Question

Consider the following fusion reaction: ${ }_{1} H ^{2}+{ }_{1} H ^{2} \rightarrow$ ${ }_{2} He ^{4}$. If $20\, MeV$ of energy is released per fusion reaction and mass of $_1H ^{2}$ consumed per day is $0.1\, g$, then what is the power of the reactor (in $MW$)? (Give answer in integer value)

Answer 1

Let $P$ be the power.
Energy produced $=P \times 24 \times 36 \times 100$
Energy released per fusion reaction
$=20 MeV =32 \times 10^{-13} J$
Energy released per ${ }_{1} H ^{2}$ atom $=32 \times 10^{-13} J$
Number of ${ }_{1} H ^{2}$ atoms used is
$N=\frac{P \times 24 \times 36 \times 100}{32 \times 10^{-13}}$
Mass of $6 \times 10^{23}$ atoms $=2\, g$
$\frac{2}{6 \times 10^{23}} \times \frac{P \times 24 \times 3600}{32 \times 10^{-13}}$
$=0.1\, g$
$P=1\, MW$