Calibrate exponential decay

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 xx grams is the mass of the substance present at time tt hours, write down a differential equation relating xx and tt. mark

(i) Show that x=200e(ln⁡0.9)t/40x=200e^{(\ln0.9)t/40}. marks

(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. marks

(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. marks

[modified 2015/YJC/Promo/9]

Total marks: 9

Solution

Differential equation

Modelling

A radioactive substance decays at a rate proportional to the mass present. Write down a differential equation relating xx (grams) and tt (hours).

Hint

  1. Rate of change is proportional to xx, and it is a decrease.

Answer · Differential equation

x′=−kxx\prime=-kx, k>0k>0.

(i)

Separating variables

Initially 200 g; after 40 hours 180 g. Show that x=200e(ln⁡0.9)t/40x=200e^{(\ln0.9)t/40}.

Explore this idea

Hint

  1. Separate: ∫dxx=−k∫dt\int\frac{dx}x=-k\int dt.

  2. Use the two readings to find AA and kk.

Answer · (i)

x=200e(ln⁡0.9)t/40x=200e^{(\ln0.9)t/40}.

Explore the idea

The decaying mass

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of x at t = 0200
Read the solution at t40
Try an example
x at t = 40
180
Rate dx/dt there
-0.474
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

Every solution is the same exponential shape, scaled by the starting mass.

Ideas to try
  1. Read the mass at t=40t=40 hours: it should be 180 g.
  2. Change the starting mass: curves are scaled copies of each other.
(ii)

Half-life

Find the half-life to the nearest hour.

Hint

  1. Solve 12=e(ln⁡0.9)T/40\frac12=e^{(\ln0.9)T/40} for TT.

Answer · (ii)

263263 hours.

(iii)

Long-term behaviour

Sketch the curve and explain what eventually happens.

Hint

  1. ln⁡0.9<0\ln0.9<0, so e(ln⁡0.9)t/40→0e^{(\ln0.9)t/40}\to0.

Radioactive decay
Radioactive decayScroll across to see the full diagram.

Answer · (iii)

Decays from 200 toward 0, remaining positive at every finite time.