volcanic eruptions and industrial smoke emit only12C (from tissue older than 100,000 years)
into the atmosphere and decrease the ratio, while nuclear testing has resulted in a 100% increase in the ratio in certain parts of the northern hemisphere! These are now factored into dating calculations (see Borelli and Coleman, 1996).
As previously mentioned, radiocarbon dating is inaccurate for the recent past and for very long time periods. Thus the lead isotope with a relatively short half-life (≃ 22 years) is used for the dating of paintings in our recent history, while at the other extreme, radioactive substances (such as Uranium-238) with half-lives of billions of years are used to date the Earth.
Summary of skills developed here:
• Understand the concept of modelling with compartments and the balance law. • Find the constant of proportionality from the half-life for radioactive elements. • Generate a solution, using Maple or analytical methods, to simple separable
differential equations.
2.3
Case Study: Detecting art forgeries
This case study is based on an article in Braun (1979).
After World War II, Europe was in turmoil. Some well-known artworks had disappeared, others had turned up miles from the galleries to which they belonged, and some new ones had appeared! The Dutch Field Security uncovered a firm that, acting through a banker, had been involved in selling artwork to the Nazi leader, Goering. In particular in the case of a painting by the seventeenth century artist Jan Vermeer, “Woman Taken in Adultery,” the banker claimed that he had been acting on behalf of Van Meegeren, a third-rate painter, who was duly imprisoned for the crime of collaboration. However, shortly after his arrest, Van Meegeren announced that he had never sold this painting to Goering and that it was in fact a forgery. He claimed that, together with “Disciples at Emmaus” and several others, he himself had painted them. In order to prove this claim, he began to forge a further work by Vermeer while in prison. When he learned that his charge had been changed from that of collaboration to one of forgery, he refused to finish the work as it would provide evidence against himself and reveal the means by which he had ‘aged’ the paintings.
A team of art historians, chemists and physicists was employed to settle the dispute. They took x-rays to establish what paintings might lie underneath, analysed pigments in the paints and looked for chemical and physical signs of old age authenticity. They found traces of modern pigment in the cobalt blue, although Van Meegeren had been very careful, in all other cases, to use pigments that the artists themselves might have used. Further examination indicated Van Meegeren had scraped old canvases that he had used as a base for the forgeries. The discovery of phenoformaldehyde, which he had cleverly mixed into the paints and then baked in the oven so that it would harden to bakelite as was typical of old paint, finally sealed his fate and he was convicted in October 1947 and sentenced to a year in prison. (He died of a heart attack while in prison in December of that year.)
But the controversy was not over. Art lovers claimed that “Disciples at Emmaus” was far too beautifully painted to be a forgery and was in quite a different class from the other forgeries. They demanded more scientific evidence to prove it a fake. Besides, a noted art historian A. Bredius had previously certified it as authentic, and it had been bought by the Rembrandt Society for a sizeable sum of money. Reputations and money were at stake here.
A scientific study was undertaken with the results based on the notion of radioactive decay. The atoms of certain elements are unstable, or radioactive, and within a given time (which depends on the element involved) a number of the atoms disintegrate. This rate of disintegration has been shown to be proportional to the number of atoms present at that time, N (t). So, starting from N (t0) = n0, where n0is the original amount of material,
dN
dt = −λN, with N = n0e
−λ(t−t0)
, where λ, the decay constant, is chosen positive.
From here, if the initial formation of the substance was at time t0, then its age is
(1/λ) ℓn(n0/N ). N can be measured, but n0, the original amount of the material, is harder
to find.
A useful quantity to know is the half-life of an element, the time taken for the number of atoms present at a particular time to halve. Now if N = 1
2n0, then rearranging the above
gives
half-life = ℓn 2 λ ≃
0.69 λ .
The half-lives of many radioactive substances are well known: for Uranium-238 it is 4.5 billion years, for Carbon-14 is 5,568 years, for Lead-210 it is 22 years and for Polonium-214 it is less than 1 second.
All paintings contain small amounts of radioactive Lead-210, which we know to decay with a half-life of 22 years, as this is contained in lead white which has been used by artists as a pigment for over 2,000 years. Lead white contains lead metal (which is smelted from rocks), which in turn contains small amounts of Lead-210 and extremely small amounts of Radium-226 (mostly removed in the smelting process). There is no further supply of lead due to the absence of Radium-226, which decays to Lead-210 in a series of steps, and so Lead-210 initially decays rapidly. Radium-226 has a half-life of 1,600 years and since it exists in very small amounts in lead white, eventually the process of lead decay stabilises when the amount disintegrating balances with the amount produced by radium disintegration.
Let N (t) be the amount of Lead-210, and then, as before, dN
dt = −λN + R(t), n0= N (t0),
where R(t) is the rate of disintegration of Radium-226 per minute per gram of lead white. Since we are interested in a time period of 300 years at the most, and the half-life of Radium- 226 is 1,600 years, and the amount present is so small, we can consider R(t) = R to be a constant. Solving the equation, using an integrating factor of eλt, we conclude that
N (t) = R λ 1 − e−λ(t−t0) + n0e−λ(t−t0).
R and N can be measured easily and λ is known; however, n0 is more difficult and is dealt
with in the following way. The above equation can be rearranged to give λn0= λN eλ(t−t0)− R(eλ(t−t0)− 1),
2.4 Scenario: Pacific rats colonise New Zealand 19