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radioactive mathematical headache


Gian

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Excuse my complete ignorance, but I'm new to this.

I have a question from a GCSE physics book as follows;

"Cobalt-60 is a radioisotope made by placing cobalt in a nuclear reactor. It has a half-life of 5 years. The activity of a piece of Cobalt-60 is 32.0 kBq. How long would it take to fall to (a)16.0kBq (b)1.0 kBq?"

I guess question (a) would be five years, or one half-life; but what maths/ formulae should I use to calculate (b)?

Of course I guess I can get it by just counting down multiplying by 0.5 each time, but I wondered if there's an algebraic formula I should use for this?

(kBq is kilo-bequerels)

cheerz

GIAN:)

Edited by Gian
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There is something called an Arrhenius equation which models rates of decay like this. I do not know if they would have introduced you to this or not at GCSE level, so they are probably expecting you to work it out from halving each time like you suggested. If they wanted you to use the Arrhenius equation then they would have taught it to you in classes already I would have thought.

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It's an exponential decay. N(t)=N0 e-at  (which is derived from the decay rate being dN/dt = -aN and integrating it)

a is the decay constant (often lambda is used for it), which is ln(2)/t1/2  That can be found by setting N(t) to 1/2 * N0  

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26 minutes ago, swansont said:

It's an exponential decay. N(t)=N0 e-at  

That's the one - Arrhenius equation type.  As I said though I don't think it is taught at GCSE level. I could be wrong - it has been a while.

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3 hours ago, DrP said:

That's the one - Arrhenius equation type.  As I said though I don't think it is taught at GCSE level. I could be wrong - it has been a while.

Possibly not (I wouldn't know). The question can be solved just by dividing by two a number of times, so the formula is not necessary.

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5 hours ago, Gian said:

Of course I guess I can get it by just counting down multiplying by 0.5 each time, but I wondered if there's an algebraic formula I should use for this?

Did you have logarithms at school?

e.g. if you would like to calculate number of the all possible values of the single byte you should use 2^8 = 256. reverse of it is log2(256)=8

e.g. 32 bit unsigned integer: 2^32 = 4294967296, log2(4294967296) = 32

ps. You can use Wolfram Alpha to calculate such things e.g. http://www.wolframalpha.com/input/?i=log2(32)

Edited by Sensei
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