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Fundamentals Of Physics
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Short Answer

After a brief neutron irradiation of silver, two isotopes are present: 108Ag(T1/2=2.42min) with an initial decay rate of 3.1×105/s,and role="math" localid="1661598035621" 110Ag(T1/2=24.6s) with an initial decay rate of. Make a semilog plot similar to Fig. 42-9 showing the total combined decay rate of the two isotopes as a function of time t = 0 from until t = 10min .We used Fig. 42-9 to illustrate the extraction of the half-life for simple (one isotope) decays. Given only your plot of total decay rate for the two-isotope system here, suggest a way to analyze it in order to find the half-lives of both isotopes.

The plot similar to the total combined decay rate of the two isotopes as a function of time from t = 0 to t = 10min is plotted.

See the step by step solution

Step by Step Solution

Step 1: The given data

  1. The half life of 108Ag,T1/2108=2.42min or 145.2s
  2. Decay rate of 108Ag,R108=3.1×105/s
  3. The half life of 110Ag,T1/2110=24.6s
  4. Decay rate of 110Ag,R110=4.1×106s

Step 2: Determine the concept of combined decay

The total combined decay rate of two-isotope system is as follows:

In R=InR0e-λt+R0'e-λt …… (i)

The disintegration constant is as follows:

λ=In2T12 …… (ii)

Here, T12 is the half-life of the substance.

Step 3: Plot the decay graph of the combined isotope model

From the given data R0=3.1×105/s and R0=4.1×106/s equation (i), the combined decay rate of the isotopes, the plot is made accordingly for disintegration constants using equation (ii) as:


The plot is given below:

Note that the magnitude of the slope for small is λ'(the disintegration constant for 110Ag ) , and for large t is λ (the disintegration constant for role="math" localid="1661598893886" 108Ag ).

Most popular questions for Physics Textbooks

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 Emmaus by 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 Emmaus was 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, Emmaus certainly 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 U238 and ends with stable PB206.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:

Th23075.4 kyRa2261.60 kyPb21022.6 yPb206

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 Pb210 nuclei changes is given by


where N210 and N226 are the numbers of Pb210 nuclei and Ra226 nuclei in the sample and λ210 and λ226 are the corresponding disintegration constants. Because the decay series has been active for billions of years and because the half-life of Pb210 is much less than that of role="math" localid="1661919868408" Ra226, the nuclides Ra226 and Pb210 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 R226R210 of the activities of these nuclides in the sample of lead ore? (c) What is the N226N210 ratio of their numbers? When lead pigment is refined from the ore, most of the radium Ra226 is eliminated. Assume that only 1.00% remains. Just after the pigment is produced, what are the ratios (d) R226R210 and (e) N226N210? Keisch realized that with time the ratio R226R210 of the pigment would gradually change from the value in freshly refined pigment back to the value in the ore, as equilibrium between the Pb210and the remaining Ra226 is established in the pigment. If Emmaus were 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 Emmaus were 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 Emmaus a Vermeer?


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