The radioactive decay of a substance is given by $$R=1000e^{-ct}, \, t \geq 0.$$ where $R$ is the number of atoms at time $t$ years and $c$ is a positive constant - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 6

It is given that it takes 5730 years for half of the substance to decay:

$R(5730) = \frac{1000}{2} = 500.$
Substituting into the decay formula:

$500 = 1000e^{-5730c}.$
Dividing both sides by 1000 gives:

$0.5 = e^{-5730c}.$
Taking the natural logarithm on both sides:

$\ln(0.5) = -5730c.$
Thus, we have:

$c = \frac{-\ln(0.5)}{5730}.$
Calculating this gives:

$c \approx 0.000121.$
The value of c to 3 significant figures is 0.000121.

Using the value of $c$ and substituting $t = 22920$ into the formula:

$R = 1000e^{-0.000121 \cdot 22920}.$
Calculating the exponent:

$R \approx 1000e^{-2.77} \approx 62.5.$
Thus, the number of atoms left when $t = 22920$ is approximately 62.5.

To sketch the graph of $R$ against $t$:

• The graph should start at $R=1000$ when $t=0$.
• As time progresses, $R$ should decay towards $0$.
• The curve should be exponential and smoothly approach the horizontal axis without touching it.
• Indicate significant points, such as $R=500$ at $t=5730$ and $R \approx 62.5$ at $t=22920$.