# Finding the formula for the number of half-lives of a substance, given the formula for its decay $Q(t) = Q_0e^{-ln(\sqrt[25]{2})t}$

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?

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

We know that

$$Q(t)=Q_0 e^{-\frac{\ln2}{25}t}$$

If $t=25n$:

$$Q(25n)=Q_0 e^{-\frac{\ln2}{25}25n}=Q_0 e^{-n\ln2}$$

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...

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 $n \in \mathbb N_0$