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    (Original post by sam2109)
    hi people im not sure if im doing these right or wrong, inputted will be appreciated:
    The half life of caseium is 42 years,
    calculate the K for Caseium
    i got Ln0.5 / 42 = 0.0165? is it right

    the next questionalculate the percentage of the original caseium that remains after 100 years?

    i got 5.21



    next question is calculate the time it takes for the amount of casium to fall 5% of its original value.
    i done ln0.05 / 42 - 181.56 hrs

    can anyone correct me please
    t1/2 = 0.693/k

    k = 0.693/42 = 0.0165

    Nt = N0e-kt

    So, when t = 100

    proportion remaining = e-kt = 0.192

    100 * Nt/N0 = 0.192 * 100 = 19.2%

    checksum -------------------------------------------------------------------

    One half life decreases count by 50%

    42 year half life in 100 years makes 2 and a bit half lives = between 12.5 and 25% - therefore reasonable answer.

    ---------------------------------------------------------------------------------

    If it falls to 5% then Nt/N0 = 0.05

    e-kt = 0.05

    -kt = -3

    therefore t = 3/0.0165 = 181.6 years

    checksum -------------------------------------------------------------------

    5% falls between four and five half lives (50%, 25%, 12.5%, 6.25%, 3.125%)

    four half lives = 4 x 42 = 168 years
    five half lives = 5 x 42 = 210 years

    Therefore 181.6 years is reasonable
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    (Original post by sam2109)
    thanks very much for helping identify my answer.
    i would like to know how u got 19.2%?, show me the procedure, once again thnaks for ur help
    I've shown you the working!

    You start with the decay equation:

    Nt = N0e-kt

    This says that the activity at time = t is equal to the activity at time = 0 (the initial activity) multiplied by the term e-kt

    You simply rearrange this and substitute in the values for t and k.

    You are told that time t = 100

    You have already worked out k = 0.0165

    You assume that N0 is 100% activity (it's the initial value)

    So you can work out Nt
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    (Original post by sam2109)
    i get a different answer
    0.693*2.718^-0.0165*100 =13.3???????????
    I don't know where you get this from! Why have you multiplied by 0.693 (ln2) - it doesn't appear in the equation.

    If you work out 'kt' first you get 0.0165 x 100 = 1.65

    Therefore e-1.65 = Nt/N0

    0.192 = Nt/N0

    So if N0 is taken as 100% then Nt = 19.2%

    ------------------------------------------------------------------------------------------
    Do you understand my checksum?

    If you do one for your answer (13.3%)

    Half lives give a percentage remaining according to the number of half-lives passed:

    1 half-life = 50%
    2 half-lives = 25%
    3 half-lives = 12.5%

    Your value of 13.3 is very close to 12.5 and so you would expect just less than three half lives to have passed. But 3 x 42 = 126, which is much larger than 100 (more than a half half-life bigger). It makes your answer very improbable (wrong)

    It is always worthwhile looking for a way of estimating the probability of an answer being reasonable (checksum) to give confidence in your answer and to make sure you haven't made a mistake in your assumptions or calculations.
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    Just a shame there isn't a Cs isotope with a 42 year half life!
 
 
 
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