Counting Time and Relative Uncertainty

Question

A detector records $10{,}000$ counts from a steady radioactive source in $5.0\,\text{minutes}$ . The counts obey Poisson statistics, so a total of $N$ counts carries an uncertainty $\sqrt{N}$ . Neglecting background and dead time, how long must the detector run to cut the relative uncertainty in the total count to half its $5.0$ -minute value?

a. $7.1$ minutes
b. $10$ minutes
c. $20$ minutes
d. $80$ minutes

Correct answer

C. $20$ minutes

Full reasoning

1Find the starting relative uncertainty: $\sigma_N = \sqrt{10{,}000} = 100$ counts, so $\sigma_N/N = 100/10{,}000 = 0.0100$ , or $1.00\%$ .
2Write the target: half of that is $0.50\%$ , and the relative uncertainty of a Poisson total is $\sigma_N/N = \sqrt{N}/N = 1/\sqrt{N}$ .
3Solve $1/\sqrt{N} = 0.0050$ , so $\sqrt{N} = 200$ and $N = 40{,}000$ counts — four times the original total, because the relative uncertainty falls only as the square root.
4Convert counts back to time. The source is steady, so $N = Rt$ with $R = 10{,}000/5.0 = 2000$ counts per minute, giving $t = 40{,}000/2000 = 20$ minutes.
5Check the result: $\sqrt{40{,}000} = 200$ counts of uncertainty on $40{,}000$ counts is exactly $0.50\%$ , half the original $1.00\%$ .

Why each choice is right or wrong

Choice A

This is $5.0\sqrt{2}$ , obtained by taking the square root of the improvement factor instead of squaring it. It inverts the rule: the counts must grow by the square of the factor by which the relative uncertainty falls, giving only about $0.84\%$ here.

Choice B

This assumes the relative uncertainty falls in direct proportion to the counting time, as $1/N$ , so halving it would need twice the data. Poisson statistics give $1/\sqrt{N}$ , and $20{,}000$ counts leave the relative uncertainty at about $0.71\%$ .

Choice C

Correct. The relative uncertainty is $1/\sqrt{N}$ , so halving it needs $4\times$ the counts: $N = 40{,}000$ . At a steady rate the count total is proportional to time, so the run must last $4(5.0) = 20$ minutes.

Choice D

This applies the square-root rule twice, treating the relative uncertainty as falling like $N^{-1/4}$ and demanding $2^4 = 16$ times the counts. One square root separates counts from their relative uncertainty, not two.

Related formula

Poisson counting uncertainty

\sigma_N = \sqrt{N}, \qquad \frac{\sigma_N}{N} = \frac{1}{\sqrt{N}}

Assumptions: Events arrive independently at a steady average rate. Background counts and detector dead time are negligible.

Counts accumulated at a steady rate

N = Rt

Assumptions: The source activity does not decay appreciably over the run. Because N is proportional to t, quadrupling the counts means quadrupling the time.

Sources

Authored for this public explainer registry. It has no database question ID and is not copied from a protected or official exam bank.

Introductory Statistics 2e, Section 4.6: Poisson Distribution — OpenStax. Accessed 2026-08-15. OpenStax textbook content is CC BY 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 1, Section 1.6: Significant Figures — OpenStax. Accessed 2026-08-15. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

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