Logarithms and Exponentials: Applications 3

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Question statement

The mass $m$ of a radioactive material is modelled by the equation

$m = m_0 e^{-kt},$ where

$t$ is time, measured in seconds, and

$m_0$ is the initial mass of the material.

The half-life (the time taken for the mass of the material to halve) is $\var{n}$ minutes.

Give any introductory information the student needs.

			Name	Type	Generated Value

n	integer	2
d	integer	20
k	number	0.0058
k5	number	0.0057762
kfull	number	0.0057762265
tdecay	decimal	dec("279")
tdecay2	decimal	dec("278.63")
tdecayrounded	decimal	dec("277")

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..7)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

d
k
kfull

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Mathematical expression
  Add
Part b) Gap-fill
1. Gap Mathematical expression
  1. Alt Unnamed alternative
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Calculate the value of $k$ , giving your answer to 2 significant figures.

$k=$ Unnamed gap

Advice

Part a:

If the half-life of the material is $\var{n}$ minutes, this means that when $t=\var{60*n}$ , $m=\frac{m_0}{2}$ .

Substituting these into the equation and rearranging, we are able to calculate $k$ :

$\begin{split} m &\,= m_0 \, e^{-kt} \\ \frac{m_0}{2} &\,= m_0 \, e^{-\var{60*n}k} \\ \frac{1}{2} &\,= e^{-\var{60*n}k} \\ \implies -\var{60*n}k &\,= \ln(0.5) \\ k&\,= \var{k} \text{ (2 s.f.)}. \end{split}$

Part b:

To find the time taken for the material to decay to $\var{d}\%$ of its initial mass, we want use the value of $k$ found in part (a) (the non-rounded answer!) and set $m=\var{d/100}\,m_0$ , then rearrange to calculate $t$ :

$\begin{split}\var{d/100}\,m_0 &\,= m_0 \, e^{-\var{k5}t} \\ \var{d/100} &\,= e^{-\var{k5}t} \\ \implies -\var{k5}\,t &\,= \ln(\var{d/100}) \\ t&\,= \var{tdecay2} \text{ seconds (2 d.p.)} \end{split}$

Therefore, to the nearest second, it takes $\var{tdecay}$ seconds for the material to decay to $\var{d}\%$ of its initial mass.

(Note: If you have used the rounded value for $k$ in part (b) then you will get an answer which differs by a few seconds. Either will be marked as correct in this case, but it is important to remember to use unrounded answers when carrying values from one calculation to the next to reduce the amount of error.)

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Taxonomy: mathcentre

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Taxonomy: Context

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Christian Lawson-Perfect 2 years, 7 months ago

Ben McGovern 3 years, 8 months ago

Ben McGovern 3 years, 9 months ago

Error in variable testing condition

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Generated value: `integer`

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Logarithms and Exponentials: Applications 3

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 2 years, 7 months ago

Ben McGovern 3 years, 8 months ago

Ben McGovern 3 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`