# Radioactive Decay and Neanderthal Art

The oldest cave art provides a window into the mind of not just our species but those of another hominid. This insight came from a method in archaeological dating called uranium-thorium dating. The paintings in question weren’t made with charcoal which would allow carbon dating. Instead, archaeologists relied on calcium carbonate deposits laid down by mineral-rich water that flowed over the paintings. These deposits must have appeared after the painting was made so by dating them we can set a lower limit for the age of the art beneath.

The age of calcium carbonate deposits on top of cave paintings in the North, Centre and South of Spain were found to be 64,800 years when the inhabitants of what is now Spain were Neanderthals. Modern humans are thought to have arrived in Spain 20,000 years later. The images show what looks to be symbolic art: for example at La Pasiega in Cantabria you can see a ladder with dots and paintings of animals inside the panels of the ladder. The white crust you see above the painting, which consists of calcite deposits, was used for uranium-thorium dating.

## Uranium Thorium Dating

This dating method involves two decays rather than one. Naturally occurring water contains the uranium isotope $${}^{234}_{92}U$$ which decays into thorium which does not normally appear in groundwater. This thorium then decays to radium.

$${}^{234}_{92}U \rightarrow {}^{230}_{90}Th + {}^4_2\alpha \rightarrow {}^{226}_{88}Ra + {}^4_2\alpha$$

The method relies on the differing half-lives of the two reactions.

• 245,000 years for $${}^{234}U \rightarrow {}^{230}Th$$ decay
• 75,000 years for $${}^{230}Th \rightarrow {}^{226}Ra$$ decay

The half-life of the thorium isotope is three times shorter than the uranium isotope. After a certain amount of time the level of thorium builds up and its rate of decay eventually approaches its rate of production so that the total amount of thorium relative to the amount of uranium reaches equilibrium. We know that the rate of decay of uranium if we have $$N_U$$ atoms and the decay constant is $$\lambda_U$$, is
$$\frac{dN_U}{dt}=-\lambda_U N_U$$

The decay rate of uranium is the production rate of thorium, but thorium also decays at rate $$-\lambda_T N_T$$ which means the rate of change of the amount of thorium is

$$\frac{dN_T}{dt}=\lambda_U N_U – \lambda_T N_T$$

The system reaches equilibrium, which means that $$\frac{dN_T}{dt}=0$$ when

$$\lambda_U N_U = \lambda_T N_T$$

or rearranging

$$\frac{\lambda_U}{\lambda_T} = \frac{N_T}{N_U}$$

We can convert the half-lives of the isotopes into decay constants using the formula $$\lambda=\frac{\ln(2)}{t_\frac{1}{2}}$$.

$$\lambda_U=\frac{\ln(2)}{245000}=2.83\times 10^-6$$ $$\lambda_T=\frac{\ln(2)}{75000}=9.24\times 10^-6$$

At equilibrium

$$\frac{N_T}{N_U} = \frac{2.83\times 10^-6}{9.24\times 10^-6} = 0.31$$

## Solving the double decay integral

It turns out that we can solve these equations using a nice little integration trick which you may find useful. It relies on the product rule:

$$d(uv)=v du + u dv$$

It’s as if we have water draining slowly out of one tank (uranium) into another tank (thorium) which drains quickly. Let’s start out with the simple bit which is the decay of uranium which feeds the production of thorium.

$$\frac{dN_U}{dt} = -\lambda_U N_U$$

We can separate the two integrals by moving $$N_U$$ over to the left hand side and integrating

$$\int_{N_U^0}^{N_U^t}\frac{dN_U}{N_U} = -\lambda_U \int_0^t dt$$

We can solve these two easy integrals

$$\left.\ln N_U\right|_{N_0}^{N_t} = \left.-\lambda_U t \right|_{0}^t$$

and the limits give

$$\ln \frac{N_U^t}{N_U^0} = -\lambda_U t$$

Take exponentials of both sides

$$\frac{N_U^t}{N_U^0} = e^{-\lambda_U t}$$

And we come up with the familiar equation for radiactive decay

$$N_U^t = N_U^0 e^{-\lambda_U t}$$

Now the hard bit which is solving both the uranium and thorium decay equations. Let’s start with the rate equation for thorium, remember the first term is the production of thorium (positive) through decay of uranium and the second term (negative) is the consequent decay of thorium

$$\frac{dN_T}{dt}=\lambda_U N_U – \lambda_T N_T$$

We solved the first term above so we can substitute that here

$$\frac{dN_T}{dt}=\lambda_U N_U^0 e^{-\lambda_U t} – \lambda_T N_T$$

We rearrange to put all the thorium terms on the left hand side and all the uranium terms on the right hand side

$$\frac{dN_T}{dt} + \lambda_T N_T=\lambda_U N_U^0 e^{-\lambda_U t}$$

The game here is to make the left hand side look like the derivative of a product. We multiply everything by $$e^{\lambda_T t}$$ to get

$$\frac{dN_T}{dt} e^{\lambda_T t} + \lambda_T N_T e^{\lambda_T t} = \lambda_U N_U^0 e^{-\lambda_U t + \lambda_T t}$$

Magically the left hand side looks like the derivative of a product which gives us the left hand side integral straight away

$$\frac{d}{dt} N_T e^{\lambda_T t} = \frac{dN_T}{dt} e^{\lambda_T t} + \lambda_T N_T e^{\lambda_T t}$$

The integral on the right hand side is easy so we put them together

$$N_T e^{\lambda_T t} = \int \lambda_U N_U^0 e^{-\lambda_U t + \lambda_T t} = \frac{\lambda_U N_U^0 e^{-\lambda_U t + \lambda_T t}}{\lambda_T – \lambda_U} + C$$

We can get rid of the $$e^{\lambda_T t}$$ term by multiplying by $$e^{-\lambda_T t}$$ to give

$$N_T = \frac{\lambda_U N_U^0 e^{-\lambda_U t}}{\lambda_T – \lambda_U} + C e^{-\lambda_T t}$$

The integration constant $$C$$ can be found by setting $$t=0$$ when we know that the amount of thorium is some initial value $$N_T = N_T^0$$ and

$$N_T^0 = \frac{\lambda_U N_U^0 }{\lambda_T – \lambda_U} + C$$

This can be rearranged

$$C = N_T^0 – \frac{\lambda_U N_U^0 }{\lambda_T – \lambda_U}$$

Substituting this back into our equation for $$N_T$$ gives us

$$N_T = \frac{\lambda_U N_U^0 e^{-\lambda_U t}}{\lambda_T – \lambda_U} + N_T^0 e^{-\lambda_T t} – \frac{\lambda_U N_U^0 }{\lambda_T – \lambda_U} e^{-\lambda_T t}$$

Which can be rearranged as

$$N_T = \frac{\lambda_U N_U^0 (e^{-\lambda_U t} – e^{-\lambda_T t})}{\lambda_T – \lambda_U} + N_T^0 e^{-\lambda_T t}$$

The second term is simply the decay of the initial amount of thorium, which is usually zero because thorium is not usually present in groundwater. The first term is the interesting one.

If we look at the abundance of thorium and uranium numerically we can see that uranium decays exponentially as we would expect. Thorium is not present initially but is created as uranium decays, its abundance peaks then falls as the abundance of its parent decreases and slows its production rate and decay into radium exceeds its rate of production.

The ratio of thorium to uranium resulting from decay is a steadily increasing amount that stabilises at 0.44.

Notice the equilibrium value we calculated (0.31) is the value of this ratio when the amount of thorium reaches its maximum. It’s easy to find when the amount of thorium peaks by differentiating the equation for thorium assuming the initial amount of thorium is zero

$$\frac{d N_T}{dt} = \frac{\lambda_U N_U^0 (-\lambda_U e^{-\lambda_U t} + \lambda_T e^{-\lambda_T t})}{\lambda_T – \lambda_U}$$

The numerator is zero when

$$\lambda_U e^{-\lambda_U t} = \lambda_T e^{-\lambda_T t}$$

If we rearrange

$$e^{(\lambda_T – \lambda_U) t} = \frac{\lambda_T}{\lambda_U}$$

Take logs

$$(\lambda_T – \lambda_U) t = \ln \frac{\lambda_T}{\lambda_U}$$

And

$$t = \frac{\ln \frac{\lambda_T}{\lambda_U}}{\lambda_T – \lambda_U}$$

Using the decay constants for uranium and thorium the amount of thorium peaks after about 185,000 years. We can also see from the graph above that the ratio is sensitive up to around 250,000 years which means it works well for archaeological remains from hominids.

#### ramin

Ramin is a science and maths tutor. He has a first class degree and a doctorate in Physics. He has written two finance books and also teaches adults about investment through his company PensionCraft.