# Determining Half-lifes from a graph

So I was studying for an exam I have thursday when I came across a question regarding half-life. I had previously thought that the definition of a half-life is the time it takes for the amount of material to half in its decay process. However, this question had a y-axis of Disintegration Rate, and used the same process.

They gave an answer of 8 days for the half life of this radioactive element(so they treated it as normal)

So my question is, what exactly does the half in half life represent? A halving of rate or an actual halving of decay of the nuclei?

• the y axis in the graph given is just decayconstant*N(where N is number of nuclei of radioactive element half life represents actual halving of decay of the nuclei(as muchas i know) May 13, 2020 at 5:40

At any time, the number of decays per reasonable$$^1$$ time unit is proportional to the amount of the isotope present. (This is expressed by the first-order rate equation for a one-decay process.) Therefore, it does not matter whether one looks at the amount or the number of decays. It may be easier to measure the number of decays by e.g. a Geiger counter and prefer it for that reason.

It now matters what you understand to be the rate: the number of decays per second or the percentage of nuclei decaying at any given time. The first becomes less over time, the second one is a constant for each isotope.

$$^1$$ You must have a reasonable number of decays within your time unit. If the half-life is long, it makes little sense to look at millisecond slices.

• It is a bit quirky, but the reasonable time unit should be much smaller than the half life, but long enough to get a decent number of counts. Radioactive decay follows the Poisson distribution. For more than a 100 counts the square of the number of counts is a very good estimator of the standard deviation. So $\sqrt{8000} = 89$ and the %RSD = 1.1%.
– MaxW
May 13, 2020 at 8:20

My explanation relates to the underlying math upon examination of the exponential decay graph.

I start with the cited statistical model given by:

$$y = \alpha \ e^{\beta t}$$

Taking the natural log of both sides of the equation, we have the following equivalent equation:

$$\ln y = \ln{\alpha} + {\beta t}$$

This equation has the form of a linear regression model where the rate of change in a natural log scale is a constant rate of $${\beta}$$ per unit of time. In the current test question, no need for a regression, just checking some $${\Delta ln y}$$ and $${\Delta t}$$ appears to suggest a constant slope of $${(ln(4,000)-ln(2,000))/(24-16)}$$, which equals $${ln(2)/8}$$ as the value for the slope $${\beta}$$.

Next, we have to know for the Exponential distribution, an expression for the half-life relating to $${\beta}$$, namely, $${T = \ln (2)/\beta}$$ (see the derivation at this reference). So, the respective half-life per the chart is 8 days.

As such, to answer the question, "what exactly does the half in half life represent", mathematically due to the employment of a natural log transform, its represents a $${\ln (2)}$$ change in natural log of decay between two points, which when scaled by the rate of decay (represented by $${\beta}$$), produces the half-life, as was demonstrated in the cited reference.

While another answer nicely explains in words what follows, and as an additional bonus points out some subtleties related to the condition of statistical sampling, for some reason it doesn't explicitly show equations, and sometimes it's nice to just look at equations, in the same way that a picture can be worth many words, so at the risk of redundancy I post this answer.

If the amount or population of something (call it $$y$$) decays exponentially in time, then we can write

$$y(t)=y(0)\exp(-kt)$$

where we define $$y(0)$$ as the initial concentration (at time $$0$$).

Then from the property of the exponential

$$\frac{dy}{dt}=-ky(0)\exp(-kt)$$

It follows, defining the decay rate as $$\rho=|\frac{dy}{dt}|$$, that

$$\rho(t)=ky(0)\exp(-kt)=\rho(0)\exp(-kt)$$

As explained in another answer, it follows that the decay rate also decreases exponentially and according to the same rate constant $$k$$. The time dependencies of the normalized functions $$y(t)/y(0)$$ and $$\rho(t)/\rho(0)$$ - described by $$\exp(-kt)$$ - are therefore identical. Since the half-life depends only on $$k$$ (as explained in another answer), it follows that both the original property and its rate of change havethe same half-life.

The others have explained the decay process of radioactive material very well. Therefore, I'm not going to elaborate the same thing again, but want to point out certain thing you seemingly do not understand clearly. In your question, you states that:

I had previously thought that the definition of a half-life is the time it takes for the amount of material to half in its decay process.

That statement is not quite to the point. The decay doesn't mean it vanish (or disappear) to air. It is not mass decay (kind of, theoretically but some mass remains, e.g., as $$\ce{^{206}Pb}$$, which is stable and not radioactive). The process is complicated one. For instance, see the total decay process for $$\ce{^{238}_{92}U -> ^{206}_{82}Pb}$$:

$$\ce{^{238}U ->[t_{1/2} = 4.4 \cdot 10^9 y] ^{234}Th ->[t_{1/2} = 24.1 d] ^{234}Pa ->[t_{1/2} = 46.69 h] ^{234}U ->[t_{1/2} = 2.455 \cdot 10^5 y] ^{230}Th \\ ->[t_{1/2} = 7.54 \cdot 10^4 y] ^{226}Ra ->[t_{1/2} = 1599 y] ^{222}Rn ->[t_{1/2} = 3.82 d] ^{218}Po ->[t_{1/2} = 3.04 min] ^{214}Pb ->[t_{1/2} = 27 min] ^{214}Bi\\ ->[t_{1/2} = 19.9 min] ^{210}Po ->[t_{1/2} = 160 \mu s] ^{206}Pb}$$

Therefore, for novices, what half-life simply means is the original radioactivity of given material to become half of its initial value (Refer to TAR86's answer). Thus, I decide to explain this process using your graph:

Radioactive decay of any active material is a spontaneous process, which follows first order kinetics:

$$\alpha = \alpha_\circ e^{-\beta t} \tag{1}$$

where $$\alpha$$ is activity of the material at any time $$t$$ and $$\alpha_\circ$$ is activity of the material at time you started to measure, $$t=0$$. The constant $$\beta$$ depends on several factors including decay process (e.g., $$\beta$$ is not same for $$\ce{U}$$ and $$\ce{Po}$$). We can simplify this as:

$$\frac{\alpha}{\alpha_\circ } = e^{-\beta t} \Rightarrow \ln \left(\frac{\alpha}{\alpha_\circ }\right) = -\beta t \Rightarrow \ln \alpha = \ln \alpha_\circ -\beta t \tag{2}$$

This is an equation for straight line, the slope of which is equal to $$\beta$$ and $$y$$-interception is $$\ln \alpha_\circ$$. By definition, $$t_{1/2}$$ is the time when $$\alpha = \frac{1}{2} \alpha_\circ$$. Applying this to equation $$(2)$$ gives:

$$\ln \frac{\alpha_\circ}{2} = \ln \alpha_\circ -\beta t_{1/2} \quad \Rightarrow \quad \therefore \; t_{1/2} = \frac{\ln 2}{\beta} \tag{3}$$

Thus, you can find $$t_{1/2}$$ by just getting $$\beta$$ from above straight line (Note that $$t_{1/2}$$ is independent of $$\alpha_\circ$$). Unfortunately, you don't have that straight line here. But still, you can find $$t_{1/2}$$ by analysing the given graph.

The equation of your graph is equation $$(1)$$. According to your graph, at $$t=0$$, activity has been measured as $$\pu{16000 decays/min}$$, which is your $$\alpha_\circ$$. Thus, $$\frac{1}{2} \alpha_\circ$$ should be $$\pu{8000 decays/min}$$ (see above graph). Accordingly, time taken to decay $$\pu{16000 decays/min \rightarrow 8000 decays/min}$$ is apparently $$\pu{8 d}$$. Therefore, $$t_{1/2}$$ is $$\pu{8 d}$$. If you uncertain of the value, you can check the next half-time by finding how much time it takes to decay $$\pu{8000 decays/min \rightarrow 4000 decays/min}$$. Not surprisingly, it is also $$\pu{8 d}$$ and so on (Note: If you choose $$\alpha_\circ = \pu{12000 decays/min}$$, you'd see time taken to decay $$\pu{12000 decays/min \rightarrow 6000 decays/min}$$ is still $$\pu{8 d}$$).

To go to an extra mile, now you can calculate the constant $$\beta$$ for this process. From the eqtation $$(2)$$:

$$\beta = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{\pu{8 d}} = \pu{0.087 d-1}$$