Question

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

Answer 1

Answer: 120

Total energy used in nuclear reactor per day $E=Pt$
$=\left(200\times 10^6\right)\times \left(24\times 3600\right)$
$=1.72\times 10^{12}J$
Energy used in reactor per reaction
$=\frac{25}{100}\left(2\times 2.0141-4.0026\right)\times 931MeV$
$=5.9584MeV$
$=9.5334\times 10^{-13}J$
The mass of deuterium used per reaction
$=2\times 2.0141=4.0282$ amu
$=\frac{4.0282}{6.02\times 10^{23}}g=0.6691\times 10^{-23}g$
Thus mass of deuterium required to produce $1.72\times 10^{12}J$ of energy
$=\frac{\left(0.6691\times 10^{-23}\right)\times \left(1.72\times 10^{12}\right)}{9.5334\times 10^{-13}}$
$=120g$