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The disintegration of a radioactive substance is well modeled by a law of the type:

$$Q(t) = Q_0\cdot e^{-kt}$$

being $Q(t)$ the remaining mass, at instant $t$, of an initial quantity $Q_0$, that corresponds to the instant $t=0$.

Substance A has an half-life time of 25 years.

Write the law for the disintegration of substance A, in function of $Q_0$, for the number $n$ of half-lives

Earlier I found that for substance A, $k = \frac{-\ln(0.5)}{25}$ (my book says this is correct)

My book says the solution is: $$Q(n) = Q_0e^{ln(0.5n)}$$

I don't understand this problem. Shouldn't a substance have only one half-life? Otherwise it wouldn't be half, would it? Can someone explain this problem and my books solution?

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  • $\begingroup$ Are you sure about $k$? $k=-\frac{\ln(\frac{1}{2})}{25}=\frac{\ln(2)}{25}$ $\endgroup$
    – MattG88
    Commented Nov 14, 2016 at 1:59
  • $\begingroup$ @MattG88 My mistake, I'll edit it $\endgroup$
    – Mark Read
    Commented Nov 14, 2016 at 2:16
  • $\begingroup$ @AliceIsDead The solution must be $Q(n)=Q_0e^{ln(0.5)\cdot n}$.$ n$ is outside of the ln-function. $\endgroup$ Commented Nov 14, 2016 at 3:33
  • $\begingroup$ @AliceIsDead Did you get my comment ? $\endgroup$ Commented Nov 14, 2016 at 17:08
  • $\begingroup$ @callculus yes. $\endgroup$
    – Mark Read
    Commented Nov 16, 2016 at 15:37

2 Answers 2

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We know that

\begin{equation} Q(t)=Q_0 e^{-\frac{\ln2}{25}t} \end{equation}

If $t=25n$:

\begin{equation} Q(25n)=Q_0 e^{-\frac{\ln2}{25}25n}=Q_0 e^{-n\ln2} \end{equation}

25 is the half-live time i.d. the time we need to have half of a substance, so if we have 100 atoms, after 25 years we'll have 50 atoms, after others 25 years we'll have 25 atoms and so on...

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Write the law for the disintegration of substance $A$, in function of $Q_0$, for the number $n$ of half-lives.

$n$ is the number of half lives. If $n=1$ then 25 years have passed. Thus

$Q(n)=Q_0\cdot e^{ln(0.5)\cdot n}=Q_0\cdot 0.5^n$

After $100$ years it exists $Q(4)=0.5^4 Q_0 =\frac{1}{16}Q_0$

Remark

$Q(n)$ is only equal to $Q(t)$ if $25n=t$ and $n \in \mathbb N_0$

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