Edexcel C3 2008 January — Question 5 9 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2008
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Functions
TypeExponential growth/decay model setup
DifficultyModerate -0.8 This is a straightforward exponential decay question requiring only direct substitution (part a), solving a simple exponential equation using logarithms (part b), substitution again (part c), and sketching a standard decay curve (part d). All parts are routine applications of standard techniques with no problem-solving insight required, making it easier than average.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context

5. The radioactive decay of a substance is given by $$R = 1000 \mathrm { e } ^ { - c t } , \quad t \geqslant 0 .$$ where \(R\) is the number of atoms at time \(t\) years and \(c\) is a positive constant.
  1. Find the number of atoms when the substance started to decay. It takes 5730 years for half of the substance to decay.
  2. Find the value of \(c\) to 3 significant figures.
  3. Calculate the number of atoms that will be left when \(t = 22920\).
  4. In the space provided on page 13, sketch the graph of \(R\) against \(t\).

Question 5:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(1000\)B1
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(1000e^{-5730c} = 500\)M1
\(e^{-5730c} = \frac{1}{2}\)A1
\(-5730c = \ln\frac{1}{2}\)M1
\(c = 0.000121\)A1 cao
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(R = 1000e^{-22920c} = 62.5\)M1 A1 Accept 62–63
Part (d):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Correct decreasing exponential shapeB1
\(R\)-intercept at \(1000\) labelledB1
# Question 5:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $1000$ | B1 | |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $1000e^{-5730c} = 500$ | M1 | |
| $e^{-5730c} = \frac{1}{2}$ | A1 | |
| $-5730c = \ln\frac{1}{2}$ | M1 | |
| $c = 0.000121$ | A1 | cao |

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = 1000e^{-22920c} = 62.5$ | M1 A1 | Accept 62–63 |

## Part (d):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct decreasing exponential shape | B1 | |
| $R$-intercept at $1000$ labelled | B1 | |

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5. The radioactive decay of a substance is given by

$$R = 1000 \mathrm { e } ^ { - c t } , \quad t \geqslant 0 .$$

where $R$ is the number of atoms at time $t$ years and $c$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Find the number of atoms when the substance started to decay.

It takes 5730 years for half of the substance to decay.
\item Find the value of $c$ to 3 significant figures.
\item Calculate the number of atoms that will be left when $t = 22920$.
\item In the space provided on page 13, sketch the graph of $R$ against $t$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C3 2008 Q5 [9]}}