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 7 - 2008 - Paper 6

Given that it takes 5730 years for half of the substance to decay, we have:

$R = 1000e^{-5730c} = 500.$

Dividing both sides by 1000 gives:

$e^{-5730c} = \frac{1}{2}.$

Taking the natural logarithm of both sides, we arrive at:

$-5730c = \ln\left(\frac{1}{2}\right).$

Solving for $c$:

$c = -\frac{\ln\left(\frac{1}{2}\right)}{5730} \approx 0.000121.$

Thus, the value of $c$ to three significant figures is 0.000121.

To find the number of atoms remaining after $t = 22920$, substitute this value into the decay formula:

$R = 1000e^{-0.000121 \cdot 22920}.$

Calculating the exponent:

$-0.000121 \cdot 22920 \approx -2.77.$

Now calculating $R$:

$R = 1000e^{-2.77} \approx 62.5.$

Thus, the number of atoms left when $t = 22920$ is approximately 62.5.

To sketch the graph of $R$ against $t$, the graph will start at $R = 1000$ when $t = 0$ and will approach zero as $t$ increases, showing an exponential decay. The general shape should depict a steep decline in the initial years that plateaus as time progresses.