Question

Assuming that about $20MeV$ of energy is released per fusion reaction ${}_1{H}^2+{}_1{H}^3\to {}_0{n}^1+{}_{2}H{e}^4,$ then the mass of ${}_1{H}^2$ consumed per day in a fusion reactor of power $1MW$ will approximately be

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

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
$=20MeV$
$=20\times 10^6\times 1.6\times 10^{-19}$
$=32\times 10^{-13}J$
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 $=2g$
$\therefore$ Mass of $27\times 10^{21}$ atoms
$=\frac{2}{6\times 10^{23}}\times 27\times 10^{21}$
$\approx 0.1g$