Question

The ratio of the amounts of energy released as a result of the fusion of $1kg$ hydrogen ($E_1$) and fission of $1kg$ of ${}_{92}{U}^{236}$ ($E_2$) will be:

• A. $1.28$
• B. $3.28$
• C. $5.28$
• D. $7.28$

Answer 1

Answer: (D)

$7.28$

Energy released in fission of one uranium nucleus, $E=200$ MeV per nucleus

$\therefore$ Energy released in fusion 1 kg $U^{236}$, $E^{\prime }=200\times \frac{1000}{236}\times 6.023\times {10}^{23}=5.1\times {10}^{26}$ MeV

$⟹$ $E^{\prime }=5.1\times {10}^{26}\times 1.6\times {10}^{-13}=8.16\times {10}^{13}$ J

Energy released during fusion of 1 kg hydrogen, $E_{H_2}=6.4\times {10}^{14}$ J

Thus ratio of energies, $\frac{E_{H_2}}{E^{\prime }}=\frac{6.4\times {10}^{14}}{8.16\times {10}^{13}}\approx 7.28$