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 U²³⁵ in a fission reactor.

Solution

a.) In sun, a He nucleus is formed by the fusion four hydrogen nucleii by releasing an amount of 26 MeV energy.

Therefore,

Energy released by fusion of 1 kg of hydrogen,

= \frac{6 \times 10^{23} \times 26}{4} \times 10^{3} MeV

E_{1} = 39 \times 10^{26} MeV

b.)
As energy released in fission of one atom of

{}_{92}U^{235}

= 200 MeV

∴

Energy released in fission of 1 kg of

{}_{92}U^{235}

= \frac{6 \times 10^{23} \times 1000}{235} \times 200 MeV

E_{2} = 5.1 \times 10^{26} MeV

Thus,

\frac{E_{1}}{E_{2}} = \frac{39 \times 10^{26}}{5.1 \times 10^{26}} = 7.65

i.e., energy released in fusion is 7.65 times the energy released in fission.

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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 U²³⁵ in a fission reactor.

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Wave Optics

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 U235 in a fission reactor.

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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 U²³⁵ in a fission reactor.