Answer
E Total = $ 4.54 \times 10^{26} MeV$
Work Step by Step
Fusion energy for one $^{239}Pu $ = 180 MeV.
To find the total energy, we first must calculate the number of particle
$ N = \frac{M_{sample} }{M_{Pu}} NA $
where
sample Mass = $1.0 kg = 1000g$
Mass of Plutonium = $ 239 g/mol $
Number Avogadro = $ 6.02 \times 10^{23} /mol $
substitute into formula
\frac{x}{y}
$ N = \frac{1000g}{ 239 g/mol} (6.02 \times 10^{23} /mol) $
$ N = 2.5 \times 10^{24} $
to find total energy, total number of particle should be multiplied by 180 MeV
E Total = $ (180 MeV) (2.5 \times 10^{24}) $
E Total = $ 4.54 \times 10^{26} MeV$
