The decay of a certain radioactive isotope can be modelled by the exponential equation $N = N_0 e^{-at}$ after $t$ weeks, where $N$ represents the amount of radioactive isotope, $N_0$ and $a$ are constants.
A sample of this radioactive isotope has a mass of 100.9 g initially.
After 2 weeks, it is found that the amount of this sample left is 84.6 g.

1. Substitute the given initial and final values into the model

Initially, $t = 0$, so $N_0 = 100.9$.
When $t = 2$, $N = 84.6$.
Substitute these into $N = N_0 e^{-at}$:
$84.6 = 100.9 e^{-2a}$

2. Isolate the exponential term and take the natural logarithm

$e^{-2a} = \frac{84.6}{100.9}$
$-2a = \ln\left(\frac{84.6}{100.9}\right)$

3. Solve for $a$ and evaluate

$a = -\frac{1}{2} \ln\left(\frac{84.6}{100.9}\right)$
$a \approx 0.088012$
$a = 0.0880 \text{ (to 3 s.f.)}$

1. Find the value of $a$ from the given initial conditions (from part a)

When $t = 0$, $N_0 = 100.9$.
When $t = 2$, $N = 84.6$.
$84.6 = 100.9 e^{-2a}$
$e^{-2a} = \frac{84.6}{100.9}$
$-2a = \ln\left(\frac{84.6}{100.9}\right)$
$a = -\frac{1}{2} \ln\left(\frac{84.6}{100.9}\right) \approx 0.088011$

2. Calculate the amount of isotope remaining after 5 weeks

Substitute $t = 5$ and $a \approx 0.088011$ into the model:
$N = 100.9 e^{-5(0.088011)}$
$N \approx 64.980 \text{ g}$

3. Calculate the percentage of the sample that has decayed

Amount decayed $= 100.9 - 64.980 = 35.920 \text{ g}$
Percentage decayed $= \frac{35.920}{100.9} \times 100\% \approx 35.6\%$

1. Find the value of $a$ from the given information

From part (a):
$84.6 = 100.9 e^{-2a}$
$e^{-2a} = \frac{84.6}{100.9}$
$-2a = \ln\left(\frac{84.6}{100.9}\right)$
$a = -\frac{1}{2} \ln\left(\frac{84.6}{100.9}\right) \approx 0.088011$

2. Set up the equation for the amount decayed

Amount decayed = Initial amount - Amount left
Amount decayed = $100.9 - 100.9 e^{-0.088011t}$
We want the time when the amount decayed is $60$ g:
$100.9 - 100.9 e^{-0.088011t} = 60$

3. Solve for $t$ and round to the nearest week

$100.9 e^{-0.088011t} = 100.9 - 60 = 40.9$
$e^{-0.088011t} = \frac{40.9}{100.9}$
$-0.088011t = \ln\left(\frac{40.9}{100.9}\right)$
$t = \frac{\ln(40.9/100.9)}{-0.088011} \approx 10.259$
Rounding to the nearest week, $t = 10$.