Consider the fusion reaction ${}_{\mathbf{1}}{}^{\mathbf{2}}\mathit{H}\mathbf{+}{}_{\mathbf{1}}{}^{\mathbf{2}}\mathit{H}\mathbf{\to }{}_{\mathbf{2}}{}^{\mathbf{3}}\mathit{H}\mathit{e}\mathbf{+}{}_{\mathbf{0}}{}^{\mathbf{1}}\mathit{n}$

(a) Estimate the barrier energy by calculating the repulsive electrostatic potential energy of the two ${}_{\mathbf{1}}{}^{\mathbf{2}}\mathit{H}$nuclei when they touch.

(b) Compute the energy liberated in this reaction in MeV and in joules.

(c) Compute the energy liberated per mole of deuterium, remembering that the gas is diatomic, and compare with the heat of combustion of hydrogen, about $\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$.

The electric potential energy of two point charges $q_1$and $q_2$separated by a distance r is given by:

$U=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_2}{r}$ ..................(i)

The radius of a nucleus with mass number A is given by:

$R=R_0{A}^{1/3}$ ..................(ii)

Where $R_0=1.20\times {10}^{-15}m$.

If initial particles A and B interact to produce final particles C and D, the reaction energy Q is defined as:

$Q=[M_A+M_B-M_C-M_D]\times 931.5MeV/u$ ..................(iii)

The atomic mass of ${}_1{}^{2}H$is $M\left({}_1{}^{2}H\right)=2.014102u$.

The atomic mass of ${}_2{}^{3}He$is $M\left({}_2{}^{3}He\right)=3.016029u$.

The mass of a neutron is $m_n=1.008665u$.

Let's consider the two ${}_1{}^{2}H$nuclei as point charges

The charge of each nucleus equals the charge of the proton inside; Thus,

$$\begin{gathered} q_1=q_2 \\ =e \\ =1.60\times {10}^{-19}C \end{gathered}$$.

When they are in touch, the separation distance is the sum of their radii, r = 2R.

So, we substitute our value of A = 2 for ${}_1{}^{2}H$into equation (ii), so we get the radius of the nucleus:

localid="1664169404610" $$\begin{gathered} R=(1.20\times {10}^{-15}m){\left(2\right)}^{1/3} \\ =1.51\times {10}^{-15}m \end{gathered}$$

So,

$$\begin{gathered} r=2\times (1.51\times {10}^{-15}m) \\ =3.02\times {10}^{-15}m \end{gathered}$$

Now, we plug our values for $q_1,q_2$, 2 and r into equation (i), so we get the barrier energy:

$$\begin{gathered} U=(8.988\times {10}^{9}N.m^2/c^2)(\frac{{(1.60\times {10}^{-19}C)}^2}{3.02\times {10}^{-15}m} \\ =7.61\times {10}^{-14}JU \\ =(7.61\times {10}^{-14}J\left)\right(\frac{1.00MeV}{1.60\times {10}^{-13}J}) \\ =0.476MeV \end{gathered}$$

First, we rewrite the fusion equation as given:

${}_1{}^{2}H+{}_1{}^{2}H\to {}_2{}^{3}He+{}_0{}^{1}n$

The mass decrease is the sum of the masses of the two ${}_1{}^{2}H$nuclei minus the sum of the masses of the ${}_2{}^{3}He$nucleus and the neutron.

And we have the atomic masses ${}_1{}^{2}H$and ${}_2{}^{3}He$which include the orbiting electrons

But since the number of electrons is equal on both sides, we can use the atomic masses to calculate the mass decrease since the electron masses will cancel

So, we substitute our values for $M\left({}_1{}^{2}H\right)$, $M\left({}_2{}^{3}He\right)$and $m_n$into equation (iii), so we get the energy released per fusion event:

$$\begin{gathered} Q=[2\times 2.014102u-3.016029u-1.008665u]931.5MeV/u \\ =[0.003510u]931.5MeV/u \\ =3.270MeV \\ Q=(3.270MeV)\frac{(1.60\times {10}^{-13}J)}{1.00MeV} \\ =5.23\times {10}^{-13}J \end{gathered}$$

Each fusion event consumes two deuterium atoms, and each gas molecule contains two atoms; So, we can say that each fusion event consumes one molecule.

In one mole of a gas, there is Avogadro's number of molecules, where Avogadro's number equals $N_A=6.022\times {10}^{23}molecules$.

Thus, in one mole, we have $6.022\times {10}^{23}$fusion events.

Therefore, the energy liberated when one mole of deuterium undergoes fusion is:

$$\begin{gathered} E=Q{N}_A \\ =(5.23\times {10}^{-13}J/event)(6.022\times {10}^{23}event/mole) \\ E=3.15\times {10}^{11}J/mole \end{gathered}$$

Compared with the heat of combustion of hydrogen, about $2.9\times {10}^{5}J/mole$.

$\frac{3.15\times {10}^{11}J/mole}{2.9\times {10}^{5}J/mole}=1.09\times {10}^6$

Therefore, the energy released when one mole of deuterium undergoes fusion is larger by $1.09\times {10}^6$times the heat of combustion of hydrogen.