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Half Lives - Physics

A substance has a half-life of 100s, and starts with 10^20 unstable nuclei. Calculate the initial activity, and from this work out the time taken for all of the nuclei to decay if the activity did not decrease with time (instead of remaining proportional to the number of remaining unstable nuclei as it actually should). Give your answer to 2 significant figures.
If you need help with the question I'd recommend posting your working thus far? I can't really see whether you're asking for it or if you simply wish for an answer since it isn't provided :_;

To start with this, remember the formulae that associates half-life and the decay constant, lambda, i.e. const. decay = (ln(2))/T(half life) and then look back at the definition of the decay constant: that'll let you solve the first part.

The second is a question regarding decay with a constant rate, i.e. analogous to moving a distance at constant velocity and finding the time.
Reply 2
Original post by Callicious
If you need help with the question I'd recommend posting your working thus far? I can't really see whether you're asking for it or if you simply wish for an answer since it isn't provided :_;

To start with this, remember the formulae that associates half-life and the decay constant, lambda, i.e. const. decay = (ln(2))/T(half life) and then look back at the definition of the decay constant: that'll let you solve the first part.

The second is a question regarding decay with a constant rate, i.e. analogous to moving a distance at constant velocity and finding the time.


Hey, sure thing. I managed to do the first part but what has stumped me is the part where activity stays the same which would be weird since that would mean the decay constant would no longer stay constant.

Here's the workings I did to calculate activity:

Decay Constant = ln(2) / 100
= 6.93e-3 s^-1

Initial Activity = Decay Constant x Initial no. of nuclei in sample
= 6.93e-3 x 10^20
= 6.9315e17 Bq

Attempt at calculating the time for all nuclei to decay:

Since Activity is constant I'd imagine the graph of no. of nuclei against time to now be negatively proportional as opposed to inversely proportional. This would mean it would be a straight line where when the line intersects the x-axis at time t, this should be my time for all nuclei to decay :biggrin:

The gradient of the line should hence be the Activity as this is the no. of nuclei that decay every second.

y = mx + c

y = no. of unstable nuclei in sample, N
m = -A
x = Time, t
c = Initial total no. of nuclei, I

N = -At + I

Crosses the x - axis when N = 0

At = I
t = I/A
= (10^20) / (6.93e17)
= 144 s


144s is the wrong answer and isn't nearly large enough to be correct
Original post by S.H.Rahman
Hey, sure thing. I managed to do the first part but what has stumped me is the part where activity stays the same which would be weird since that would mean the decay constant would no longer stay constant.

Here's the workings I did to calculate activity:

Decay Constant = ln(2) / 100
= 6.93e-3 s^-1

Initial Activity = Decay Constant x Initial no. of nuclei in sample
= 6.93e-3 x 10^20
= 6.9315e17 Bq

Attempt at calculating the time for all nuclei to decay:

Since Activity is constant I'd imagine the graph of no. of nuclei against time to now be negatively proportional as opposed to inversely proportional. This would mean it would be a straight line where when the line intersects the x-axis at time t, this should be my time for all nuclei to decay :biggrin:

The gradient of the line should hence be the Activity as this is the no. of nuclei that decay every second.

y = mx + c

y = no. of unstable nuclei in sample, N
m = -A
x = Time, t
c = Initial total no. of nuclei, I

N = -At + I

Crosses the x - axis when N = 0

At = I
t = I/A
= (10^20) / (6.93e17)
= 144 s


144s is the wrong answer and isn't nearly large enough to be correct


The first part seems correct... as for the activity remaining constant, the answer should be correct? Perhaps the question is flawed, that's the only way this wouldn't be the answer.
Reply 4
Original post by Callicious
The first part seems correct... as for the activity remaining constant, the answer should be correct? Perhaps the question is flawed, that's the only way this wouldn't be the answer.


I doubt it, it's from Isaac Physics which I haven't really seen make mistakes like that on answers. I think if anything's wrong in my answer it's probably assuming the gradient was -A. I'll ask my teacher and see what he has to say about it. Would you like me to let you know what the answer is once I find out?
Original post by S.H.Rahman
I doubt it, it's from Isaac Physics which I haven't really seen make mistakes like that on answers. I think if anything's wrong in my answer it's probably assuming the gradient was -A. I'll ask my teacher and see what he has to say about it. Would you like me to let you know what the answer is once I find out?


Yeah; based on the wording I can't see any flaws in how we answered the question.
Reply 6
Original post by Callicious
Yeah; based on the wording I can't see any flaws in how we answered the question.


Just read the part b) to the question and it states:

Calculate what fraction of the nuclei in question J3.19 remain after the time calculated as the answer to question J3.19. Give your answer as a decimal to 2 significant figures.

(J3.19 is the original question I posted)

It said 'work out the time taken for all of the nuclei to decay' so why would there be any of that nuclei left? Unless it's talking about the nuclei it decays into or something?? Any clue? Hoping you have some sort of Eureka moment solve it all lmao
The second part wants you to find the answer at t=144s using the A = A0e^-lambdat / the exponential formula and compare the results, I think.
Original post by S.H.Rahman
Hey, sure thing. I managed to do the first part but what has stumped me is the part where activity stays the same which would be weird since that would mean the decay constant would no longer stay constant.

Here's the workings I did to calculate activity:

Decay Constant = ln(2) / 100
= 6.93e-3 s^-1

Initial Activity = Decay Constant x Initial no. of nuclei in sample
= 6.93e-3 x 10^20
= 6.9315e17 Bq

Attempt at calculating the time for all nuclei to decay:

Since Activity is constant I'd imagine the graph of no. of nuclei against time to now be negatively proportional as opposed to inversely proportional. This would mean it would be a straight line where when the line intersects the x-axis at time t, this should be my time for all nuclei to decay :biggrin:

The gradient of the line should hence be the Activity as this is the no. of nuclei that decay every second.

y = mx + c

y = no. of unstable nuclei in sample, N
m = -A
x = Time, t
c = Initial total no. of nuclei, I

N = -At + I

Crosses the x - axis when N = 0

At = I
t = I/A
= (10^20) / (6.93e17)
= 144 s


144s is the wrong answer and isn't nearly large enough to be correct


144 s is 3 s.f. but the question asks for 2 s.f. :smile:
Reply 9
Original post by Eimmanuel
144 s is 3 s.f. but the question asks for 2 s.f. :smile:


Lmao exactly what my teacher emailed back with. Oh how sig figs have caused me hours of struggle D:

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