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Additional Exercises · 13.30

Q.Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of 235U^{235}\text{U} in a fission reactor.

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Compute total energy from 1 kg of each fuel using the mass-defect (fusion, 4H→4He4\text{H}\rightarrow{}^4\text{He}) and per-fission (fission, 200 MeV/event) methods respectively. Fusion releases roughly 7.8 times more energy per kilogram than fission — consistent with fusion's much higher energy release per nucleon.

  1. Fusion of 1.0 kg of hydrogen The Sun's net proton-proton chain effectively converts 4 hydrogen nuclei into 1 helium-4 nucleus (plus positrons/neutrinos/gammas whose net effect is already captured in the atomic mass defect):

    Δm=4 mH−m(4He)=4(1.007825)−4.002603=4.031300−4.002603=0.028697 u\Delta m = 4\,m_H - m(^4\text{He}) = 4(1.007825) - 4.002603 = 4.031300 - 4.002603 = 0.028697\ \text{u}

    Qper event=0.028697×931.5=26.73 MeV per 4 H atoms fusedQ_{\text{per event}} = 0.028697 \times 931.5 = 26.73\ \text{MeV per 4 H atoms fused}

    Number of hydrogen atoms in 1.0 kg:

    NH=1000 g1.007825 g/mol×6.023×1023=5.976×1026 atomsN_H = \frac{1000\ \text{g}}{1.007825\ \text{g/mol}} \times 6.023\times10^{23} = 5.976\times10^{26}\ \text{atoms}

    Number of fusion events (4H→1He4\text{H}\rightarrow1\text{He}):

    Nevents=NH4=1.494×1026N_{\text{events}} = \frac{N_H}{4} = 1.494\times10^{26}

    Total energy:

    Efusion=1.494×1026×26.73 MeV=3.994×1027 MeV=3.994×1027×1.6×10−13 JE_{\text{fusion}} = 1.494\times10^{26} \times 26.73\ \text{MeV} = 3.994\times10^{27}\ \text{MeV} = 3.994\times10^{27}\times1.6\times10^{-13}\ \text{J}

    Efusion≈6.39×1014 JE_{\text{fusion}} \approx 6.39\times10^{14}\ \text{J}

  2. Fission of 1.0 kg of U-235 Number of U-235 atoms in 1.0 kg:

    NU=1000 g235 g/mol×6.023×1023=4.2553 mol×6.023×1023=2.563×1024 atomsN_U = \frac{1000\ \text{g}}{235\ \text{g/mol}} \times 6.023\times10^{23} = 4.2553\ \text{mol} \times 6.023\times10^{23} = 2.563\times10^{24}\ \text{atoms}

    Using the standard ≈200\approx200 MeV released per U-235 fission (as used elsewhere in this chapter): Efission=2.563×1024×200 MeV=5.126×1026 MeV=5.126×1026×1.6×10−13 JE_{\text{fission}} = 2.563\times10^{24} \times 200\ \text{MeV} = 5.126\times10^{26}\ \text{MeV} = 5.126\times10^{26}\times1.6\times10^{-13}\ \text{J} …

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