The energy produced by the fusion of a \(1.00 - kg\) mixture of deuterium and tritium was found in Example Calculating Energy and Power from Fusion. Approximately how many kilograms would be required to supply the annual energy use in the United States?

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 reaction is as follows –

\(^2H{ + ^3}H{ \to ^4}He + n\)

Where\(17.59{\rm{ }}MeV\)is released per reaction. Through problem\(32.2\)it is known that\(1{\rm{ }}kg\)of deuterium and tritium through the nuclear reaction is releasing an amount of –

\(E = 3.37 \times {10^{14}}J\)

Also, it is known that the estimated yearly energy consumption of the US is around\(105 \times {10^{18}}{\rm{ }}J\).

So simply dividing the yearly need of energy consumption relative to the energy given from one kilogram of the mixture of deuterium and tritium, the need of the mixture in total for the US needs in one year can be obtained –

\(Energy = \frac{{\left( {1.05 \times {{10}^{20}}J} \right)}}{{\left( {3.37 \times {{10}^{14}}J} \right)}}\)

Which gives the result as –

\(Energy = 3.116 \times {10^5}kg\)

Therefore, the value for energy is obtained as \(Energy = 3.116 \times {10^5}kg\).