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).
The magic nucleon numbers are: 2, 8, 20, 28, 50, 82, and 126.
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
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.
A worker at a breeder reactor plant accidentally ingests 2.5 mg of 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 and ends with stable .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:
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 nuclei changes is given by
where are the numbers of nuclei and nuclei in the sample and are the corresponding disintegration constants. Because the decay series has been active for billions of years and because the half-life of is much less than that of role="math" localid="1661919868408" , the nuclides 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 of the activities of these nuclides in the sample of lead ore? (c) What is the ratio of their numbers? When lead pigment is refined from the ore, most of the radium is eliminated. Assume that only 1.00% remains. Just after the pigment is produced, what are the ratios (d) and (e) ? Keisch realized that with time the ratio of the pigment would gradually change from the value in freshly refined pigment back to the value in the ore, as equilibrium between the and the remaining 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?
The radionuclide has a half-life of 57.0 min. At t = 0, the counting rate of a sample of it is above the general background activity, which is . 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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