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Homework Help: Need help with half-life of C-14 question (exponents unit)

  1. Jun 29, 2012 #1
    Question: Carbon-14 has a half-life of 5730 years. A bone is found in which [itex]\frac{7}{8}[/itex] of the original [16g of] carbon-14 has decayed. How old is the bone?

    Formula: C(t)=Co([itex]\frac{1}{2}[/itex])[itex]\frac{t}{H}[/itex] (it's hard to see but it says [itex]\frac{1}{2}[/itex] to the power of [itex]\frac{t}{H}[/itex])

    C(t) - the number at time t
    Co - the number or concentration at time 0
    2-1 or ([itex]\frac{1}{2}[/itex]) - because the amount is halving
    t - represents the elapsed time since measuring started (same units as H)
    H - the half life

    What I did was :

    C(t) = [itex]\frac{1}{8}[/itex] (since [itex]\frac{7}{8}[/itex] had already decayed) so i did
    I worked that out and got 40 110, which is wrong, but that doesn't matter right now. My problem is that the text worked it out this way:

    they did everything the same as me except, after figuring the [itex]\frac{1}{8}[/itex] from the [itex]\frac{7}{8}[/itex] they multiplied it by 16 for some reason getting 2 for C(t). So they start off with 2=16([itex]\frac{1}{2}[/itex])[itex]\frac{t}{5730}[/itex] and worked it out the same way I did. I can't figure out why they did [itex]\frac{1}{8}[/itex] * 16 ? Wouldn't they have to multiply the other side of the equation by another 16 too?
    They then go on to say C(t)=Co=([itex]\frac{1}{2}[/itex])[itex]\frac{t}{H}[/itex] but there is no way that can be true! Can someone please explain firstly why the multiply the [itex]\frac{1}{8}[/itex] by 16 in the first place, and secondly why they equate C(t) & Co & ([itex]\frac{1}{2}[/itex])[itex]\frac{t}{H}[/itex] ??
    Last edited: Jun 29, 2012
  2. jcsd
  3. Jun 29, 2012 #2
    From what I gather, the multiplication is due to an attempt to divide a fraction by another fraction:


    This would cancel out the 16 on the RHS and leave you with:


    Also from what I understand, the last part must be false.

    If [itex]c(t)=c_{0}[/itex] then [itex]t=0[/itex] and [itex]\frac{t}{H}=0[/itex], which means that [itex]\frac{1}{2}^\frac{t}{H}=1[/itex].
  4. Jun 29, 2012 #3
    You got the wrong idea. The initial condition is a constant. Since it says the original mass is 16g, that will be your initial mass. 1/8 of the 16grams is c(t) so c(t)=16(1/8)

    put that into the formula: 16(1/8)= 16(1/2)^(t/H) and solve for t

    C(t) is the mass after a time has passed. If 16grams is the original amount than after some time t it will have decayed. the problem tells you that you have 1/8 of the initial mass so you're told to find out how long it took for the 16grams to decay into 1/8 the original amount. Is that clear enough?
  5. Jun 29, 2012 #4
    if i understand what your saying, the C(t) = 16(1/8) so C=16 and t=1/8 but t on the RHS of the equation isn't equal to 1/8 or is it? other than that i get what your saying
  6. Jun 29, 2012 #5
    no what im trying to say is the function is equal to 16/8 which is 2.. you see c(t) is the mass after a certain time period. the question asks you for the time given the c(t) value which is 1/8 the original amount
  7. Jun 29, 2012 #6


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    Wrong, because on the LHS (left hand side), you're dealing with a fraction (of material left), whereas on the RHS, you're dealing with mass (of material you started with).

    To stay consistent, you should deal either with fractions on both sides, or masses on both sides.

    To get the mass left after decay, you take the fraction left (which is 1/8) and multiply it by the mass you originally started with (which is 16g). Hence you get (1/8) * 16 = 2g left after the decay.

    Why? The C0 on the RHS refers to the initial mass of 16g. It's not a fraction. Why would you multiply that again by 16?

    With every half-life (H), the amount of radioactive material will half. That's the nature of exponential radioactive decay. So after one half-life, your material goes to ½ the original, after another, you go to 1/4, etc. Remember that these are fractions of radioactive material remaining.

    So the fraction of the original remaining after n half-lives will be ½n. It's simple to see this is correct e.g. if n=0 (at start), the fraction is 1 (all the material remains), at n=1 (one half-life), the fraction is ½ (half remains), etc.

    n is the number of half-lives, which is just the time elapsed divided by the half-life, which is (t/H).

    As I mentioned above, the problem can be worked out either with mass (as they've done), or with fractions of material. The latter is actually quicker. If 1/8 of the original material remains at time t, then 3 half-lives (½3 = 1/8) have elapsed. That means t/H = 3, and hence t = 3*5730 years = 17,190 years.
  8. Jun 30, 2012 #7
    C0=16 grams

    If you only have 1/8 for the material left at the end, then C(t) must be 2 grams.
  9. Jun 30, 2012 #8


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    But the actual amount is irrelevant. 1/8 was left so you need to solve
    [tex]\left(\frac{1}{2}\right)^{t/5730}= \frac{1}{8}[/tex]

    8 is 2 to what power?
  10. Jun 30, 2012 #9
    I was only trying to point out to the OP what C0 and C(t) signify in his original equation. He seemed to be confused about this. Obviously, in a problem like this, the only thing that really matters is the ratio of C(t) to C0.
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