| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Half-life and doubling time |
| Difficulty | Moderate -0.8 This is a straightforward exponential decay question with standard bookwork. Part (i) is a routine sketch, (ii) is direct substitution, (iii) is a standard derivation that appears in textbooks, and (iv) is simple calculator work. All parts follow predictable patterns with no problem-solving or novel insight required. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Graph showing correct shape, passing through \((0, M_0)\) | B1 B1 | Correct shape, Passes through \((0, M_0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{M}{M_0} = e^{-0.00121 \times 5730} = e^{-0.0033} \approx \frac{1}{2}\) | M1 E1 | substituting \(k = -0.00121\) and \(t = 5730\) into equation (or in eqn) showing that \(M \approx \frac{1}{2}M_0\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{M}{M_0} = e^{-kT} = \frac{1}{2}\) | M1 | substituting \(M/M_0 = \frac{1}{2}\) into equation (oe) taking lns correctly |
| \(\Rightarrow \ln \frac{1}{2} = -kT\) | M1 | |
| \(\Rightarrow \ln 2 = kT\) | ||
| \(\Rightarrow T = \frac{\ln 2}{k}\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = \frac{\ln 2}{2.88 \times 10^{-4}} \approx 24000 \text{ years}\) | B1 [8] | \(24,000\) or better |
**(i)**
Graph showing correct shape, passing through $(0, M_0)$ | B1 B1 | Correct shape, Passes through $(0, M_0)$
**(ii)**
$\frac{M}{M_0} = e^{-0.00121 \times 5730} = e^{-0.0033} \approx \frac{1}{2}$ | M1 E1 | substituting $k = -0.00121$ and $t = 5730$ into equation (or in eqn) showing that $M \approx \frac{1}{2}M_0$
**(iii)**
$\frac{M}{M_0} = e^{-kT} = \frac{1}{2}$ | M1 | substituting $M/M_0 = \frac{1}{2}$ into equation (oe) taking lns correctly
$\Rightarrow \ln \frac{1}{2} = -kT$ | M1 |
$\Rightarrow \ln 2 = kT$ |
$\Rightarrow T = \frac{\ln 2}{k}$ | E1 |
**(iv)**
$T = \frac{\ln 2}{2.88 \times 10^{-4}} \approx 24000 \text{ years}$ | B1 [8] | $24,000$ or better6 The mass $M \mathrm {~kg}$ of a radioactive material is modelled by the equation
$$M = M _ { 0 } \mathrm { e } ^ { - k t } ,$$
where $M _ { 0 }$ is the initial mass, $t$ is the time in years, and $k$ is a constant which measures the rate of radioactive decay.\\
(i) Sketch the graph of $M$ against $t$.\\
(ii) For Carbon $14 , k = 0.000121$. Verify that after 5730 years the mass $M$ has reduced to approximately half the initial mass.
The half-life of a radioactive material is the time taken for its mass to reduce to exactly half the initial mass.\\
(iii) Show that, in general, the half-life $T$ is given by $T = \frac { \ln 2 } { k }$.\\
(iv) Hence find the half-life of Plutonium 239, given that for this material $k = 2.88 \times 10 ^ { - 5 }$.
\hfill \mbox{\textit{OCR MEI C3 2006 Q6 [8]}}