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Half-life exercise, radioactive/non-radioactive ratio

  1. Feb 26, 2012 #1
    Hey guys,
    I've been given a worksheet with a couple of questions concerning radioactivity, half-life etc. It is high school level and for me, it seems like the first year doing physics as our current teacher is the first one who really knows something about the subject. :) So please be patient with me. :)

    1. The problem statement, all variables and given/known data
    "Consider having one kilogram of radioactive nitrogen 13 isotope – half-life is 10 minutes. How much nitrogen will you have after 30 minutes? When is the ratio between radioactive and non-radioactive material 1:31? Estimate when the ratio is 1:10."
    m0 = 1 kilogram
    mt (final mass) = x
    t = 30 minutes
    tHL (half-life) = 10 minutes

    2. Relevant equations
    mt = m0/2^x

    3. The attempt at a solution
    mt = 1/2^3 = 0.125 kg (since 3 H-Ls have occurred in the 30mins.) But that is everything I got to. I'm completely lost with the ratios. Could you help me, please?

    Next problem:
    1. The problem statement, all variables and given/known data
    "How many times do you have to cut your fingernail in half to separate atoms from each other? How many times to get to atom’s nucleus? (suppose the fingernail is 0.5 cm long)

    2. Relevant equations
    Well, as far as I know, the size of an atom should be equal to 10^-10 m while the size of a nucleus is 10^-14 m.

    3. The attempt at a solution
    I think I need to convert the fingernail's size to a negative power so that it is easier to compute the quotient. So, 0.5 cm = 0.005 m = 5x10^-3 m. And I'm done.. :/ Could someone please explain the process in detail?

    Thanks so much!
  2. jcsd
  3. Feb 26, 2012 #2


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    This might help

    In one half-life, approximately half of a radionuclide will decay. Assuming that a radio nuclide decays to a non-radioactive (inert or stable) nuclide, in one half-life, the ratio of radioactive to nonradioactive is 1:1. In two half-lives, one-quarter (1/4) of a radio nuclide remains, which means that 3/4 of the atoms would be non-radioactive, so the ratio (r:nr) = 1:3. What fraction (or proportion) of r atoms remain such that r:nr = 1:31.

    The cutting in half is a similar concept. One halves the length, with each cut, so a cut is treated like a half-life.
  4. Feb 26, 2012 #3
    Oh, thanks! So, if I understand it correctly the answer to the first one will be after 5 half-lives, am I right?

    Because 1/2^5 = 1/32 -> 1/32 remains radioactive while 31/32 have decayed, i.e. they are not radioactive -> the ratio is 1:31. Correct?

    To the second exercise - I understand the concept is the same but the negative powers keep confusing me. I know that I can rewrite 0.005 m as 5x10^-3 m, but what about the rest? The next half will be 0.0025 m - and I don't know how to rewrite it so that I can easily compare it to the size of an atom at the end. Any help?
  5. Feb 26, 2012 #4


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    Answer to first part of the first problem is correct. Now one has to estimate when the ratio is 1:10.

    In the second problem, let x be the remaining length after cutting (in half) N times. Starting with L, after one cut, one has x=L/2, after 2 cuts one has x=L/4, or (L/2)2, now find the expression for x after N cuts.
  6. Feb 26, 2012 #5
    Ok - so the 1:10 ratio will be obtained somewhere between 3 and 4 half-lives, right? 2^3 = 8, 2^4 = 16... I would say something like 3.3 half-lives...?

    To the second problem - if I know the final size has to be 10^-10 meters (the size of an atom), the equation should be as follows:
    10^(-10) = 0.005 (size of the fingernail) /2^x
    Now solve for 2^x:
    2^x = 0.005/ 10^(-10) = 50,000,000 half-lives
    So, we would have to cut it 50,000,000 times in order to match the size of an atom... Is it right? The figure seems to be quite huge :D
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