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# half life

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1. 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
Attached Images

2. 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
3. (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
4. 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
5. (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?
6. (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...
7. (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?
8. (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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