Using Radioactive-Decay Equations | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Use halving when the ratio is simple; otherwise use the exponential law

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Information in the question	Best route	Reason
A whole number of half-lives or a fraction such as 1/8	Repeated halving	Fast and transparent
Two values at arbitrary times	Exponential equation	The ratio is not a simple power of 1/2
A decay graph	Read two halving points	Their horizontal separation is one half-life

x=x_0e^{-\lambda t}

Symbols

$x$ = value after elapsed time t (same as x₀)
$x_0$ = initial value (none, Bq or s⁻¹)
$\lambda$ = decay constant (s⁻¹)
$t$ = elapsed time (s)

Here $x$ may be the number of undecayed nuclei, the activity or the corrected count rate. Use the same quantity for $x$ and $x_0$ .

The calculator function $\ln$ undoes an exponential. Use it only after forming the later-value divided by earlier-value ratio.

Subtract background first if the data are detector readings.
Form the remaining fraction by dividing the later value by the earlier value.
Use consistent time units in the exponential.
Check that the result decreases with time and remains positive.

Preserve the ratio before using logarithms

Worked example

Activity data from a 2025 paper

Activity falls from 180 Bq to 120 Bq in 8.4 min (9702/41/M/J/25 Q9(c)).

Substitute the ratio into the exponential law.
$\frac{120}{180}=e^{-8.4\lambda}$
Use a natural logarithm to isolate the decay constant.
$\lambda=\frac{\ln(180/120)}{8.4}$
$\lambda=0.0483\ \mathrm{min}^{-1}$
Convert the decay constant to half-life.
$t_{1/2}=\frac{0.693}{0.0483}=14.4\ \mathrm{min}$

Answer

0.048\ \mathrm{min}^{-1};\ 14\ \mathrm{min}

Worked example

A ratio that reveals five half-lives

After 6.0 s, the ratio decayed : undecayed is 31 : 1 (9702/42/F/M/24 Q8(b)(iv)).

Convert decayed : undecayed into the fraction that remains.
$\text{total parts}=31+1=32$
$\text{fraction remaining}=\frac{1}{32}$
Recognise the repeated halving.
$\frac{1}{32}=\left(\frac12\right)^5$
Six seconds is five half-lives.
$t_{1/2}=\frac{6.0}{5}=1.2\ \mathrm{s}$

Answer

1.2\ \mathrm{s}

Your turncalculation check

A corrected count rate falls from 160 s⁻¹ to 40 s⁻¹ in 12 min. Find the half-life.

Show worked answer

The rate falls to one quarter, so two half-lives have passed. The half-life is 12 ÷ 2 = 6 min.

Choose the quantity and method before calculating. Repeated halving is best for simple powers of one half; the exponential law is best for an arbitrary ratio.

Use halving when the ratio is simple; otherwise use the exponential law

Swipe left or right to read every column →

Information in the question	Best route	Reason
A whole number of half-lives or a fraction such as 1/8	Repeated halving	Fast and transparent
Two values at arbitrary times	Exponential equation	The ratio is not a simple power of 1/2
A decay graph	Read two halving points	Their horizontal separation is one half-life

x=x_0e^{-\lambda t}

Symbols

$x$ = value after elapsed time t (same as x₀)
$x_0$ = initial value (none, Bq or s⁻¹)
$\lambda$ = decay constant (s⁻¹)
$t$ = elapsed time (s)

Here $x$ may be the number of undecayed nuclei, the activity or the corrected count rate. Use the same quantity for $x$ and $x_0$ .

The calculator function $\ln$ undoes an exponential. Use it only after forming the later-value divided by earlier-value ratio.

Subtract background first if the data are detector readings.
Form the remaining fraction by dividing the later value by the earlier value.
Use consistent time units in the exponential.
Check that the result decreases with time and remains positive.

Preserve the ratio before using logarithms

Worked example

Activity data from a 2025 paper

Activity falls from 180 Bq to 120 Bq in 8.4 min (9702/41/M/J/25 Q9(c)).

Substitute the ratio into the exponential law.
$\frac{120}{180}=e^{-8.4\lambda}$
Use a natural logarithm to isolate the decay constant.
$\lambda=\frac{\ln(180/120)}{8.4}$
$\lambda=0.0483\ \mathrm{min}^{-1}$
Convert the decay constant to half-life.
$t_{1/2}=\frac{0.693}{0.0483}=14.4\ \mathrm{min}$

Answer

0.048\ \mathrm{min}^{-1};\ 14\ \mathrm{min}

Worked example

A ratio that reveals five half-lives

After 6.0 s, the ratio decayed : undecayed is 31 : 1 (9702/42/F/M/24 Q8(b)(iv)).

Convert decayed : undecayed into the fraction that remains.
$\text{total parts}=31+1=32$
$\text{fraction remaining}=\frac{1}{32}$
Recognise the repeated halving.
$\frac{1}{32}=\left(\frac12\right)^5$
Six seconds is five half-lives.
$t_{1/2}=\frac{6.0}{5}=1.2\ \mathrm{s}$

Answer

1.2\ \mathrm{s}

Your turncalculation check

A corrected count rate falls from 160 s⁻¹ to 40 s⁻¹ in 12 min. Find the half-life.

Show worked answer

The rate falls to one quarter, so two half-lives have passed. The half-life is 12 ÷ 2 = 6 min.

Choose the quantity and method before calculating. Repeated halving is best for simple powers of one half; the exponential law is best for an arbitrary ratio.