Question

It is proposed to use the nuclear fusion reaction ${}_1{H}^2{+}_1{H}^2{⟶}_{2}H{e}^4$ in a nuclear reactor with an electrical power rating of $200MW$. If the energy form the above reaction is used with $25$ percent efficiency in the reactor, how many grams of deuterium fuel will be needed per day? (The masses of ${}_1{H}^2$ and ${}_2{He}^4$ are $2.0141amu$ and $4.0026$ amu respectively.)

Answer 1

Mass defect occurring in one fusion reaction
$△m=(2\times 2.0141-4.0026)$ amu$=0.0256$ amu
Energy released$=0.256\times 931MeV$
Energy used in reactor per fusion reaction
$=\frac{25}{100}\times 0.0256\times 931MeV$
$=5.9584MeV=9.5334\times {10}^{-13}J$
Total energy required per day$=\left(200MW\right)\times (24\times 60\times 60\sec .)$
Mass of deuterium fuel needed per reaction$=2\times 2.0141$ amu
$=\frac{4.0282}{6.02\times {10}^{23}}gm=0.6691\times {10}^{-23}gm$
Masses of deuterium required $=\frac{0.6691\times {10}^{-23}\times 200\times {10}^6\times 24\times 60\times 60}{9.5334\times {10}^{-13}}=120gm$.