(a) Calculate the number of grams of deuterium in a \(80,000 - L\) swimming pool, given deuterium is \(0.0150\% \) of natural hydrogen.

(b) Find the energy released in joules if this deuterium is fused via the reaction\(^2H{ + ^2}H{ \to ^3}He + n\).

(c) Could the neutrons be used to create more energy?

(d) Discuss the amount of this type of energy in a swimming pool as compared to that in, say, a gallon of gasoline, also taking into consideration that water is far more abundant.

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.

• Capacity of pool is:\(80,000 - L\).
• Amount of natural hydrogen in deuterium is:\(0.0150\% \).
• Fusion reaction of deuterium is: \(^2H{ + ^2}H{ \to ^3}He + n\).

Firstly, calculate the mass of one deuterium molecule, it has a mass number of\(2\), and

dividing it with Avogadro number –

\({m_d} = \frac{M}{{{N_A}}}\)

Plugging in the values –

\({m_d} = \frac{{2{\rm{ }}g/mol}}{{6.022 \times {{10}^{23}}\;mo{l^{ - 1}}}}\)

And the result is obtained as –

\({m_d} = 3.32 \times {10^{ - 24}}g\)

The amount of water and the mass of the deuterium is known. Since the ratio of deuterium within the hydrogen inside the water is known, now calculate the number of water molecules. One water molecule has a molar mass of\(18\;g/mol\)–

\(\begin{array}{c}n = \frac{m}{M} = \frac{{8 \times {{10}^7}\;g}}{{18\;g/mol}}\\n = 4.44 \times {10^6}\;mol\end{array}\)

To get the absolute number of water molecules, we simply multiply the number of moles with Avogadro number –

\(N = 4.44 \times {10^6}\;mol \times 6.022 \times {10^{23}}\;mo{l^{ - 1}}\)

And the number of water molecules is –

\(N = 2.67 \times {10^{30}}{\rm{ }}molecules\)

Only\(0.0150\% \)of calculated water molecules is possibly having deuterium inside, so the number of deuterium will be –

\({N_d} = 0.00015 \times 2.67 \times {10^{30}}{\rm{ }}molecules\)

Which also gives –

\({N_d} = 4.01 \times {10^{26}}\)

And knowing the mass of the deuterium molecule we know that the total mass of all deuterium in whole pool will be –

\({m_{all}} = 1331.53\;g\)

Therefore, the value for mass is obtained as\(m = 1331.53\;g\).

The energy release is happening due to the different total mass of nuclei before and after the nuclear reaction. 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 –

\(^2H{ + ^2}H{ \to ^3}He + n\)

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

\(\begin{array}{c}^2H = 2.014102{\rm{ }}u\\^4He = 4.002603{\rm{ }}u{\rm{ }}\\neutron{\rm{ }} = 1.008665{\rm{ }}u\end{array}\)

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

\(\Delta m = \left( {{m_{{N^{14}}}} + {m_{{H^1}}} - {m_{{O^{15}}}}} \right)\)

Plugging in the values –

\(\begin{array}{c}\Delta m = \left( {2 \times 2.01355{\rm{ }}u - 3.016049{\rm{ }}u - 1.008665{\rm{ }}u} \right)\\\Delta m = 0.002412{\rm{ }}u\end{array}\)

So, from the energy relation it is obtained that –

\(\begin{array}{c}E = \Delta m{c^2}\\ = 0.002412 \cdot \frac{{931.5{\rm{ }}MeV}}{{{c^2}}} \cdot {c^2}\\E = 2.247{\rm{ }}MeV\end{array}\)

Hence, the result in joules is obtained as –

\(E = 3.595 \times {10^{ - 13}}J\)

Therefore, the energy value for the reaction is obtained as \(E = 3.595 \times {10^{ - 13}}J\).

In specific case neutrons could be used to have a better reaction and generate more energy, as it can induce the chain reaction, we also have a creation of deuterium, when neutron merges with hydrogen, which gives a lot of energy.

Therefore, more energy can be created.

The gasoline would be having much more molecular weight\({C_3}{H_{18}}\), which would give less hydrogen, especially because we are comparing a full pool of water to the gallon of gasoline, which is less than\(4L\). With the pool, there is more hydrogen which later fuses.

Therefore, energy is more in swimming pool as compared to a gallon of gasoline.