Two fusion reactions mentioned in the text are

\(n{ + ^3}He{ \to ^4}He + \gamma \)

and

\(n{ + ^1}H{ \to ^2}H + \gamma \).

Both reactions release energy, but the second also creates more fuel. Confirm that the energies produced in the reactions are \(20.58\) and\(2.22{\rm{ }}MeV\), respectively. Comment on which product nuclide is most tightly bound, \(^4He\) or\(^2H\).

The power kept in the centre of an atom is known as nuclear energy. All stuff in the cosmos is made up of tiny particles called atoms. The centre of an atom's nucleus is typically where most of its mass is concentrated. Neutrons and protons are the two subatomic particles that make up the nucleus. Atomic bonds that hold them together carry a lot of energy.

• The two fusion reactions are:\(\begin{array}{c}n{ + ^3}He{ \to ^4}He + \gamma \\n{ + ^1}H{ \to ^2}H + \gamma \end{array}\).
• Energies that should be produced in these reactions are:\(20.58{\rm{ }}MeV\)and\(2.22{\rm{ }}MeV\).

Energy difference can be calculated by the free mass-energy formula –

\(E = \Delta m{c^2}\)

To get these mass differences take parent nuclei and subtract from them daughter nuclei and masses of all other by-products of the nuclear reaction. Since all atomic masses will be expressed with the unified atomic unit, remember how to convert it to \(eV\) –

\(lu = \frac{{931.5{\rm{ }}MeV}}{{{c^2}}}\)

Now analyse the reaction –

\(n{ + ^3}He{ \to ^4}He + \gamma \)

Looking in the Appendix for the atomic masses the following is obtained–

\(\begin{array}{c}^3He = 3.016049{\rm{ }}u\\^4He = 4.002603{\rm{ }}u{\rm{ }}\\neutron{\rm{ }} = 1.008665{\rm{ }}u\end{array}\)

Subtracting the masses of parent and daughter nuclei, it is obtained –

\(\Delta m = 1.008665{\rm{ }}u + 3.016049{\rm{ }}u - 4.002603{\rm{ }}u\)

And the result is obtained as –

\(\Delta m = 0.022091{\rm{ }}u\)

So, from the energy relation it is obtained that –

\(\begin{array}{c}E = \Delta m{c^2}\\ = 0.022091 \cdot \frac{{931.5MeV}}{{{c^2}}} \cdot {c^2}\end{array}\)

Hence, the result is obtained as –

\(E = 20.58{\rm{ }}MeV\)

Therefore, the energy value for the helium reaction is obtained as \(E = 20.58{\rm{ }}MeV\).

Energy difference can be calculated by the free mass-energy formula –

\(E = \Delta m{c^2}\)

To get these mass differences take parent nuclei and subtract from them daughter nuclei and masses of all other by-products of the nuclear reaction. Since all atomic masses will be expressed with the unified atomic unit, remember how to convert it to \(eV\) –

\(E = \Delta m{c^2}\)

Now analyse the reaction –

\(n{ + ^1}H{ \to ^2}H + \gamma \)

Looking in the Appendix for the atomic masses the following is obtained–

\(\begin{array}{c}^1H = 1.007825{\rm{ }}u\\^2H = 2.014087{\rm{ }}u{\rm{ }}\\neutron{\rm{ }} = 1.008665{\rm{ }}u\end{array}\)

Subtracting the masses of parent and daughter nuclei, it is obtained –

\(\Delta m = 1.008665{\rm{ }}u + 1.007825{\rm{ }}u - 2.014087{\rm{ }}u\)

And the result is obtained as –

\(\Delta m = 0.002383{\rm{ }}u\)

So, from the energy relation it is obtained that –

\(\begin{array}{c}E = \Delta m{c^2}\\ = 0.002383 \cdot \frac{{931.5{\rm{ }}MeV}}{{{c^2}}} \cdot {c^2}\end{array}\)

Hence, the result is obtained as –

\(E = 2.22{\rm{ }}MeV\)

The more energy release we have from the nucleus it means it is more tightly bound. Since the first reaction gives a lotmore energy, it means that helium nuclei are more tightly bound than the deuterium.

Therefore, the energy value for the hydrogen reaction is obtained as \(E = 2.22{\rm{ }}MeV\).Helium nuclei is more tightly bound from the deuterium.