Ncert Exercises & Solutions

13.31 Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ²³⁵U to be about 200MeV.

Hint: The energy released per fission is given by \(E=\Delta mc^2\).

Step 1: Find the amount of energy converted per fission.

Amount of electric power to be generated, P=

2 \times 10^{5}

10% of this amount has to be obtained from nuclear power plants.

∴ A m o u n t o f n u c l e a r p o w e r, p_{1} = \frac{10}{100} \times 2 \times 10^{5} = 2 \times 10^{4} \times 10^{6} J / s = 2 \times 10^{10}

\times 60 \times 60 \times 24 \times 365 J / y

H e a t e n e r g y r e l e a s e d p e r f i s s i o n o f a {}_{235}U n u c l e u s, E = 200 M e V

E f f i c i e n c y o f a r e a c t o r = 25 %

H e n c e, t h e a m o u n t o f e n e r g y c o n v e r t e d i n t o t h e e l e c t r i c a l e n e r g y p e r f i s s i o n i s c a l c u l a t e d a s :

\frac{25}{100} \times 200 = 50 M e V

= 50 \times 1.6 \times 10^{- 19} \times 10^{6} = 8 \times 10^{- 12} J

Step 2: Find the amount of uraniumm per year.
The number of atoms required for fission per year:

\frac{2 \times 10^{10} \times 60 \times 60 \times 24 \times 365}{8 \times 10^{- 12}} = 78840 \times 10^{24} a t o m s

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

∴ M a s s o f 6.023 \times 10^{23} a t o m s o f U^{235} = 235 g = 235 \times 10^{- 3} k g

∴ M a s s o f 78840 \times 10^{24} a t o m s o f U^{235}

= \frac{235 \times 10^{- 3}}{6.023 \times 10^{23}} \times 78840 \times 10^{24}

= 3.076 \times 10^{4} k g

H e n c e, t h e m a s s o f u r a n i u m n e e d e d p e r y e a r i s 3.076 \times 10^{4} k g .

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