Question

Assuming that about 20 MeV of energy is released per fusion reaction,
${}_1{H}^2+$ ${}_1{H}^3$ $⟶$ ${}_0{n}^1+$ ${}_{2}H{e}^4.$
then the mass of $1_H^1$ 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. 100 g

Answer 1

The correct option is D 100 g

$U=Pt$

$={10}^6\times 24\times 36\times {10}^{12}$

$=24\times 36\times {10}^{8}J$

Energy released per fusion reaction

$=20MeV$

$=20\times {10}^6\times 16\times {19}^{-19}$

$=32\times {10}^{-13}J$

Energy released per atom of ${}_1{H}^2=32\times {10}^{-13}J$

Number of ${}_1{H}^2$ atom of used

$=\frac{24\times 36\times {10}^8}{32\times {10}^{-13}}$

$=27\times {10}^{21}$

Mass of $6\times {10}^{23}$ atom $=2g$

$=\frac{2}{6\times {10}^{23}}\times 27\times {10}^{21}$

$90=100g$