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Q.Calculate the amount of energy released in joules when 1 kg of 92235U^{235}_{92}U undergoes fission reaction.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2023Subjective· 3mImportance★★★★★
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Fissioning 1 kg of 92235U^{235}_{92}U (with each fission releasing about 200 MeV) liberates approximately 8.2×1013 J8.2\times10^{13}\,J of energy.

1. Number of atoms in 1 kg of 235U^{235}U. The molar mass of 92235U^{235}_{92}U is 235 g/mol235\,g/mol. Number of moles in 1000 g1000\,g:

n=1000235=4.255 moln = \dfrac{1000}{235} = 4.255\,\text{mol}

Using Avogadro's number NA=6.022×1023 mol−1N_A = 6.022\times10^{23}\,\text{mol}^{-1}, the number of 235U^{235}U nuclei is

N=n×NA=4.255×6.022×1023≈2.56×1024 nucleiN = n\times N_A = 4.255\times6.022\times10^{23} \approx 2.56\times10^{24}\ \text{nuclei}

2. Energy released per fission. Each fission of a 235U^{235}U nucleus (e.g., on absorbing a slow neutron) releases approximately

E1≈200 MeVE_1 \approx 200\,\text{MeV}

Converting to joules, using 1 MeV=1.6×10−13 J1\,\text{MeV}=1.6\times10^{-13}\,J:

E1=200×1.6×10−13=3.2×10−11 JE_1 = 200\times1.6\times10^{-13} = 3.2\times10^{-11}\,J

3. Total energy released. If every nucleus in the 1 kg sample undergoes fission, the total energy is …

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