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

Answer: (B)

Energy produced $U=Pt$
$=10^{6}\times 24\times 36\times 10^{2}$
$=24\times 36\times 10^{8}J$
The energy released per fusion reaction
$=20\,MeV$
$=20\times 10^{6}\times 1.6\times 10^{- 19}$
$=32\times 10^{- 13}J$
The 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 $6\times 10^{23}$ atom $=2\,g$
$\therefore $ Mass of $27\times 10^{21}$ atoms
$=\frac{2}{6 \times 10^{23}}\times 27\times 10^{21}$
$\approx0.1\,g$