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Nuclear Fusion (Worksheet) - Solutions

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    Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.

    Rest masses in amu: electron, 0.00055, positron 0.00055, 1H, 1.007825; n, 1.008665; 2H, 2.0140; 3He, 3.01603; 3T, 3.01605; 4He, 4.00260; 12C, 12.000000, 16O, 15.99491.

    1. The burning of hydrogen in stars is usually represented by the reaction:
    2. \[4 1H ® 4He + 2 b+ + Q\]
      where Q is the energy of the reaction, includes g-ray energy. This reaction is not balance in terms of charge, but most literature gives the reaction equation this way. Write the balanced reaction equation and explain subsequent reactions of the positrons. How do these reactions contribute to the energy of the reaction, Q?

      Calculate the amount of energy Q in MeV (actually MeV/molecule) for each 4He atom formed. Also evaluate the energy Q in J per mole of 4He gas formed.

      If all particles are counted, we have the balanced reaction equation:

      1. 4 1H ® 4He + 2 (b+ + e-) + Q
        2 (b+ + e-) ® 4 g (photons)
        Q = (4*1.007825 - 4.00260) * 931.5 MeV
        = 26.7 MeV / reaction equation
        = 26.7 MeV * 6.02E23 /mol
        = 1.61E25 MeV/mol * (1.602E-13 J/MeV)
        = 2.58E12 J/mol

      The energy includes the annihilation energy of positrons and electrons.
    3. Calculate the energy Q in J released in each of the following hypothetical processes.
      1. 3 4He2 ® 12C6 + Q
      2. 6 1H1 + 6 1n ® 12C6 + Q
      3. 6 2D1 ® 12C6 + Q
      Discuss your results.


      1. Qa = 3 * 4.0026 - 12.000) amu * (1.4924E-10 J/amu)
        = 1.17E-12 J
      2. Qb = (6*(1.007825 + 1.008665) - 12.00000) amu * (1.4924E-10 J/amu)
        = 1.476E-11 J
      3. Qc = 6*2.014102 - 12.00000 amu * (1.4924E-10 J/amu)
        = 1.263E-11 J


      Relative energy content
      | 6 D
      | | 3 He
      1.476 1.263 0.117
      Fusion of He to give C releases the least amount of energy, because the fusion to produce He has released a large amount.

      The difference between the second and the third is the binding energy of deuterium.

      The conservation of mass-and-energy is well illustrated in these calculations. On the other hand, the calculation is based on the conservation of mass-and-energy.

    4. Hint...
      Each D, T fusion reaction releases 17.6 MeV
      (1.6021E-13 J/MeV) = 2.819E-12 J/(D, T, or He)

      In order to produce 1016 J, we carry out the calculation this way

      1. (1E16 J / 2.819E-12 J) (2 g of D2) / (6.022E23)
        = 11785 g
        = 11.8 kg D.
        Mass of T required is
        11.8 kg (3/2) = 17.7 kg.

      Since we are calculating the mass of D2, a mass of 2 rather than the precise rest mass of D is adequate. Why 2 g of D2 rather than 4 g of D2 is used?

    5. The heat of combustion for propane, C3H8, is 2202 J per mole,
      1. C3H8 + 5 O2 = 3 CO2 + H2O + 2202 J.
      The molar masses of propane and oxygen are 44.1 and 32.0 g/mol respectively. Calculate the weight in kg of propane and the weight of oxygen required to provide 1016 J energy. Comment on the contrast of weights between this and the previous problem.


              1 mol C3H8  (3*12.011+8*1.008) g C3H8
      10E16 J ----------  ----------------------
               2202 J        1 mol C3H8
        = 2.08E14 g
        = 2.1E11 kg of propane.


      Actually the heat of combustion for propane is 2202 kJ/mole. There are certain conditions to be met for this value, and for more accurate estimates, the conditions must be specified. In this simple exercise, you are only interested in developing a method to estimate. Thus the weight of propane is 2.1E8 kg (200 milion kg) as opposed to a total mass of 29 kg of T and D. Of course, a lot of physics, chemistry, and engineering is involved in the making of H-bomb.

    6. Hint...
      The electrolysis did show a large amount of energy released. The possible fusion reactions are those given in the text, namely,

      1. 2D + 2D = 3He + n + 3.3 MeV
        2D + 2D = 3T + p + 4.3 MeV
        and a hypothetical reaction,
        2D + 2D = 4He + ?.? MeV
      Thus, any of the product 3He, 4He, 3T, n, and p should be present. Gasses 2He, 2He, and 2T can be detected when the sample is subjected to chemical (mass spectrometry) analysis. The presence of proton is hard to confirm, because water is always a component in the experiment. However, the product neutron should induce some nuclear reactions, and the electrode should be radioactive as a result.

      The questions should provoke you to think deeper when you have just learned of a new discovery. Only when all your questions are satisfactory answered, you can accept the results.

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