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I Matter conversion during fission

  1. Jul 21, 2017 #1
    Hi all,

    This is probably a newbie question - I'm a non-physicist but I did do some physics at university.

    My question is, during fission, when matter is converted into energy, what is the matter/particle that is actually converted into this energy? I would imagine that the particle(s) converted into energy probably varies depending on the fission feedstock. So if we take uranium 235, for example - can anyone tell me what particle(s) are no longer present after the fission process?

    Thanks,
    Craig
     
  2. jcsd
  3. Jul 21, 2017 #2

    Bandersnatch

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    Hi, helmetheid. Welcome to PF!

    There are no disappearing particles. The energy comes from the difference in binding energy between one larger blob of protons and neutrons, and two smaller blobs made up the same total number of nucleons.
    The two smaller blobs have all their components closer together, which allows them to be better held by the strong nuclear force, while having less positively charged protons, which additionally means the nucleus is less pulled apart by electrostatic repulsion.
     
  4. Jul 21, 2017 #3
    Hi,
    Thanks for that reply - I think I follow it.

    Can I ask, if the energy released comes from the difference in binding energy, where is the mass lost from after fission?

    It's my understanding that after the fission process the total mass is less than the total mass before fission. And I thought the reduction in mass is what produced the energy released through e=mc^2
     
  5. Jul 21, 2017 #4

    Bandersnatch

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    Yes, that's correct. But the difference in mass = the difference in binding energy. It's the same thing.

    Two more tightly bound systems have less energy = are less massive than one less tightly bound system made up of the same components.
     
  6. Jul 21, 2017 #5
    Ok, thanks. That's really interesting. So does it follow that the binding energy has some kind of mass component itself, despite not being constituted of matter? So that a reduction in the binding energy means a reduction in mass
     
  7. Jul 21, 2017 #6

    Bandersnatch

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    Here, I think I know what might be throwing you off - the sort of general common sense one acquires is that mass is something only particles ('matter') have.

    Mass is intuitively understood here as one of the things that cause gravity (other being stress and momentum, but these are not important here), and the thing that resists changes in motion (inertia).

    The first thing to realise is that mass as described above and energy are the same thing. One might just as well use the two interchangeably.

    The second thing to realise is that mass=energy is not only a property of particles, but also of their arrangement and motion.

    The last thing needed is that there is a property some particles have called 'rest mass', which is the energy 'encoded' in an isolated, motionless particle. This rest mass is just one of the components of the gravitational and inertial mass/energy described above. So as not to confuse mass and rest mass, it's best to stick to calling the former one energy (we'll do that from now on).

    Now, let's take U-235 fission process (the first step only), where it spontaneously decays into a thorium-231 nucleus and an alpha particle.
    The total energy (remember, that's the conventional idea of what 'mass' is) of a U-235 nucleus, ##E_{total}## is the sum of rest masses of its nucleons:
    $$m_{U235, rest, total} = 92*m_{rest, proton} + 143*m_{rest, neutron}$$
    plus their binding energy ##E_{binding}##. Binding energy in attractive potential is negative, so adding it makes the ##E_{total}## lower. The more tightly bound the nucleus, the more negative the energy.
    So, the total energy balance in the fusion reaction is:
    $$(m_{U235, rest, total} + E_{binding, U235}) = (m_{Th231, rest, total} + E_{binding, Th231}) + (m_{alpha, rest, total} + E_{binding, alpha}) + E_{released}$$
    Where the rest energies of all components (protons+neutrons) of U235 equal the sum of rest energies of components of Th231 and the alpha particle (which is made of 2 protons and 2 neutrons). That is to say, both sides of the equation have the same number of protons and the same number of neutrons.
    But the energy of U235 is reduced by its binding energy ##E_{binding, U235}##, which is less negative than the sum of binding energies of Thorium-231 and the alpha particle:
    $$E_{binding, U235}>E_{binding, Th231} + E_{binding, alpha}$$
    (remember higher binding energy means less negative)
    The excess energy coming from the difference shown above is the released energy, or in other words - the mass defect. That energy is the missing mass.
     
  8. Jul 22, 2017 #7
    Hi,

    Thanks a lot for that explanation - it was really clear. I hadn't encountered rest mass and relative mass before, so have just been reading up on them. New to me, and really, really interesting
     
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