The Student Room Group
Reply 1
Which part are you confused about? Tell us where you're stuck and we can help with specific points :smile:
Reply 2
i'd start off by writing out what you have, what are the atomic numbers for all the particles involved etc, just be sure the equation balances
Reply 3
yhsa
Which part are you confused about? Tell us where you're stuck and we can help with specific points :smile:



Oh i'm realy sorry i forgot to mention that ive done i) and ii). It is iii) that i'm not sure of.
Reply 4
Well, you have the half-life, so you can work out what the decay constant is. You can then use the nuclear decay law (equation) to work out initial activity :smile:

Hint: A=dNdt=λNA = -\frac{dN}{dt} = \lambda N

So you can get the decay law in terms of A rather than N :smile:
Reply 5
what formula relates, half life, time, activity and initial activity?

Spoiler

Reply 6
yhsa
Well, you have the half-life, so you can work out what the decay constant is. You can then use the nuclear decay law (equation) to work out initial activity :smile:

Hint: A=dNdt=λNA = -\frac{dN}{dt} = \lambda N

So you can get the decay law in terms of A rather than N :smile:


I worked out λ \lambda to be = 1.80 X 10^11
Reply 7
wizz_kid
I worked out λ \lambda to be = 1.80 X 10^11

λ\lambda is usually very small.
Reply 8
yhsa
λ\lambda is usually very small.


half life X lambda = 069

so
Unparseable latex formula:

\lambda = \frac {069} {5730 \times 365 \times 24 \times 60^2} = 3.82 \times 10^-^1^2

. Sorry calculation error!

So once i've worked out the decay constant, how do i work out the initial activity?
Reply 9
wizz_kid
half life X lambda = 069

so
Unparseable latex formula:

\lambda = \frac {069} {5730 \times 365 \times 24 \times 60^2} = 3.82 \times 10^-^1^2

. Sorry calculation error!

So once i've worked out the decay constant, how do i work out the initial activity?

Well, N=NoeλtN = N_o e^{-\lambda t}, right?

And A=λN A = \lambda N

so A=Aoeλt A = A_o e^{-\lambda t}.

You have A, λ\lambda and t :tongue:
Reply 10
yhsa
Well, N=NoeλtN = N_o e^{-\lambda t}, right?

And A=λN A = \lambda N

so A=Aoeλt A = A_o e^{-\lambda t}.

You have A, λ\lambda and t :tongue:



This is from phy3 where that equation is not given. Isnt there a simpler way? :p:
Doing it ur way, do i sub in 12000 years as t? My calculator gives an error that way.
Reply 11
wizz_kid
This is from phy3 where that equation is not given. Isnt there a simpler way? :p:
Doing it ur way, do i sub in 12000 years as t? My calculator gives an error that way.

It's just one of those equations you're expected to know, I think. You should see the kind of thing I have to remember :tongue:

And really? Mine doesn't... Got your units right?

I get about 1384 Bq, though I'm not really paying attention so there could be a calculator mistyping in there somewhere :tongue:
Reply 12
yhsa
It's just one of those equations you're expected to know, I think. You should see the kind of thing I have to remember :tongue:

And really? Mine doesn't... Got your units right?

I get about 1384 Bq, though I'm not really paying attention so there could be a calculator mistyping in there somewhere :tongue:


Thanks for replying. The mark scheme (for some reason) says that the answer is 5124 Bq :




Thanks again.
Reply 13
wizz_kid
Thanks for replying. The mark scheme (for some reason) says that the answer is 5124 Bq :




Thanks again.

Fail, well, there was that calc error I was talking about :tongue: I'd trust the mark scheme over me.
Reply 14
yhsa
Fail, well, there was that calc error I was talking about :tongue: I'd trust the mark scheme over me.


From the markscheme, once you divide 12000 by 5730 , what do u do next to get 5124Bq ?

Also for the synoptic, do we need to know the equations that u just quoted?

Thanks.