm{mol}\end{array}$