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Homework Help: Halflife question

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data

    An archaeologist finds an oak wine cask in one of her digs. Testing the activity from the radioactive carbon-14 in the cask reveals that it is only one-quarter that of the activity coming from the modern sample of the same type of oak. How old is the sample?

    2. Relevant equations

    3. The attempt at a solution
    I tried finding the halflife of carbon-14 (which is 5.73EXP3a) from a chart given in my textbook.
    after that, i tried filling int he blanks for my equation above, but i just didnt know how to formulate it.
  2. jcsd
  3. Feb 16, 2010 #2


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    Gold Member

    You could use your equation to solve this one. The amount of carbon 14 based radioactivity is proportional to the amount of Carbon 14 present.

    [tex] A = A _0 (0.5)^{\frac{t}{t_{1/2}}} [/tex]

    Here, [tex] A_0 [/tex] can be considered the activity of the modern sample. Since it is being used as the reference of comparison, you can set it equal to 1. Then [tex] A [/tex] becomes the activity of a sample relative to a modern sample (0.25 in this case). [tex] t _{1/2} [/tex] is the halflife. Then just solve for [tex] t [/tex].

    There's also an easier way to do this particular problem. If the activity is 1/2 of that compared to a modern sample, then the age is one halflife. If the activity is 1/2 of 1/2, then the age is two halflives, and so on.
  4. Feb 16, 2010 #3
    Thank you so much, you helped me a lot.
    Just a question, is there another way to formulate this question using log?
  5. Feb 16, 2010 #4


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    There certainly is! Just take the logarithm of both sides of the equation, and note that

    [tex] Log(x^a) = aLog(x) [/tex]

    or more specifically in this case,

    [tex] Log \left( \left(0.5 \right) ^\frac{t}{t_{1/2}} \right) = \frac{t}{t_{1/2}}Log(0.5) [/tex]

    You'd have to do that anyway when solving for t. So in other words, take the logarithm of both sides (noting the above identity), and solve for t.

    Btw, I used the common logarithm, but you can use the natural logarithm if you'd like or logarithms of any base for that matter. Whatever you prefer.
  6. Feb 16, 2010 #5
    again, thanks so much!
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