Question

Sketch the graph to show the relationship between tand mfor t≥ 0 days.Show that the value of A is 120.Show that the value of k is 0.02877, given to 4 significant figures.Using the value of k as 0.02877, find the mass of the substance after 16 days,giving the answer to 3 significant figures,Using the value of k as 0.02877, find the day during which the mass of thesubstance first reaches 50 milligrams.

Answer 1

Given the mass of a radioactive substance after time t days is given by

`m(t)=Ae^(-kt)`

The initial mass of the substance is 120 milligrams, which means m(0)=120. So,

`\begin{gathered} m(0)=120 \\ \Rightarrow Ae^0=120 \\ \Rightarrow A=120 \end{gathered}`

The mass of the substance is 90 milligrams after 10 days, which means m(10)=90. So,

`\begin{gathered} m(10)=90 \\ \Rightarrow120e^(-10k)=90 \\ \Rightarrow e^(-10k)=(3)/(4) \end{gathered}`

Now, take natural logarithm on both the side and use the property

`\ln (e^n)=n`

So, the above equation will become

`\begin{gathered} \ln (e^(-10k))=\ln ((3)/(4)) \\ \Rightarrow-10k=-0.2877 \\ \Rightarrow k=(-0.2877)/(-10) \\ \Rightarrow k=0.02877 \end{gathered}`

Therefore, the relation between the mass of the substance at time t days is given by t

`m(t)=120e^(-0.02877t)`

The mass of the substance after 16 days will be

`m(16)=120e^(-0.02877(16))=120(0.631)=75.729\text{ milligrams}`

The day during which the mass of the substance reaches 50 milligrams can be obtained as follows:

`\begin{gathered} m(t)=50 \\ 120e^{\mleft\{-0.02877t\mright\}}=50 \\ \Rightarrow e^(-0.02877t)=(50)/(120) \\ \Rightarrow-0.02877t=\ln ((50)/(120)) \\ \Rightarrow t=(-0.8754)/(-0.02877) \\ \Rightarrow t=30.43\text{ days} \end{gathered}`

The graph that shows the relationship between m and t is given below: