Question

Assuming that about $20.\,MeV$ of energy is released per fusion reaction ${ }_{1} H ^{2}+{ }_{1} H ^{3} \rightarrow{ }_{0} n^{1}+{ }_{2} He ^{4}$ then the mass of ${ }_{1} H ^{2}$ consumed per day in a fusion reactor of power $1\, MW $ will approximately be

• A. 0.001 g
• B. 0.1 g
• C. 10.0 g
• D. 1000 g

Answer 1

The Correct Option is B

Energy produced,
$U=P t$
$=10^{6} \times 24 \times 36 \times 10^{2}$
$=24 \times 36 \times 10^{8} \,J$
Energy released per fusion reaction
$=20\, MeV$
$=20 \times 10^{6} \times 1.6 \times 10^{-19} $
$=32 \times 10^{13} \,J$
Energy released per atom of ${ }_{1} H^{2}$
$=32 \times 10^{-13}\, J$
Number of $_{1} H^{2}$ atoms used
$=\frac{24 \times 36 \times 10^{8}}{32 \times 10^{-13}}$
$=27 \times 10^{21}$
Mass of $6 \times 10^{23}$ atoms $=2\, g$
$\therefore$ Mass of $27 \times 10^{21}$ atoms
$=\frac{2}{6 \times 10^{23}} \times 27 \times 10^{21}=0.1 \,g$