In a science laboratory the mass of a sample of a radioactive material is measured. The mass, \(m\) grams, of the sample at time \(t\) hours after the first...

The model \(m = 60e^{-kt}\) is exponential decay. At \(t = 0\) the exponential equals \(1\), so \(60\) grams is the initial mass, and the positive constant \(k\) fixes how fast the sample decays. Every part of the question is solved by the same two-step technique: isolate the exponential on one side, then take natural logarithms, using the fact that \(\ln\left(e^{A}\right) = A\). Natural logarithms are the right choice because the base of the exponential is \(e\).

(a) Show that \(k = \dfrac{1}{5}\ln\dfrac{4}{3}\), and find \(k\) to 3 s.f. [3]

Substitute the given data point, \(m = 45\) when \(t = 5\), and isolate the exponential by dividing by \(60\):

\[45 = 60e^{-5k} \quad\Rightarrow\quad e^{-5k} = \frac{45}{60} = \frac{3}{4} \qquad \textbf{[M1]}\]

The [M1] is for the substitution and the isolation, and it survives an arithmetic slip. Now take natural logarithms of both sides:

\[-5k = \ln\frac{3}{4} \qquad\text{so}\qquad 5k = -\ln\frac{3}{4} = \ln\frac{4}{3} \quad\Rightarrow\quad k = \frac{1}{5}\ln\frac{4}{3} \qquad \textbf{[A1]}\]

The [A1] is for reaching the printed form, and the step that makes it work is \(-\ln\dfrac{3}{4} = \ln\dfrac{4}{3}\), because \(-\ln M = \ln M^{-1}\). Since the target is printed, this reciprocal manipulation is the substance of the answer, and a candidate who merely writes the given result earns nothing.

\[k = \frac{1}{5}(0.287682\ldots) = 0.0575364\ldots = 0.0575 \text{ (3 s.f.)} \qquad \textbf{[A1]}\]

The final [A1] is for the decimal value. Three significant figures on a number beginning \(0.0\) means the leading zeros do not count, so \(0.0575\) is right and \(0.058\) is only two significant figures. Keep the unrounded value in the calculator for parts (b) and (c).

(b) Mass when \(t = 12\) [2]

\[m = 60e^{-12k} = 60e^{-0.690437\ldots} \qquad \textbf{[M1]}\] \[m = 60 \times 0.501357\ldots = 30.0814\ldots = 30.1 \text{ grams (1 d.p.)} \qquad \textbf{[A1]}\]

The [M1] is for the substitution of \(t = 12\) into the model, the [A1] for the value to one decimal place. Use the stored value of \(k\) rather than the rounded \(0.0575\); the rounded version gives \(30.09\), which still rounds to \(30.1\) here but will not always be so forgiving. The frequent error is to forget the minus sign in the index, which gives \(60e^{+0.6904} = 119.7\) grams, a mass larger than the original sample and therefore obviously wrong for a decaying quantity.

(c) Time at which the mass is 20 grams [3]

Now the mass is known and the time is not, so the same isolation is followed by a logarithm and then a division:

\[20 = 60e^{-kt} \quad\Rightarrow\quad e^{-kt} = \frac{20}{60} = \frac{1}{3} \qquad \textbf{[M1]}\] \[-kt = \ln\frac{1}{3} \quad\Rightarrow\quad kt = \ln 3 \quad\Rightarrow\quad t = \frac{\ln 3}{k} = \frac{1.098612\ldots}{0.0575364\ldots} \qquad \textbf{[M1]}\] \[t = 19.0942\ldots = 19.1 \text{ hours (1 d.p.)} \qquad \textbf{[A1]}\]

The first [M1] is for isolating the exponential, the second for taking logarithms and rearranging for \(t\); both are method marks and so survive slips. The [A1] requires the value to one decimal place with the unit. Again \(-\ln\tfrac{1}{3} = \ln 3\).

The mistake this part is built to catch is dividing \(20\) by \(60\) the wrong way round and using \(3\) instead of \(\tfrac{1}{3}\), which gives a negative time. A negative \(t\) is impossible when the mass is falling below its starting value, so the sign of the answer is itself a check on the working.

Check: the results must be consistent with each other and with the physics. The mass falls from \(60\) to \(45\) grams in \(5\) hours, so the half-life is \(\dfrac{\ln 2}{k} = \dfrac{0.6931}{0.05754} \approx 12.0\) hours, which is why the mass at \(t = 12\) is almost exactly half of \(60\), namely \(30.1\) grams, agreeing with part (b). Since \(20\) grams is a further reduction beyond half, the answer to (c) must exceed \(12\) hours, and \(19.1\) does. Substituting back also confirms it: \(60e^{-0.0575364 \times 19.0942} = 60 \times 0.33333 = 20.0\) grams.