Question

It is proposed to use the nuclear fusion reaction ${}_1^{2}H+{}_1^{2}H\to {}_2^{4}He$ in a nuclear reactor with an electrical power rating of 200 MW. If the energy from 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 ${}_1^{2}H$ and ${}_2^{4}He$ are $2.0141\text{amu}$ and $4.0026\text{amu}$ respectively.)

• A. $290\text{gm}$
• B. $100\text{gm}$
• C. $130\text{gm}$
• D. $230\text{gm}$

Answer 1

The correct option is C $130\text{gm}$

Mass defect occuring in one fusion reaction,
$\Delta m=(2\times 2.0141-4.0026)amu=0.256amu$

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 $=\left(200MW\right)\times (24\times 60\times 60)s$

Mass of deuterium fuel needed per reaction $=2\times 2.0141amu=\frac{4.0282}{6.023\times {10}^{23}}g=0.6691\times {10}^{-23}g$

Masses of deuterium required = $\frac{0.6691\times {10}^{-23}\times 200\times {10}^6\times 24\times 60\times 60}{9.5334\times {10}^{-13}}=130g$