Calculate and compare the energy released by
$(a)$ fusion of $1.0 \; kg$ of hydrogen deep within the Sun and
$(b)$ the fission of $1.0 \; kg$ of $^{235} U$ in a fission reactor.

(A) Amount of hydrogen,$m = 1 \; kg = 1000 \; g$. One mole,i.e.,$1 \; g$ of hydrogen $(^1_1 H)$ contains $6.023 \times 10^{23}$ atoms.
Therefore,$1000 \; g$ of $^1_1 H$ contains $6.023 \times 10^{26}$ atoms.
Within the Sun,four $^1_1 H$ nuclei combine to form one $^4_2 He$ nucleus. In this process,$26 \; MeV$ of energy is released. Hence,the energy released from the fusion of $1 \; kg$ of $^1_1 H$ is:
$E_1 = \frac{6.023 \times 10^{26}}{4} \times 26 \; MeV = 39.15 \times 10^{26} \; MeV$.
$(b)$ Amount of $^{235}_{92} U = 1000 \; g$. One mole,i.e.,$235 \; g$ of $^{235}_{92} U$ contains $6.023 \times 10^{23}$ atoms.
Therefore,$1000 \; g$ of $^{235}_{92} U$ contains $\frac{6.023 \times 10^{23} \times 1000}{235}$ atoms.
It is known that the energy released in the fission of one atom of $^{235}_{92} U$ is $200 \; MeV$.
Hence,energy released from the fission of $1 \; kg$ of $^{235}_{92} U$ is:
$E_2 = \frac{6.023 \times 10^{23} \times 1000}{235} \times 200 \; MeV \approx 5.13 \times 10^{26} \; MeV$.
Comparing the two:
$\frac{E_1}{E_2} = \frac{39.15 \times 10^{26}}{5.13 \times 10^{26}} \approx 7.63 \approx 8$.
Therefore,the energy released in the fusion of $1 \; kg$ of hydrogen is nearly $8$ times the energy released in the fission of $1 \; kg$ of uranium.