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Alpha decay

  1. Feb 16, 2016 #1
    1. The problem statement, all variables and given/known data

    Th-232 ---> He-4 + Ra-228 + energy

    How much is the energy?

    2. Relevant equations

    3. The attempt at a solution

    my solution is this:

    [m(Th-232)-90m(electron)] - m(He-4) - [8m(Ra-228)-88m(electron)]

    My book skipped the electrons, why?
     
  2. jcsd
  3. Feb 16, 2016 #2

    SteamKing

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    What's the mass of an electron compared to the mass of say, a proton?
     
  4. Feb 16, 2016 #3
    veeeeeeery small
     
  5. Feb 16, 2016 #4

    SteamKing

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    So if you took 88 veeery small electrons and put them together, would you have something which approached the mass of even 1 proton?
     
  6. Feb 16, 2016 #5
    no :P but my friend is saying that they skipped the electrons because there are 88 electrons before the decay and equally as much after. but if that was the case then I should be able to write the mass of the electrons and then cross them off because they cancel out each other
     
  7. Feb 17, 2016 #6
    Can someone tell me if my friend is right?
     
  8. Feb 17, 2016 #7
    Firstly, I think there should not be a 8(the blue one)here.

    Secondly, I doubt if the red 88 is true. In alpha decay, a positive helium nucleus without electrons is emitted and the 90 electrons in Th-232 should be retained in the Ra-228 formed. Thus, they are fully cancelled out.

    By the way, from what I have learnt, we focus on the mass defect inside the nucleus when calculating the energy released from nuclear reactions, so we don't consider the electrons.

    Correct me if I was mistaken though.
     
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