The fission properties of $_{94}^{239} P u$ are very similar to those of $_{92}^{235} U$ . 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 $_{94}^{239} P u$ undergo fission?

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Solution

Number of atoms in 1 kg of pure Pu= $\frac{6.023 \times 10^{23}}{239} \times 1000 = 2.52 \times 10^{24}$

As average energy released in fission is $180 M e V$ , the total energy released is
$\Rightarrow 2.52 \times 10^{24} \times 180 M e V$

$\Rightarrow 4.53 \times 10^{26} M e V$ .

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Similar questions

Q. The fission properties of 23994 Pu are very similar to those of 23592 U . Theaverage energy released per fission is 180 MeV. How much energy,in MeV, is released if all the atoms in 1 kg of pure 23994 Pu undergofission?

A reactor is generating energy at the rate of $1500 k W$ . How many atoms of $U_{235}$ undergo fission per second, if $180 M e V$ energy is released per fission?

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The fission properties of 23994Pu are very similar to those of 23592U. 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 23994Pu undergo fission?

Number of atoms in 1 kg of pure Pu=6.023×1023239×1000=2.52×1024As average energy released in fission is 180MeV, the total energy released is⇒2.52×1024×180MeV⇒4.53×1026MeV.

The fission properties of $_{94}^{239} P u$ are very similar to those of $_{92}^{235} U$ . 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 $_{94}^{239} P u$ undergo fission?

Number of atoms in 1 kg of pure Pu= $\frac{6.023 \times 10^{23}}{239} \times 1000 = 2.52 \times 10^{24}$

As average energy released in fission is $180 M e V$ , the total energy released is
$\Rightarrow 2.52 \times 10^{24} \times 180 M e V$

$\Rightarrow 4.53 \times 10^{26} M e V$ .