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A measurement of the energy Eof an intermediate nucleus must be made within the mean lifetime tof the nucleus and necessarily carries an uncertainty Eaccording to the uncertainty principle

Et=h.

(a) What is the uncertainty Ein the energy for an intermediate nucleus if the nucleus has a mean lifetime of10-22s? (b) Is the nucleus a compound nucleus?

Short Answer

Expert verified
  1. The uncertainty in the energy for an intermediate nucleus is 6.6 MeV .
  2. The nucleus is not a compound nucleus.

Step by step solution

01

Given data

Mean lifetime of the nucleus,tavg=10-22s

02

Understanding the concept of uncertainty principle  

The uncertainty principle given by Heinsberg states that an electron's position and velocity can be measured. At the same time, not even in theory.

A compound nucleus is an unstable nucleus formed by the coalescence of an atomic nucleus with a captured particle.

Formula:

The energy-time uncertainty relation,Et=horEt=h/2........1

where, h is the Planck鈥檚 constant.

03

a) Calculation of the uncertainty in energy

Using the given data in equation (1), we can get the uncertainty in energy for an intermediate nucleus as follows:

Eh/2tavg=6.62610-34J.s/21.610-19J.eV1.010-22s6.6106eV=6.6MeV

Hence, the uncertainty in energy is 6.6MeV.

04

b) Calculation for checking if the nucleus is a compound nucleus or not

In order to fully distribute the energy in a fairly large nucleus, and create a compound nucleus鈥 equilibrium configuration, about 10-15sis typically required. A reaction state that exists no more than about 10-22sdoes not qualify as a compound nucleus.

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Most popular questions from this chapter

The isotope U238decays to P206bwith a half-life of4.47109Y. Although the decay occurs in many individual steps, the first step has by far the longest half-life; therefore, one can often consider the decay to go directly to lead. That is,U238P206b+variousdecayproducts

A rock is found to contain 4.20mgofU238and 2.135mgofP206b. Assume that the rock contained no lead at formation, so all the lead now present arose from the decay of uranium. How many atoms of (a)U238and (b)P206bdoes the rock now contain? (c) How many atoms ofU238did the rock contain at formation? (d) What is the age of the rock?

A radioactive nuclide has a half-life of 30.0y. What fraction of an initially pure sample of this nuclide will remain undecayed at the end of (a) 60.0 yand (b) 90.0y?

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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鈥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鈥檚 choice of pigments; he had then signed Vermeer鈥檚 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鈥檚 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 unstableU238and ends with stablePB206.To follow the spirit of Keisch鈥檚 analysis, focus on the following abbreviated portion of that decay series, in which intermediate, relatively short-lived radionuclides have been omitted:

Th23075.4kyRa2261.60kyPb21022.6yPb206

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 ofPb210nuclei changes is given by

dN210dt=226N226-210N210,

whereN210andN226are the numbers ofPb210nuclei and Ra226nuclei in the sample and210and226are the corresponding disintegration constants. Because the decay series has been active for billions of years and because the half-life of Pb210is much less than that of role="math" localid="1661919868408" Ra226, the nuclidesRa226andPb210are in equilibrium; that is, the numbers of these nuclides (and thus their concentrations) in the sample do not change. (b) What is the ratioR226R210of the activities of these nuclides in the sample of lead ore? (c) What is the N226N210ratioof their numbers? When lead pigment is refined from the ore, most of the radiumRa226 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 ratioR226R210of the pigment would gradually change from the value in freshly refined pigment back to the value in the ore, as equilibrium between thePb210and the remainingRa226is 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?

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