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    Hi Guys,
    i got stuck on the part 2(a)(ii) of the question in the attachment . should i use the formula for activity to calculate t? but i don't have enough information to use the formula
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    iirc the fomula sheet gives you

    N=N0 e-λt (or equivalent)

    which can be rearranged by dividing both sides of the equals by N0 to get

    N/N0= e-λt

    and then taking the natural log gets you
    ln (N) - ln (N0) = -λt

    which should be useful
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    (Original post by Joinedup)
    iirc the fomula sheet gives you

    N=N0 e-λt (or equivalent)

    which can be rearranged by dividing both sides of the equals by N0 to get
    Yeah but my problem is i dont know how to find out the value of N and N0

    N/N0= e-λt

    and then taking the natural log gets you
    ln (N) - ln (N0) = -λt

    which should be useful
    Yeah but my problem is i dont know how to find out the value of N and N0
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    you don't need to give a monkeys about what the actual number of atoms is... but you ARE interested in the ratio (which is in the question)


    so try
    N0 = 1
    N = 0.375

    though any values where N=0.375 N0 will give the exact same answer
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    (Original post by Joinedup)
    you don't need to give a monkeys about what the actual number of atoms is... but you ARE interested in the ratio (which is in the question)


    so try
    N0 = 1
    N = 0.375

    though any values where N=0.375 N0 will give the exact same answer
    How did you realise that? The question only mentioned that the new Ca has 0.375 as many Ca-14 atoms as an equal mass of living wood? Did you get it from this part of the question?
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    (Original post by Alen.m)
    How did you realise that? The question only mentioned that the new Ca has 0.375 as many Ca-14 atoms as an equal mass of living wood? Did you get it from this part of the question?
    well it's a disguised ratio - only thinly disguised, but the examiner is hoping you'll bottle it if you see something that's not exactly what you're used to.

    just practice I suppose...
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    (Original post by Joinedup)
    well it's a disguised ratio - only thinly disguised, but the examiner is hoping you'll bottle it if you see something that's not exactly what you're used to.

    just practice I suppose...
    So basically if varaible A has 0.375 as many atom as varaible B of the same mass , the atom's ratio of A over B is 1/ 0.375?
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    (Original post by Alen.m)
    So basically if varaible A has 0.375 as many atom as varaible B of the same mass , the atom's ratio of A over B is 1/ 0.375?
    other way round I think; B over A is 1/0.375 (there's more atoms in B than A)
 
 
 
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