Ncert Exercises & Solutions

13.17 The fission properties of \({ }_{94}^{239} \mathrm{Pu}\) are very similar to those of \({ }_{92}^{235} \mathrm{U}\). The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure \({ }_{94}^{239} \mathrm{Pu}\) undergo fission?

Hint: One atom of \({ }_{94}^{239} \mathrm{Pu}\) gives 180 MeV of energy.
Step 1: Find the number of atoms in 1kg of \({ }_{94}^{239} \mathrm{Pu}\).
Average energy released per fission of

{}_{94}^{239}P u, E_{a v} = 180 M e V

Amount of pure Atomic mass of

{}_{94}^{239}P u, m = 1 k g = 1000 g

N_{A} = A v o g a d r o n u m b e r = 6.023 \times 10^{23}

Mass number of

{}_{94}^{239}P u

=239 g

1 mole of

{}_{94}^{239}P u

contains

N_{A}

atoms.

∴ m g o f {}_{94}^{239}P u c o n t a i n s (\frac{N_{A}}{M a s s n u m b e r} \times m) a t o m s

∴ 1000 g o f {}_{94}^{239}P u c o n t a i n s (\frac{N_{A}}{M a s s n u m b e r} \times 1000) a t o m s = 2.52 \times 10^{24} a t o m s

Step 2: Find the total energy released.
Total energy released during the fission of 1 kg of

{}_{94}^{239}P u

is calculated as:

E = E_{a v} \times 2.52 \times 10^{24}

= 180 \times 2.52 \times 10^{24} = 4.536 \times 10^{26} M e V

Hence,

4.536 \times 10^{24}

MeV has released if all the atoms in 1 kg of pure

{}_{94}^{239}P u

undergo fission.

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