## A surprising application of Differential Equations — The Van Meegeren art forgeries Part II

However, we can still use equation 3 to distinguish between a 17th century painting and a modern forgery. The basis for this statement  is the simple observation if  the paint is very old compared to the 22 year half-life of lead, then the amount of radioactivity from the lead-210 in the paint will be nearly equal to  the amount of radioactivity from the radium in the paint. On the other hand, if the painting is modern (approximately 20 years old or so), then the amount of radioactivity from the lead-210 will be much greater than the amount of radioactivity from the radium.

We make this argument precise in the following manner. Let us assume that the painting in question is either very new or about 300 years old. Set $t-t_{0}=300$ years in equation 5. Then, after some simple algebra, we see that

$\lambda y_{0}=\lambda y(t)e^{3000}-r(e^{3000}-1)$ equation VI

If the painting is indeed a modern forgery, then $\lambda y_{0}$ will be absurdly large. To determine what is an absurdly high disintegration rate we observe that if the lead-210 decayed originally (at the time of manufacture) at the rate of 100 disintegrations per minute per gram of white lead, then the ore from which it was extracted had a uranium content of approximately 0.014 percent. This is a fairly high concentration of uranium since the average amount of uranium in rocks of earth’s crust is about 2.7 parts per million. On the other hand, there are some very rare ores in the Western Hemisphere whose uranium content is 2-3 percent. To be on the safe side, we will say that a disintegration rate of lead-210 is certainly absurd if it exceeds 30000 disintegrations per minute per gram of white lead.

To evaluate $\lambda y_{0}$, we must evaluate the present disintegration rate, $\lambda y(t)$ of the lead-210, the disintegration rate r of the radium-226, and $e^{3000}$. Since the disintegration rate of polonium-210 equals that of lead-210 after several years, and since it is easier to measure the disintegration rate of polonium-210, we substitute these values for those of lead-210. To compute $e^{3000}$, we observe from equation 3 that $\lambda = (ln 2/22)$. Hence,,

$e^{300 \lambda}=e^{(300/22)ln 2}=2^{(150/11)}$

The disintegration rates of polonium-210 and radium-226 were measured for the “Disciples at Emmaus” and various other alleged forgeries have been calculated. For the “Disciples at Emmaus”, the disintegration rate of polonium-210 is 8.5 per minute per gram of white lead and the disintegration rate of radium-226 is 0.8 per minute per gram of white lead.

If we now evaluate $\lambda y_{0}$ from equation 6 for the white lead in the painting “Disciples at Emmaus”we obtain that

$\lambda y_{0}=(8.5)2^{150/11}-0.8(2^{150/11}-1)=98.050$

which is unnecessarily large.

Thus, this painting must be a modern forgery.

More later…

— Nalin Pithwa

### 4 thoughts on “A surprising application of Differential Equations — The Van Meegeren art forgeries Part II”

1. Joy Xu February 22, 2020 / 9:55 am

Hi Nalin,
This was so interesting!! I was just wondering, is there a more specific range for the number being “unnecessarily large”, like what counts as unnecessarily large?
Thanks,
Joy

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• Nalin Pithwa February 22, 2020 / 3:05 pm

I am not sure. I would request you to pick up the reference and go into the details mentioned therein. Regards, NP

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• Joy Xu March 2, 2020 / 11:24 am

Sorry, where am I able to find the references? I’m not the best with websites sorry!! Thank you, Joy

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2. Nalin Pithwa March 2, 2020 / 2:21 pm

What I meant was the print copy (of the text which I had used; I think I have mentioned that book in the article). If I can find some time to look into it again, I will post here as the details….please give me some time…Regards NP

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