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Nucleus Decay Question

  1. May 3, 2012 #1
    A proton collides with a nucleus of nitrogen-14 (atomic number 7). This collision produces a nucleus of carbon-11 (atomic number 6) and what else?


    My attempt:
    14/7N + proton -----> 11/6C + alpha particle?
    I know it cannot be gamma decay, but I'm not sure if it's alpha or beta decay and what particle is produced (alpha particle, proton, beta particle, or neutron). Thanks for any help.
     
  2. jcsd
  3. May 3, 2012 #2

    tms

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    Look at what must be conserved. Could charge, for instance, be conserved if a beta particle were emitted?
     
  4. May 3, 2012 #3
    B- wouldn't work, but couldn't B+ decay work in that case? If it's B+ decay though. I don't know what happens with the lost protons from the Nitrogen atom.
     
  5. May 3, 2012 #4

    tms

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    The initial charge is plus 7 for the nitrogen and plus 1 for the proton, so the final charge must also be plus 8. And charge is not the only thing conserved.
     
  6. May 3, 2012 #5
    So would proton + 14/7N ---> 11/6 + proton + alpha particle be correct?
     
  7. May 3, 2012 #6

    tms

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    No. Add up the charges on the right hand side; you now have too much. What is an alpha?
     
  8. May 3, 2012 #7
    the alpha particle has 2+, and the carbon has 5+ no? Sorry if I'm being difficult.
     
  9. May 3, 2012 #8
    Alpha decay is when a nucleus, obviously displeased with its present state splits into one helium (2 protons, 2 neutrons) and another nucleus that has two less protons and neutrons.

    So in this case the reaction would look like this:

    14/7N + 1/1H ---> 11/6C + 4/2He (15 = 15, amazing!)

    Since a lonely proton = 1/1H (Hydrogen)
     
  10. May 3, 2012 #9
    Thanks Tobbin, I look pretty stupid right now, but thanks for the clear explanation.
     
  11. May 3, 2012 #10

    tms

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    Correct for the alpha, but carbon has 6 protons, as you wrote in your earlier equations, not 5.
     
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