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I have the problem,

A radioactive substance has a half-life of $10$ days. The initial amount of the substance is $100$ milligrams. (a) Determine the decay rate of the substance. (b) How much of the substance is left after $5$ days? (c) How long does it take for the substance to decay to $1$0 percent of its original amount?

So the decay formula is $A = A_0\cdot (\frac12)^{t/h}$. $h = 10$, $t = ?$ , $A_0 = 100$

How am I suppose to handle this if I don't have a value for A? I'm solving for t in part (a), correct?

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  • $\begingroup$ I think that half life means that A=5 if Ao was 100... $\endgroup$
    – zoli
    Commented Apr 20, 2015 at 20:32
  • $\begingroup$ Usually one takes $A(t) = A_0 e^{-t/h}$. $\endgroup$
    – Joelafrite
    Commented Apr 20, 2015 at 21:17
  • $\begingroup$ @user3032755 $A_{\circ}$ is given $\endgroup$
    – Narasimham
    Commented Apr 20, 2015 at 22:30

1 Answer 1

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The decay formula is $A(t) = A_0 2^{-\lambda t}$ with $\lambda = \dfrac{1}{h}$ the decay rate.

  1. Half-time $t_{1/2}$ being the time such that $A(t_{1/2}) = \dfrac{A_0}{2}$ one has : \begin{align} A(t_{1/2}) = \dfrac{A_0}{2} &\Leftrightarrow A_0 2^{-\lambda t_{1/2}} = \dfrac{A_0}{2}\\ &\Leftrightarrow \lambda t_{1/2} = 1\\ &\Leftrightarrow \lambda = \dfrac{1}{t_{1/2}}\\ \end{align} Or $t_{1/2} = 10\ d$, hence $\lambda = \dfrac{1}{10}=0.1\ d^{-1}$.
  2. The amount of matter after $5$ days is $A_0 2^{-5\lambda} = \dfrac{A_0}{\sqrt 2} = \dfrac{100}{\sqrt 2} \approx 70.71\ mg$.
  3. Let $t_{1/10}$ such that $A(t_{1/10}) = \dfrac{A_0}{10}$. \begin{align} A(t_{1/10}) = \dfrac{A_0}{10} &\Leftrightarrow A_0 2^{-\lambda t_{1/10}} = \dfrac{A_0}{10}\\ &\Leftrightarrow \lambda t_{1/10} = \log_2 10\\ &\Leftrightarrow t_{1/10} = t_{1/2}\log_2 10 \end{align} Since $t_{1/2} = 10\ d$ one has $t_{1/10} = 10 \log_2 10\approx 33.22\ d$.
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