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Time taken for amount of a nuclear material to remain

  1. Dec 17, 2015 #1
    1. The problem statement, all variables and given/known data
    The half life T½ of Carbon-14 is 5730 years. What is its decay constant? After what length of time will 35% of an initial sample of Carbon-14 remain?

    2. Relevant equations
    Decay constant λ= 0.693 / T½
    Where N = amount of radioactive substance,
    N=N0e-λt

    3. The attempt at a solution
    Okay, so I can get the decay constant easily,
    λ= 0.693 / 5730 = 1.2094x10-4
    but I just can't figure out an expression for the time taken for the amount of nuclear material to remain.
    I'm thinking the amount can be expressed in the formula:
    0.35=e-λt
    It's probably some simple algebra but I can't figure out how to isolate t in this equation :confused:
     
  2. jcsd
  3. Dec 17, 2015 #2

    Mister T

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    Have you reviewed the basic algebra associated with the exponential function? I'll give you a hint: logarithm.
     
  4. Dec 17, 2015 #3

    SteamKing

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    Remember, ln (ex) = x
     
  5. Dec 17, 2015 #4
    Thanks for the hint :) I haven't done any maths in over 2 years before starting my physics course so I've forgotten a lot of basic stuff which they tend to skip out on in Uni.

    logeN = -λt
    loge0.35 = -1.209x10-4t

    t = loge0.35 / -1.209x10-4 = 8683.4 years.
    Is this correct?
     
  6. Dec 17, 2015 #5

    SteamKing

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    Seems to be. Half of the C-14 disappears after 5730 years, and 75% is gone after 5730 + 5730 = 11,460 years.
     
  7. Dec 18, 2015 #6

    SammyS

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    You can also find the number of half-lifes to get 35% by solving ##\displaystyle \ 0.35=\left(\frac{1}{2}\right)^x \,,\ ## where x is the number of half-lifes .

    That gives an answer close to yours.
    8678.5 years
     
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