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Given an equation of the form $m=m_0 e^{-kt}$ to model the mass of a radioactive material, calculate the decay constant $k$ and the time taken for the material to reach a certain percentage of its initial mass. 

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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.

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

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Part a:

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

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

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\\[ \\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} \\]

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Part b:

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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 $t$:

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\\[ \\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} \\]

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

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

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$k=$[[0]]

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How long does it take for the material to decay to $\\var{d} \\%$ of its initial mass? Give your answer to the nearest second.

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[[0]] seconds

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This answer is marked as correct, but you have used the rounded answer from part (a). Remember that using a rounded answer in a further calculation will be an approximation of the true answer.

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