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Homework Help: 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:


    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.
    ... 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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