= Solution
Assume independent radioactive detections with negligible dead-time loss, so accumulated counts have a <Poisson distribution>. The reference measurement has total count $K_0=N_0t_0=2\,018\,500$ over $t_0=1000\,\mathrm s$. Since a Poisson count has <variance> equal to its mean, the rate uncertainty is
$$
\boxed{\sigma_{N_0}=\frac{\sqrt{K_0}}{t_0}=\sqrt{\frac{2018.5}{1000}}=1.42\,\mathrm{counts\,s^{-1}}.}
$$
Its relative uncertainty is $7.04\times10^{-4}$, or $0.0704\%$. For each one-second measurement, $\sigma_y\simeq\sqrt y$, giving the estimated count and rate uncertainties
$$
\boxed{\begin{array}{c|rrrrrrrr}
n&1&2&3&4&5&6&7&8\\ \hline
\sigma_y&26.9&15.6&9.75&5.92&2.83&2.24&2.45&2.24
\end{array}}
$$
in counts, numerically the same in counts per second for these one-second exposures. Their fractional uncertainties range from about $3.7\%$ at the first sheet to $35$–$45\%$ for the smallest counts. The long, high-rate reference is therefore much more precise. These are statistical errors; a longer exposure does not automatically remove calibration errors, source drift or background bias.
Back to article page
