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Determine the value of a decay constant

  1. Aug 26, 2015 #1
    1. The problem statement, all variables and given/known data

    What's up guys, I'm stuck on #2 of this worksheet,http://luc.edu/faculty/dslavsk/courses/phys111/homework/phys111-2015hw1.pdf
    The half - life of C - 14 is 5730 years. Determine the value of its decay constant (and express the answer in appropriate MKS units).

    2.The attempt at a solution
    I'm not sure if I'm doing this correctly , but I set up my formula like this , 5730=14e-(constant of decay)5730
     
    Last edited: Aug 26, 2015
  2. jcsd
  3. Aug 26, 2015 #2

    SammyS

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    Hello . Welcome to PF !

    You will get a better response if you display the question so that it can be read directly. The following is what I was able to obtain from your pdf file using Window's "Snipping Tool" & easily display here.
    upload_2015-8-26_20-13-4.png

    Better yet, you could type out the problem. I makes it easier to respond to, and it shows initiative on your part.

    As far as your attempt at a solution goes, it looks like you simply picked out some numbers at random from the problem statement and then plugged them into the given equation.
     
  4. Aug 26, 2015 #3
    I've been using this fourm. https://www.physicsforums.com/threads/decay-constant.24169/ , as a guide and I can't figure out what to do... it's the first assignment of the semester and we haven't gone over this...
     
  5. Aug 26, 2015 #4

    SammyS

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    C-14 is carbon 14. That's just an identifier for the type of nucleus being addressed. The number 14, will not be involved anywhere in the solution.

    The problem statement gives many hints, maybe even telling you much of what you need to do to solve the problem.

    If you start with N0 C-14 nuclei, then after 5730 years (one half-life), how many C-14 nuclei remain in your sample?

    That quantity is what you put in for N(5730) .
     
  6. Aug 26, 2015 #5
    Omg, you just made it so easy... so the formula would be 5370=(.5)e^(-(decay)5370) or would I set the entire thing to zero?
     
  7. Aug 26, 2015 #6

    SammyS

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    No.

    And you didn't answer the main question I asked..

    If you start with N0 nuclei, then how many will remain after one half-life has elapsed?
     
  8. Aug 26, 2015 #7
    Half of the N0 nuclei would remain.
     
  9. Aug 26, 2015 #8

    SammyS

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    Correct.

    That's what goes on the left hand side of the equation.
     
  10. Aug 26, 2015 #9
    So the formula would start looking like this .5=(.5)e^-(decay)5730
     
  11. Aug 26, 2015 #10

    SammyS

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    How did you get that?

    Where does the .5 come from on both sides?

    Isn't the left side N0/2 , and the right side N0 times the exponential ?
     
  12. Aug 26, 2015 #11
    I assumed that .5 is half, I get confused when I don't enter values in. So No/2=N0e^-(decay)5730? The thing that confuses me is what values to put in the fourmla that I have now.
     
  13. Aug 26, 2015 #12

    SammyS

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    You should use another set of parentheses to make that technically correct.

    0.5⋅N0=N0e^(-(decay)5730) .

    Just solve for the decay constant. First step is to divide both sides by N0 . Right ?
     
  14. Aug 26, 2015 #13
    Okay I divided both sides by No and then took the natural log of both sides. Afterwards I divided by -5730 to get my answer, correct?
     
  15. Aug 26, 2015 #14

    SammyS

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    Yes, but 5730 is in units of years. That gives you the decay constant in units of years-1. MKS would be in units of seconds-1.
     
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