Question

Assuming that about $20\text{MeV}$ of energy is released per fusion reaction,

${}_1{\text{H}}^2{+}_1{\text{H}}^3{\to }_0{\text{n}}^1{+}_2{\text{He}}^4$

The mass of ${}_1{\text{H}}^2$ consumed per day in a future fusion reactor of power $1\text{MW}$ would be approximately-

• A. $0.1\text{g}$
• B. $0.01\text{g}$
• C. $1\text{g}$
• D. $10\text{g}$

Answer 1

The correct option is A $0.1\text{g}$

Total energy released per day is,

$E=Pt=1\times 24\times 3600\times {10}^6=8.6\times {10}^{10}\text{J}$

$\text{No. of fusion reactions}\left(n\right)=\frac{\text{Total energy released}}{\text{Energy released per reaction}}$

$\Rightarrow n=\frac{8.6\times {10}^{10}}{20\times 1.6\times {10}^{-13}}=0.268\times {10}^{23}$

Mass of ${}_1{\text{H}}^2$ required per day is,

$m=\frac{nA}{N_A}=\frac{0.268\times {10}^{23}\times 2}{6.023\times {10}^{23}}=0.089\text{g}$

$m\approx 0.1\text{g}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.