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Q. 13.17  The fission properties of _{94}^{239}\textrm{Pu} are very similar to those of _{92}^{235}\textrm{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}\textrm{Pu}  undergo fission?

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Number of atoms present in 1 kg(w) of  _{94}^{239}\textrm{Pu} =n

\\n=\frac{w\times N_{A}}{mass\ number\ of\ Pu}\\ n=\frac{1000\times 6.023\times 10^{23}}{239}\\n=2.52\times 10^{24}

Energy per fission (E)=180 MeV

Total Energy released if all the atoms in 1 kg  _{94}^{239}\textrm{Pu} undergo fission = E \times n

=180 \times 2.52\times1024

=4.536\times1026 MeV

 

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Sayak

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