Question

The fission properties of ${}_{94}^{239}$Pu are very similar to those of ${}_{92}^{235}$U. the average energy released per fission is 180 Me V. if all the atoms in 1 kg of pure ${}_{94}^{239}$ Pu undergo fission, then the total energy released in me V is

• A. $4.53 \times 10^{26}MeV$
• B. $2.21 \times 10^{14}MeV$
• C. $1 \times 10^{13}MeV$
• D. $6.33 \times 10^{24}MeV$

Answer 1

Answer: (A)

$4.53 \times 10^{26}MeV$

Answer 2

: Number of atoms in 1 kg of pure $239pu$

$= \frac{6.023 \times 10^{23}}{239} \times 1000 = 2.52 \times 10^{24}$

As average energy released per fission is 180 MeV

$\therefore$total energy released

$$$= 4.53 \times 10^{26}MeV$