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What is the half-life of radon-222?

  1. Feb 19, 2004 #1
    After 3 days a sample of radon-222 decayed to 58% of its original amount:
    a) what is the half-life of radon-222?
    b) how long would it take the sample to decay to 10% of its original amount?

    check my answers please:

    a) m(t)=mo*e^kt
    t = 3 and m(t)=.58mo
    0.58mo = mo*e^3k
    ln(.58) = 3k
    k = -0.1816

    1/2 = e^(-0.1816t)
    ln(1/2)= -.1816t
    t = 3.817 half life is approx 3.817 days

    b) e^(-0.1816t) = .10
    -0.1816t = ln(.10)
    t = 12.68 it will take approx 12.68 days to reach 10%

    the answer to (b) throws me. if half life is 3.817 days, 0% would be 7.634 days. where does this 12.68 days come from? i must be doing something wrong, unless (a) is wrong too.

    thanks in advance
  2. jcsd
  3. Feb 20, 2004 #2
    You have a fundamential misunderstanding of [tex]\lambda[/tex]. Half-life is the time required for 1/2 of a sample to decay thus forming another product. If 1/2 of a sample decays in x days, the other half still remains... That other half would then undergo a decay for x days leaving 1/2 of the sample (1/4 of the original sample). This process continues on an on and on dividing a sample by 1/2 each time, so simply assuming 2 half lifes equals 100% of a sample is flawed.

    Look here, I plugged the numbers you got into excel and spred the data out beginning from t=0[tex]\lambda[/tex] to t=15[tex]\lambda[/tex]. If you notice each step is approximetly 1/2 of the prevuous (this was due to in-precise rounding). It took about 11 1/2 lives after time zero for the sample to theoretically decay to an insignificant ammount which is substantially greater than 7.6 days (more along the lines of 42 days):

    t1= 50.15359446
    t2= 25.15383037
    t3= 12.61555008
    t4= 6.327151823
    t5= 3.173294066
    t6= 1.591521037
    t7= 0.798205007
    t8= 0.400328502
    t9= 0.200779133
  4. Feb 20, 2004 #3


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    Saying that the half-life of any substance is "T" is the same as saying that [tex]X(t)= X(0)\({\frac{1}{2}}\)^{\frac{t}{T}}[/tex]

    where t is the time measured in whatever units T is in.

    Since it takes 3 days to decay to 58% of the original amount, we have [tex]X(3)= X(0)\({\frac{1}{2}}\)^{\frac{3}{T}}= 0.58X(0)[/tex]
    so that [tex]\({\frac{1}{2}}\)^{\frac{3}{T}}= 0.58[/tex]
    Then [tex]\frac{3}{T}ln(\frac{1}{2})= ln .58[/tex]
    [tex]\frac{3}{T}= 0.786 [/tex]
    [tex]\frac{T}{3}= \frac{1}{0.786}= 1.27 [/tex]

    and, finally T= 3(1.27)= 3.82 days. ('cuz 0.58 is pretty close to 0.5!)

    Knowing now that [tex]X(3)= X(0)\({\frac{1}{2}}\)^{\frac{3}{3.82}}= 0.58X(0)[/tex]
    We can answer the second question by solving
    [tex]\(\frac{1}{2}\)^{\frac{t}{3.82}= 0.10 [/tex]
    [tex]\frac{t}{3.82}ln(0.5)= ln(.10) [/tex]
    [tex] t= 3.82(\frac{ln(.10)}{ln(0.5)}) [/tex]
    t= 12.68 days.

    "the answer to (b) throws me. if half life is 3.817 days, 0% would be 7.634 days. "

    No, that would be a linear function. The whole reason this has a "half-life" is that it is an exponential function. If it decreases to half in 3.817 days, it will decrease to half of that (that is to 1/4 of the original amount) in another 3.817 days. It decreases by the same proportion not the same amount.
    (Decreasing by the same amount is linear.)
    Last edited by a moderator: Feb 20, 2004
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