91Ó°ÊÓ

Question: At the end of World War II, Dutch authorities arrested Dutch artist Hans van Meegeren for treason because, during the war, he had sold a masterpiece painting to the Nazi Hermann Goering. The painting, Christ and His Disciples at Emmausby Dutch master Johannes Vermeer (1632–1675), had been discovered in 1937 by van Meegeren, after it had been lost for almost 300 years. Soon after the discovery, art experts proclaimed that Emmauswas possibly the best Vermeer ever seen. Selling such a Dutch national treasure to the enemy was unthinkable treason. However, shortly after being imprisoned, van Meegeren suddenly announced that he, not Vermeer, had painted Emmaus. He explained that he had carefully mimicked Vermeer's style, using a 300-year-old canvas and Vermeer’s choice of pigments; he had then signed Vermeer’s name to the work and baked the painting to give it an authentically old look.

Was van Meegeren lying to avoid a conviction of treason, hoping to be convicted of only the lesser crime of fraud? To art experts, Emmauscertainly looked like a Vermeer but, at the time of van Meegeren’s trial in 1947, there was no scientific way to answer the question. However, in 1968 Bernard Keisch of Carnegie-Mellon University was able to answer the question with newly developed techniques of radioactive analysis.

Specifically, he analyzed a small sample of white lead-bearing pigment removed from Emmaus. This pigment is refined from lead ore, in which the lead is produced by a long radioactive decay series that starts with unstable${}^{\mathbf{238}}\mathbf{U}$and ends with stable${}^{\mathbf{206}}\mathbf{PB}$.To follow the spirit of Keisch’s analysis, focus on the following abbreviated portion of that decay series, in which intermediate, relatively short-lived radionuclides have been omitted:

${}^{\mathbf{230}}\mathbf{Th}\underset{\mathbf{75}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{ky}}{\mathbf{→}}{}^{\mathbf{226}}\mathbf{Ra}\underset{\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{ky}}{\mathbf{→}}{}^{\mathbf{210}}\mathbf{Pb}\underset{\mathbf{22}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{y}}{\mathbf{→}}{}^{\mathbf{206}}\mathbf{Pb}$

The longer and more important half-lives in this portion of the decay series are indicated.

a) Show that in a sample of lead ore, the rate at which the number of${}^{\mathbf{210}}\mathbf{Pb}$nuclei changes is given by

$\frac{\mathbf{d}{\mathbf{N}}_{\mathbf{210}}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{λ}}_{\mathbf{226}}{\mathit{N}}_{\mathbf{226}}\mathbf{-}{\mathit{λ}}_{\mathbf{210}}{\mathit{N}}_{\mathbf{210}\mathbf{,}}$

where${\mathbf{N}}_{\mathbf{210}\mathbf{}}\mathbf{and}\mathbf{}{\mathbf{N}}_{\mathbf{226}}$are the numbers of${}^{\mathbf{210}}\mathbf{Pb}$nuclei and ${}^{\mathbf{226}}\mathbf{Ra}$nuclei in the sample and${\mathit{λ}}_{\mathbf{210}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{λ}}_{\mathbf{226}}$are the corresponding disintegration constants. Because the decay series has been active for billions of years and because the half-life of ${}^{\mathbf{210}}\mathbf{Pb}$is much less than that of role="math" localid="1661919868408" ${}^{\mathbf{226}}\mathbf{Ra}$, the nuclides${}^{\mathbf{226}}\mathbf{Ra}\mathbf{}\mathbf{and}\mathbf{}{}^{\mathbf{210}}\mathbf{Pb}$are in equilibrium; that is, the numbers of these nuclides (and thus their concentrations) in the sample do not change. (b) What is the ratio$\frac{{\mathbf{R}}_{\mathbf{226}}}{{\mathbf{R}}_{\mathbf{210}}}$of the activities of these nuclides in the sample of lead ore? (c) What is the $\frac{{\mathbf{N}}_{\mathbf{226}}}{{\mathbf{N}}_{\mathbf{210}}}$ratioof their numbers? When lead pigment is refined from the ore, most of the radium${}^{\mathbf{226}}\mathbf{Ra}$ is eliminated. Assume that only 1.00% remains. Just after the pigment is produced, what are the ratios (d)$\frac{{\mathbf{R}}_{\mathbf{226}}}{{\mathbf{R}}_{\mathbf{210}}}$ and (e)$\frac{{\mathbf{N}}_{\mathbf{226}}}{{\mathbf{N}}_{\mathbf{210}}}$? Keisch realized that with time the ratio$\frac{{\mathbf{R}}_{\mathbf{226}}}{{\mathbf{R}}_{\mathbf{210}}}$of the pigment would gradually change from the value in freshly refined pigment back to the value in the ore, as equilibrium between the${}^{\mathbf{210}}\mathbf{Pb}$and the remaining${}^{\mathbf{226}}\mathbf{Ra}$is established in the pigment. If Emmauswere painted by Vermeer and the sample of pigment taken from it was 300 years old when examined in 1968, the ratio would be close to the answer of (b). If Emmauswere painted by van Meegeren in the 1930s and the sample were only about 30 years old, the ratio would be close to the answer of (d). Keisch found a ratio of 0.09. (f) Is Emmausa Vermeer?

Step by step solution

01

Write the given data

The radioactive reaction with their half-lives is given as follows:

${}^{230}\mathrm{Th}\underset{75.4\mathrm{ky}}{→}{}^{226}\mathrm{Ra}\underset{1.60\mathrm{ky}}{→}{}^{210}\mathrm{Pb}\underset{22.6\mathrm{y}}{→}{}^{206}\mathrm{Pb}$

02

Determine the formula for decay as:

Formula:

The rate of decay of a radioactivity is as follows:

$R=\mathrm{λ}N=-\frac{dN}{dt}$ …… (i)

The disintegration constant is as follows:

$\mathrm{λ}=\frac{\ln 2}{T_{1/2}}$ …… (ii)

Here,$T_{\frac{1}{2}}$ is the half-life of the substance.

03

a) Calculate the activity of lead substance

From the decay series, consider that $N_{210}$, the amount of ${}^{210}Pb$nuclei, changes because of two decays: the decay from ${}^{210}\mathrm{Pb}\mathrm{into}{}^{206}\mathrm{Pb}$at the rate considering equation (i),, and the decay frominto at the rate (from equation (i)),$R_{210}={\mathrm{λ}}_{210}{N}_{210}$. The first of these decays causes$N_{210}$ to increase while the second one causes it to decrease. Thus, the required activity of lead-210 is given using equation (i) as follows:

$$\begin{gathered} \frac{d{N}_{210}}{dt}=R_{226}-R_{210} \\ ={\mathrm{λ}}_{226}{N}_{226}-{\mathrm{λ}}_{210}{N}_{210} \end{gathered}$$

Hence, in a sample of lead ore, the rate at which the number of lead ${}^{210}\mathrm{Pb}$nuclei changes is given by,$\frac{d{N}_{210}}{dt}={\mathrm{λ}}_{226}{N}_{226}-{\mathrm{λ}}_{210}{N}_{210}$where, where $N_{210}\mathrm{and}{N}_{226}$are the numbers of ${}^{210}\mathrm{Pb}$nuclei and ${}^{210}\mathrm{Ra}$nuclei in the sample, ${\mathrm{λ}}_{210}\mathrm{and}{\mathrm{λ}}_{226}$are the corresponding disintegration constants.

04

b) Calculate the ratio of activities of radium-226 and lead-210

From the above concept and decay process, we set

$$\begin{gathered} \frac{d{N}_{210}}{dt}=R_{226}-R_{210} \\ =0 \end{gathered}$$

To obtain,$\frac{R_{226}}{R_{210}}=1.00$

Hence, the value of the ratio is 1.00.

05

c) Calculate the ratio of number of nuclei of radium-226 and lead-210

From equation (i) and (ii) and the given data of half-life, we get that

$$\begin{gathered} \frac{N_{226}}{N_{210}}=\frac{{\mathrm{λ}}_{210}}{{\mathrm{λ}}_{226}}\left(âˆ\mu \frac{R_{226}}{R_{210}}=1.00\right) \\ =\frac{T_{1/2_{226}}}{T_{1/2_{210}}} \end{gathered}$$

Substitute the value and solve as:

$$\begin{gathered} \frac{N_{226}}{N_{210}}=\frac{1.60×{10}^3\mathrm{y}}{22.6\mathrm{y}} \\ =70.8 \end{gathered}$$

Hence, the value of the ratio is.

06

d) Calculate the ratio of activities of radium-226 and lead-210 for remaining 1.00%

Since only 1.00% of the 226Ra remains, the ratio$\frac{R_{226}}{R_{210}}$ is 0.00100 of that of the equilibrium state computed in part (b). Thus the ratio now becomes:

$$\begin{gathered} \frac{R_{226}}{R_{210}}=\left(0.001\right)\left(1\right) \\ =0.001 \end{gathered}$$

Hence, the value of ratio is 0.001.

07

e) Calculate the ratio of number of nuclei of radium-226 and lead-210 for remaining 1.00%

This is similar to part (d) above. Since only 1.00% of the radium-226 remains, the ratio$\frac{N_{226}}{N_{210}}$ is 1.00% of that of the equilibrium state computed in part (c). Thus, the ratio of number of nuclei becomes:

$$\begin{gathered} \frac{N_{226}}{N_{210}}=\left(0.001\right)\left(70.8\right) \\ =0.708 \end{gathered}$$

Hence, the value of the ratio is 0.708.

08

f) Calculate whether Emmaus a Vermeer

Since the actual value of$\frac{N_{226}}{N_{210}}$ is 0.09, which much closer to 0.0100 than to 1, the sample of the lead pigment cannot be 300 years old. So, Emmausis not aVermeer.

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