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Mass Defect

  1. Nov 1, 2014 #1
    1. The problem statement, all variables and given/known data

    Americium-244 is a rare isotope of Americium. What is the mass defect of Americium-244?

    Use the following values for atomic and neutron masses when calculating your answer:

    858f65e47a35265cc6aa7033b9f77cac.png
    3110f40de3ca8912c1ba39fbd209f10e.png
    6b4f52f4f36ea2c8c6ddb79df176172d.png

    2. Relevant equations


    3. The attempt at a solution

    This was what I did:

    Mass of protons = 95 x 1.007825 = 95.743375
    Isotope = 149 x 1.008665 = 150.291085

    Actual Mass Provided = 95.743375 + 150.291085 = 246.03446

    Mass Defect = 246.03446 - 244.064279 = 1.970181

    I've got that answer and its wrong. I also tried -1.97 and its still wrong. Can somebody please help me?

    Do i need to consider electron mass? I computed it to be: Mass Defect with electron mass = 244.064279 - (246.03446 + 0.052155) = 2.022336 which is - 2.022336. Is this correct?
     
    Last edited by a moderator: Apr 29, 2017
  2. jcsd
  3. Nov 1, 2014 #2

    Simon Bridge

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    The approach seems correct - computer mediated questions can be finicky about the exact form of the answer.
    Your "actual mass provided" seems to have been calculated to only 5dp, but 6dp accuracy provided ...
    I get:
    Code (Text):
    [SIZE=4]> (1.007825*95 + (244-95)*1.008665) - 244.064279[/SIZE]
    ans =  1.97018099999997
    amu for mass deficit.
    So it looks like it was fair to round up: you should provide the trailing zero to demonstrate the dp accuracy.
    http://physics.bu.edu/~duffy/sc546_notes10/mass_defect.html
    ... it may well be that you should include the electrons. Did you try? Did you check your notes?
     
  4. Nov 1, 2014 #3
    The answers should be left to 3 significant figures. Thats why I can't seem to point out where I went wrong. My lecture notes did not mention anything about it.
     
  5. Nov 1, 2014 #4
    Also, do you think the mass defect should be left as a negative answer?
     
  6. Nov 1, 2014 #5

    Simon Bridge

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    A negative mass deficit would be like a negative deceleration wouldn't it?
    If a negative surplus is a deficit then... but check how your course defines it.

    Usually you would keep the lowest sig-fig in multiplication ... the lowest would be 2 in the atomic number ... but that's an absolute number so it's really 95.0000 to 6 sig fig. The next lowest is the atomic weight - which is 3 sig fig ... since there may be different isotopes in the sample, one could argue that the sig-fig here is important but IMO that's over-thinking things: nobody uses sig-fig IRL.

    Maybe it's just a rule ... it means the computer is testing whether you can guess the format more than it tests your physics.
    Fortunately it's not an exam.
     
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