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Minimum amount of energy necessary to remove a proton from 42-Ca atom

  1. Oct 22, 2005 #1
    Q: What is the minimum amount of energy necessary to remove a proton from the nucleus of a 42-Ca atom, thereby converting it into a 41-K atom? The former has a mass of 41.958618 u, the latter 40.961825 u, and the hydrogen atom has a mass of 1.007825 u.

    My answer:

    42-Ca --> 41-K + 1-H

    Therefore change in mass (Delta m) = 0.011032 u

    I then used this mass in E = mc^2

    and got the answe E = 10.278 MeV

    Can someone else try this question and see if they get the same answer? Or tell me if i have done it correctly/incorrectly

    Thanks in advance :smile:
     
  2. jcsd
  3. Oct 22, 2005 #2
    Well you seem to have the correct answer - however the mass of a proton isn't the same as the mass of a hydrogen atom, so if you have the data for the mass of a proton you'd get a more accurate answer.
     
  4. Oct 22, 2005 #3

    Astronuc

    User Avatar

    Staff: Mentor

    The method is correct. Using atomic mass units includes the mass of the electrons. The binding energies of the electrons effectively cancel. There is perhaps a small error of a few eV or keV.
     
  5. Oct 24, 2005 #4
    Thanks for that
     
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