Calibrate exponential decay · source reference

Printed question

A certain radioactive substance is known to decay at a rate proportional to the mass present. A block of this substance has an initial mass of 200 grams. After 40 hours, its mass is reduced to 180 grams. Given that $x$ grams is the mass of the substance present at time $t$ hours, write down a differential equation relating $x$ and $t$. [1]

(i) Show that $x=200e^{(\ln0.9)t/40}$. [4]

(ii) Find the half-life of the radioactive substance, that is, the time it takes for the substance to decay to half of its mass, giving your answer correct to the nearest hour. [2]

(iii) Sketch the curve that shows the mass of the block of radioactive substance over time. Hence explain what happens to the radioactive substance eventually. [2]

[modified 2015/YJC/Promo/9]

Printed working

$dx/dt=-kx$, $k>0$. $\int dx/x=-k\int dt$; $\ln x=-kt+c$ since $x>0$; $x=e^ce^{-kt}=Ae^{-kt}$, $A=e^c$.

At $t=0,x=200$, $200=Ae^0\Rightarrow A=200$. At $t=40,x=180$, $180=200e^{-40k}\Rightarrow e^{-40k}=0.9\Rightarrow k=-\ln0.9/40$. Thus $x=200e^{(\ln0.9)t/40}$.

Let $T$ hours be the half-life. $x=200e^{(\ln0.9)t/40}$ (1) and $x/2=200e^{(\ln0.9)(t+T)/40}$ (2).

Divide (2) by (1): $1/2=e^{(\ln0.9)T/40}$; $(\ln0.9)T/40=\ln(1/2)$; $T=40\ln0.5/\ln0.9=263$ (nearest hours).

The radioactive substance will gradually decrease in mass but will never disintegrate completely in this continuous model.

Printed answer key

$dx/dt=-kx$, where $k>0$. (ii) 263 (to nearest hours). (iii) Native sketch below. The radioactive substance will gradually decrease in mass but will never disintegrate completely.

Teacher note 1

Given $dx/dt=f(x)$. Note the substance decays, so $x$ decreases as $t$ increases and $dx/dt<0$. Usually denote a negative constant as minus a positive value.

Teacher note 2

Note $\ln0.9$ is negative, so as $t\to\infty$, $e^{(\ln0.9)t/40}\to0$. Time refers to hours, so $t\ge0$; sketch only that part of the curve.

Diagram 10-07-decay

Mathematical review

Write $x'=-kx$, $k>0$. Separation gives $x=Ae^{-kt}$. Initial mass gives A=200 and $e^{-40k}=180/200=0.9$, so $k=-\ln0.9/40$ and $x=200e^{(\ln0.9)t/40}$.

For half-life T, $e^{(\ln0.9)T/40}=1/2$, so $T=40\ln(1/2)/\ln0.9\approx263.15$ hours, or 263 hours. The graph starts at (0,200), decreases convexly, and approaches x=0 for t≥0. Under this continuous model mass remains positive at every finite time; this is a model prediction rather than a literal account of individual atoms.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 2. Working page(s): 8, 9. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.