The Student Room Group

Radioactivity and Uncertainty

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that's a good question, and I think it might take me an hour to answer.

how many days are there in 5.27 years?, that's one place to start (I'm typing this with a de-fueled research reactor in view, out of the window)
Reply 2
Original post by LuigiMario
that's a good question, and I think it might take me an hour to answer.

how many days are there in 5.27 years?, that's one place to start (I'm typing this with a de-fueled research reactor in view, out of the window)


I found that A=(3.5×105±2%)e5.27×365tA = (3.5 \times 10^5 \pm 2\%)e^{5.27\times365t} but I don't know how to proceed from here. How to find % error of A?
Initial activity has a an uncertainty of 2%
Activity at a time t has an uncertainty of 10%
The change in activity corresponds to the increase in uncertainty of 8%
So the source must decay by 8%, giving 230 days
Interesting question.
Intuitively, I'd be tempted to add 2 and 10% to get an uncertainty of 12%, but I find it hard to get my head around it.
Do you know the actual answer?
Reply 5
Original post by old_teach
Interesting question.
Intuitively, I'd be tempted to add 2 and 10% to get an uncertainty of 12%, but I find it hard to get my head around it.
Do you know the actual answer?


Yes. The answer is 230 days.
Original post by esrever
I found that A=(3.5×105±2%)e5.27×365tA = (3.5 \times 10^5 \pm 2\%)e^{5.27\times365t} but I don't know how to proceed from here. How to find % error of A?

The exponent should be negative (activity is decreasing).


A=A0eλt A = A_0 e^{- \lambda t}
λ=ln2t12 \lambda = \dfrac{\ln 2}{t_{ \frac{1}{2}} }
Here the decay constant λ \lambda is 0.132 (to 3 sf) year-1 (using the formula given in the syllabus, where λ=0.693t12 \lambda = \dfrac{0.693}{t_{\frac{1}{2}}} , decay constant is 0.131 year^-1

There is no need to find the decay constant
You may use
A=A02tT A = A_0 2^{-\dfrac{t}{T}}
where T is the half-life
to find the time in the same units as the half life
then convert to unit required in the question
(edited 5 years ago)
My inuition is wrong! Always good to see new question ideas.
Remember, if it decays by 8%, it means it is now 92% of original, BTW.
Reply 8
Original post by BobbJo
The exponent should be negative (activity is decreasing).


A=A0eλt A = A_0 e^{- \lambda t}
λ=ln2t12 \lambda = \dfrac{\ln 2}{t_{ \frac{1}{2}} }
Here the decay constant λ \lambda is 0.132 (to 3 sf) year-1 (using the formula given in the syllabus, where λ=0.693t12 \lambda = \dfrac{0.693}{t_{\frac{1}{2}}} , decay constant is 0.131 year^-1

There is no need to find the decay constant
You may use
A=A02tT A = A_0 2^{-\dfrac{t}{T}}
where T is the half-life
to find the time in the same units as the half life
then convert to unit required in the question


Thank you so much :smile:

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