Oxford PAT 2012, Question 20

2012_page18

3 thoughts on “Oxford PAT 2012, Question 20

    1. Yes, you’re right: the exact answer is 0.0166. The formula above is correct, but the approximation I have used here is pretty inaccurate. There is probably a way of manipulating the formula so that you can use the approximations provided to give a less inaccurate answer — if you have found one then maybe post it here?

      1. I used slightly different approach, that allowed me to take advantage of most of the approximations provided:

        I start by working out the factor for t for an arbitrary half life: e^(a*t) = 1/2 <=> a = – ln(2) / t

        Then I plug this into the equation for r(t), where Rp is the initial relative abundance:
        r(t) = Rp * { e^[- (ln(2) / 7) * 10^(-8)] / e^[- (ln(2) / 4.5) * 10^(-9)] }

        Now I use the approximation ln 2 = 0.7 and the expression simplifies to:
        r(t) = Rp * { e^[-10^(-9)] / e^[-2 * 10^(-10)] } = Rp * e^[t * 10^(-9) * (0.2 – 1)]

        Finally I plug in the missing values (0.0072 for Rp and -10^(9) for t):
        r(0.0072) = 72 * 10^(-4) * e^[-10^(9) * 10(-9) * (0.2 – 1)] = 72 * 10^(-4) * e^[0.8]

        Now I use the approximation e^x = 1 + x for small x (as 0.8 < 1):
        r(0.0072) = 72 * 10^(-4) * (1 + 0.8) = 1.8 * 72 * 10^(-4) = (72 + 7.2) * 10^(-4) = 79.2 * 10^(-4) ≈ 0.001 ≈ 0.00166

        The approximation is not perfect (I suspect that 0.8 is not small enough for it to be) but it arrives in pretty much the same magnitude as the correct result. Would this be a valid answer?

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