Calculate and compare the energy released by
Question:

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

Solution:

(a) Amount of hydrogen, m = 1 kg = 1000 g

1 mole, i.e., $1 \mathrm{~g}$ of hydrogen $\left({ }_{1}^{1} \mathrm{H}\right)$ contains $6.023 \times 10^{23}$ atoms.

$\therefore 1000 \mathrm{~g}$ of ${ }_{1}^{1} \mathrm{H}$ contains $6.023 \times 10^{23} \times 1000$ atoms.

Within the sun, four ${ }_{1}^{1} \mathrm{H}$ nuclei combine and form one ${ }_{2}^{4} \mathrm{He}$ nucleus. In this process $26 \mathrm{MeV}$ of energy is released.

Hence, the energy released from the fusion of $1 \mathrm{~kg}{ }^{1} \mathrm{H}$ is:

$E_{1}=\frac{6.023 \times 10^{23} \times 26 \times 10^{3}}{4}$

$=39.1495 \times 10^{26} \mathrm{MeV}$

(b) Amount of ${ }_{92}^{235} \mathrm{U}=1 \mathrm{~kg}=1000 \mathrm{~g}$

1 mole, i.e., $235 \mathrm{~g}$ of ${ }_{92}^{235} \mathrm{U}$ contains $6.023 \times 10^{23}$ atoms.

$\therefore 1000 \mathrm{~g}$ of $_{92}^{235} \mathrm{U}$ contains $\frac{6.023 \times 10^{23} \times 1000}{235}$ atoms

It is known that the amount of energy released in the fission of one atom of ${ }_{2}^{235} \mathrm{U}$ is $200 \mathrm{MeV}$.

Hence, energy released from the fission of $1 \mathrm{~kg}$ of ${ }_{92}^{235} \mathrm{U}$ is:

$E_{2}=\frac{6 \times 10^{23} \times 1000 \times 200}{235}$

$=5.106 \times 10^{26} \mathrm{MeV}$

$\therefore \frac{E_{1}}{E_{1}}=\frac{39.1495 \times 10^{26}}{5.106 \times 10^{26}}=7.67 \approx 8$

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.

Administrator

Leave a comment

Please enter comment.
Please enter your name.