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