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Homework Help: Conservation of Energy fusion reaction

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the following fusion reaction
    2H+3H →4He + n
    in which deuterium and tritium fuse together to form a stable isotope of helium plus a neutron.
    In all of physics, we see experimental confirmation of Conservation of Energy, in which the total
    amount of energy in a closed system is constant. The total rest plus kinetic energy before the
    reaction must be equal to the total rest plus kinetic energy after the reaction.

    (a) Look up the masses of the three nuclei and of the neutron in kilograms, keeping at least six
    significant figures. Cite your reference. Because we have the same number of electrons before
    and after the reaction (and because atomic binding energies are tiny compared to the energies
    in this problem), it is not important whether you use nuclear masses or atomic masses.

    (b) Convert each of those masses into rest energies in electron-volts (eV). Recall that the speed of
    light is 299,792,458 m/s, and that 1 eV = 1.60217653× 10−19 J.

    (c) Assuming that the deuterium and tritium have a total of 0.05 MeV of kinetic energy before
    the reaction, use conservation of energy to calculate the total kinetic energy after the reaction.
    Recall that we handle addition and subtraction differently than multiplication and division
    when dealing with significant figures.

    (d) Assuming that all of this kinetic energy can be converted with 100% efficiency into usable
    energy, how many fusion reactions per second do we need to generate 1 MW of power?

    2. Relevant equations
    (a) Nuclei Masses:
    Deuterium: 3.32108e-27 kg
    Tritium: 4.98162e-27 kg
    Isotope of Helium: 6.64215e-27 kg
    Neutron: 1.67493e-27 kg

    3. The attempt at a solution
    (b)First I will convert the masses to eV (online conversion):
    Deuterium: 1,862,990,000 eV
    Tritium: 2,794,480,000 eV
    Isotope of Helium: 3,725,970,000 eV
    Neutron: 939,567,000 eV

    (c) I will add together the eV of 4He + neutron = 4,665,537,000 eV
    Now I will convert eV to MeV:
    4,665.537 MeV
    Can I subtract this by .05 MeV? I feel like I am on the wrong track
  2. jcsd
  3. Oct 14, 2009 #2
    do you go to msu?
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