<p>The other answers are entirely correct.  But I like graphical representations.</p>

<p>From <a href="http://en.wikipedia.org/wiki/Radioactive_decay" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Radioactive_decay</a> we see the decay formula is:</p>

<p>$$N(t) = N_0e^\frac{-t}{τ}$$</p>

<p>Where N<sub>0</sub> is the starting number of nuclides and τ is the mean lifetime.  We also see that the half-life is</p>

<p>$$t_{1/2} = τ ln(2)$$</p>

<p>Substituting for τ, we get:</p>

<p>$$N(t) = N_0e^\frac{-tln(2)}{t_{1/2}}$$</p>

<p>So for example if we have N<sub>0</sub> = 1000 and t<sub>1/2</sub> = 100 we can plot the following graph:</p>

<p><img src="https://i.sstatic.net/1Pk80.png" alt="Exponential decay plot"></p>

<p><sub>Note that the horizontal axis is the t-axis.</sub></p>

<p>We see the following:</p>

<ul>
<li>At t=100 (one half life), N(t) is 500, half of 1000, as expected</li>
<li>At t=200 (two half lives), N(t) is 250, 1/4 of 1000, as expected</li>
<li>We intersect the curve with a line at N(t) = 750.  Here we can see this occurs at about t=41.5 as predicted by the other answers.</li>
</ul>

CC BY-SA 3.0
2014-05-10T00:50:26.400Z
*Here we can see this occurs at about t=41.5 as predicted by the other answers.* If you can see, that is. This site is overly-reliant on imagery. Good answers should be accessible to those who can't see / can't see very well.
David Hammen
CC BY-SA 3.0
2014-05-09T00:54:43.730Z
2014-05-09T05:44:51.227Z
2014-05-09T05:44:51.227Z
Digital Trauma
<p>You can use simple logarithms to calculate the answer.  The number of half-lives that have elapsed can be calculated with</p>

<p>$$
 - \frac{\log{f}}{\log{2}}
$$</p>

<p>where $f$ is the fraction that remains.</p>

<p>So plugging in the numbers gives</p>

<p>$$
- \frac{\log(0.75)}{\log(2)} = 0.415 = 41.5\%
$$</p>

CC BY-SA 3.0
2014-11-03T02:02:55.843Z
It feels odd to express the number of half-lives as a percentage.
200_success
CC BY-SA 3.0
2014-05-08T15:28:28.813Z
2014-05-08T15:28:28.813Z
gerrit
<p>@gerrit provided a formula, but without stating the reasoning behind it.</p>

<p>Radioactive decay is an exponential function.  After $n$ half-lives, the amount of the original material remaining is</p>

<p>$$\textrm{amount remaining after}\ n\ \textrm{half-lives} = \left(\frac{1}{2}\right)^n$$</p>

<p>Therefore, you want to solve</p>

<p>$$\begin{align*}
\left(\frac{1}{2}\right)^n &amp;= \frac{3}{4}\\
\log_\frac{1}{2} \left(\frac{1}{2}\right)^n &amp;= \log_\frac{1}{2} \frac{3}{4}\\
n &amp;= \frac{\log \frac{3}{4}}{\log \frac{1}{2}} = \frac{\log \frac{3}{4}}{- \log 2}
   \approx 0.415
\end{align*}$$</p>

CC BY-SA 3.0
2014-05-08T16:44:59.893Z
2014-11-03T01:58:17.000Z
2014-11-03T01:58:17.000Z
200_success
<p>Radioactive element <code>A</code> radioactively decays into material <code>B</code>. If 75% of <code>A</code> and 25% of <code>B</code> are present, how many half-lives of material <code>A</code> have elapsed?</p>

<p>I was recently taught that the correct answer is "one half of a half-life has elapsed". However, due to the fact that the amount of radioactive material remaining scales exponentially (logarithmically) instead of in a linear fashion, wouldn't the answer be less than one half of a half-life?</p>

CC BY-SA 3.0
2014-05-08T15:30:26.607Z
It doesn't scale quadratically, but rather exponentially (logarithmically).
gerrit
CC BY-SA 3.0
2014-05-10T22:13:33.780Z
"I was taught that the correct answer is..." - you were taught badly (and you realize it which is good). Not only was the given answer wrong - the question was poorly phrased. In most instances element B would not be stable - and the only information you need is what percentage of A remains. Then again, determination of age often relies on ratios of isotopes present - but do realize you need to know the half life of the entire chain, not just the parent, in order to do the math correctly.
Floris
CC BY-SA 3.0
2014-05-12T21:47:21.167Z
This is a math question.
Paul
CC BY-SA 3.0
2015-02-01T18:48:24.610Z
For future visitors, this task can be facilitated on a TI-83 calculator or above using this program: http://tibasic.com/science/earth-science/
Timtech
<measurements><radioactivity>
If 75% of radioactive material remains, how many half-lives have elapsed?
Timtech
# If 75% of radioactive material remains, how many half-lives have elapsed?

(Asked by: username_0 on 2014-05-08)

Radioactive element `A` radioactively decays into material `B`. If 75% of `A` and 25% of `B` are present, how many half-lives of material `A` have elapsed?

I was recently taught that the correct answer is "one half of a half-life has elapsed". However, due to the fact that the amount of radioactive material remaining scales exponentially (logarithmically) instead of in a linear fashion, wouldn't the answer be less than one half of a half-life?

**Tags:** measurements, radioactivity

> **Comments:**
> **username_1**: It doesn't scale quadratically, but rather exponentially (logarithmically).  
> **username_2**: "I was taught that the correct answer is..." - you were taught badly (and you realize it which is good). Not only was the given answer wrong - the question was poorly phrased. In most instances element B would not be stable - and the only information you need is what percentage of A remains. Then again, determination of age often relies on ratios of isotopes present - but do realize you need to know the half life of the entire chain, not just the parent, in order to do the math correctly.  
> **username_3**: This is a math question.  
> **username_0**: For future visitors, this task can be facilitated on a TI-83 calculator or above using this program: http://tibasic.com/science/earth-science/  

---

## 3 answers

### Answer by username_6 on 2014-05-08. Score: 25 | Accepted answer

@username_1 provided a formula, but without stating the reasoning behind it.

Radioactive decay is an exponential function. After $n$ half-lives, the amount of the original material remaining is

$$\textrm{amount remaining after}\ n\ \textrm{half-lives} = \left(\frac{1}{2}\right)^n$$

Therefore, you want to solve

$$\begin{align\*}
\left(\frac{1}{2}\right)^n &= \frac{3}{4}\\
\log\_\frac{1}{2} \left(\frac{1}{2}\right)^n &= \log\_\frac{1}{2} \frac{3}{4}\\
n &= \frac{\log \frac{3}{4}}{\log \frac{1}{2}} = \frac{\log \frac{3}{4}}{- \log 2}
\approx 0.415
\end{align\*}$$


---

### Answer by username_4 on 2014-05-09. Score: 16

The other answers are entirely correct. But I like graphical representations.

From <http://en.wikipedia.org/wiki/Radioactive_decay> we see the decay formula is:

$$N(t) = N\_0e^\frac{-t}{τ}$$

Where N0 is the starting number of nuclides and τ is the mean lifetime. We also see that the half-life is

$$t\_{1/2} = τ ln(2)$$

Substituting for τ, we get:

$$N(t) = N\_0e^\frac{-tln(2)}{t\_{1/2}}$$

So for example if we have N0 = 1000 and t1/2 = 100 we can plot the following graph:

![Exponential decay plot](https://i.sstatic.net/1Pk80.png)

Note that the horizontal axis is the t-axis.

We see the following:

* At t=100 (one half life), N(t) is 500, half of 1000, as expected
* At t=200 (two half lives), N(t) is 250, 1/4 of 1000, as expected
* We intersect the curve with a line at N(t) = 750. Here we can see this occurs at about t=41.5 as predicted by the other answers.

> **Comments:**
> **username_5**: *Here we can see this occurs at about t=41.5 as predicted by the other answers.* If you can see, that is. This site is overly-reliant on imagery. Good answers should be accessible to those who can't see / can't see very well.  

---

### Answer by username_1 on 2014-05-08. Score: 15

You can use simple logarithms to calculate the answer. The number of half-lives that have elapsed can be calculated with

$$
- \frac{\log{f}}{\log{2}}
$$

where $f$ is the fraction that remains.

So plugging in the numbers gives

$$
- \frac{\log(0.75)}{\log(2)} = 0.415 = 41.5\%
$$

> **Comments:**
> **username_6**: It feels odd to express the number of half-lives as a percentage.  

---

