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Fundamentals Of Physics
Found in: Page 1302
Fundamentals Of Physics

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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Short Answer

The magic nucleon numbers for nuclei are given in Module 42-8 as 2, 8, 20, 28, 50, 82, and 126. Are nuclides magic (that is, especially stable) when (a) only the mass number A, (b) only the atomic number Z, (c) only the neutron number N, or (d) either Z or N (or both) is equal to one of these numbers? Pick all correct phrases.

The correct phrase is (d).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: The given data

The magic nucleon numbers are: 2, 8, 20, 28, 50, 82, and 126.

Step 2: Understanding the concept of magic numbers  

In nuclear physics, magic numbers are several nucleons such that the nucleons of the nuclide are arranged into complete shells within the atomic nucleus. Thus, their nuclei are more stable than in comparison to other nuclides. They are 2, 8, 20, 28, 50, 82, 126, and so on

Step 3: Calculation on the factors that give magic numbers

According to the concept, the nucleons of the nuclide are arranged into complete shells. Also, any nuclide whose proton number Z or neutron number N has one of these values (or both) turns out to have a special stability that may be made apparent in a variety of ways are called magic (or double magic) nuclides.

Thus, the concept defines that only atomic number, only neutron number or both should be a match to one of the magic numbers, 2, 8, 20, 28, 50, 82, and 126 for them to be magic nuclides. Mass number only being a magic number does not satisfy the filling condition of the nucleon shells.

Hence, according to this, option (d) is the correct pick.

Most popular questions for Physics Textbooks

The nuclide A198u, with a half-life of 2.07d, is used in cancer therapy. What mass of this nuclide is required to produce an activity of 250Ci?

A worker at a breeder reactor plant accidentally ingests 2.5 mg of P239u dust. This isotope has a half-life of 24 100y , decaying by alpha decay. The energy of the emitted alpha particles is 5.2 MeV , with an RBE factor of 13. Assume that the plutonium resides in the worker’s body for (it is eliminated naturally by the digestive system rather than being absorbed by any of the internal organs) and that 95% of the emitted alpha particles are stopped within the body. Calculate (a) the number of plutonium atoms ingested, (b) the number that decay during the 12h , (c) the energy absorbed by the body, (d) the resulting physical dose in grays, and (e) the dose equivalent in sieverts.

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:

Th230→75.4 kyRa226→1.60 kyPb210→22.6 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

dN210dt=λ226N226-λ210N210,

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?

Question: Using a nuclidic chart, write the symbols for (a) all stable isotopes with Z = 60, (b) all radioactive nuclides with N = 60, and (c) all nuclides with A = 60.

The radionuclide 49Schas a half-life of 57.0 min. At t = 0, the counting rate of a sample of it is 6000countsminabove the general background activity, which is 30countsmin. Without computation, determine whether the counting rate of the sample will be about equal to the background rate in 3h, 7h, 10h or a time much longer than 10th.
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