A sample of $\ce{^{41}Ar}$, a radioisotope used to measure the flow of gases from smokestacks, decays initially at a rate of 34,500 disintegrations/min, but the decay rate falls to 21,500 disintegrations/min after 75 min. What is the half-life of $\ce{^{41}Ar}$?

Here’s how I solved it:

$$\ln (A/A_0) = -k \cdot t$$ $$\ln (34500/21500) = -k \cdot 4500\ \mathrm{s}$$ $$k=1.05 \cdot 10^{-4}\ \mathrm{s^{-1}}$$

$$t_{1/2} = \ln(2)/k$$ $$t_{1/2} = 0.66 \cdot 10^{-4}\ \mathrm{s}$$

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    $\begingroup$ You forgot to switch the exponent's sign when performing the final division, whoops! $\endgroup$ – Nicolau Saker Neto Apr 14 '15 at 1:11

Your approach is correct, but you made small mistakes.
First, you mixed up the count rates for $A$ and $A_0$.
Then you ignored the minus sign of $-k$, and so your first mistake was cancelled.
In the final result, the minus sign of the exponent is wrong. The numerical value should be $0.66 \cdot 10^{4}$ or preferably $6.6 \cdot 10^{3}$ (not $0.66 \cdot 10^{-4}$).

The corrected calculation reads as follows:

$$\ln (A/A_0) = -k \cdot t$$ $$\ln (21500/34500) = -k \cdot 4500\ \mathrm{s}$$ $$k=1.05 \cdot 10^{-4}\ \mathrm{s^{-1}}$$ $$t_{1/2} = \ln(2)/k$$ $$t_{1/2} = 6.60 \cdot 10^{3}\ \mathrm{s} = 110\ \mathrm{min} = 1.83\ \mathrm{h}$$

By way of comparison, the reference value (taken from ICRP 107) for the half-life of Ar-41 is $109.61\ \mathrm{min}$.

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