Ncert Exercises & Solutions

13.30 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 ²³⁵U in a fission reactor.

Hint: The energy released by the nuclear reaction can be found by \(E=\Delta mc^2\).
Step 1: Find the energy released by 1kg of Hydrogen.

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

1 mole, i.e., 1 g of hydrogen

({}_{1}^{1}H) c o n t a i n s 6.023 \times 10^{23}

atoms.

∴ 1000 g o f {}_{1}^{1}H c o n t a i n s 6.023 \times 10^{23} \times 1000 a t o m s .

W i t h i n t h e s u n, f o u r {}_{1}^{1}H n u c l e i c o m b i n e a n d f o r m o n e {}_{2}^{4}H e n u c l e u s . I n t h i s p r o c e s s, 26 M e V o f e n e r g y i s r e l e a s e d .

H e n c e, t h e e n e r g y r e le a s e d f r o m t h e f u s i o n o f 1 k g {}_{1}^{1}H i s :

E_{1} = \frac{6.023 \times 10^{23} \times 26 \times 10^{3}}{4} = 39.1495 \times 10^{26} M e V

Step 2: Find the energy released by 1 kg of Uranium.

(b) A m o u n t o f {}_{92}^{235}U = 1 k g = 1000 g

1 m o l e o f, i . e ., 235 g o f {}_{92}^{235}U c o n t a i n s 6.023 \times 10^{23} a t o m s .

∴ 1000 g o f {}_{92}^{235}U c o n t a i n s \frac{6.023 \times 10^{23} \times 1000}{235} a t o m s .

I t i s k n o w n t h a t t h e a m o u n t o f e n e r g y r e l e a s e d i n t h e f i s s i o n o f o n e a t o m o f {}_{92}^{235}U i s 200 M e V .

H e n c e, the e n e r g y r e l e a s e d f r o m t h e f i s s i o n o f 1 k g o f {}_{92}^{235}U i s :

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

= 5.106 \times 10^{26} M e V

∴ \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.

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