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    an isotope has a half life of 195 years.

    How many years must elapse before the number of atoms of this parent
    isotope has decayed to 1% of the original number?

    how would you go about working this out? i guess you need to work out how many half lives it has first? and the D/P ratio?

    any advice appreciated
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    N=Ae^(-xt)

    where N is number of nuclei at time t
    where A is number of nuclei at time 0 (original nuclei)
    where t is time
    where x is decay constant

    N/A = 0.01 = e^(-xt)
    ln(0.01)=-xt
    :. t = -ln(0.01)/x

    To find out the decay constant, just use equation:
    x = ln2/half life
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    1 Minute Ago 23:07 (ogloom)
    N=Ae^(-xt)

    where N is number of nuclei at time t
    where A is number of nuclei at time 0 (original nuclei)
    where t is time
    where x is decay constant

    N/A = 0.01 = e^(-xt)
    ln(0.01)=-xt
    :. t = -ln(0.01)/x

    To find out the decay constant, just use equation:
    x = ln2/half life
    my text book is basic though. i have no idea what that equation is or how to use it. sorry :o:
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    (Original post by Flo_Ryder89)
    my text book is basic though. i have no idea what that equation is or how to use it. sorry :o:
    For ANY exponential decay (radioactive decay, capacitor discharge etc..)

    can be modelled with: N = Ae^(-lambda*t)
    If we want to how long it takes for nuceli (N) to reach 1% of the original value:
    or 1% of A. we want to know time t when N = 0.01A:

    then: 0.01A = Ae^(-lambda*t)
    the A's cancel
    then: 0.01 = e^(-lambda*t)
    log both sides
    then: ln(0.01)=-lambda*t

    lambda is otherwise known as the decay constant, where the decay constant is probabiltiy of a nucleus to decay (per second).
    Hence Activity = lambda * Total nuclei (as probability of nucelus decaying per second * total nuclei = nuclei decaying per second).
    lambda is equal to ln2/(half life), which can be proved using the equation mentioned above. We have the half life, therefore we can work out lambda. Now that we know lambda, we simply plug it in to ln(0.01)=-lambda*t to find out t. Hope this helps =]
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    ok thanks for youre help. ill have a go, see what i get
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    (Original post by Flo_Ryder89)
    ok thanks for youre help. ill have a go, see what i get
    make sure you convert years --> seconds
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    (Original post by ogloom)
    make sure you convert years --> seconds
    ok thanks
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    You dont need to convert - just work in years throughout.
 
 
 
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