Radioactivity and Radioactive Decay Quick Check
1. Which of the following transformations is an example of a β−
-decay?
31/14Si becomes 31/15 P because a neutron is converted to a proton.
2. The unstable uranium nuclide(I don't feel like typing the whole thing)
238/92U because the very long half-life means a very small rate of decay
3. The beta minus decay process 13/7N→13/6C+β−+ν¯¯¯ is proposed in which nitrogen’s proton becomes a neutron. The process is not possible because:
charge conservation is not satisfied
4. The fact that the relative abundance is greater for elements in which the total number of protons and neutrons is a multiple of two is because __________.
of the four nuclear decay processes only the α
-decay changes the baryon number and does so in increments of four
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1. The transformation in which 31/14Si becomes 31/15P because a neutron is converted to a proton is an example of β−-decay. In β−-decay, a neutron in the nucleus is converted into a proton, and an electron (β−) and an electron antineutrino (ν¯¯¯) are emitted.
2. The unstable uranium nuclide 238/92U has a very long half-life, which means that its rate of decay is very small. This characteristic is due to the large number of protons and neutrons in its nucleus, which makes it less likely for it to undergo radioactive decay.
3. The beta minus (β−) decay process proposed in which 13/7N decays into 13/6C + β− + ν¯¯¯ is not possible because it violates the principle of charge conservation. In this process, a proton in the nitrogen nucleus is supposed to be converted into a neutron. However, the conversion of a proton into a neutron would create an overall charge imbalance since protons have a positive charge and neutrons have no charge. Therefore, the process cannot occur.
4. The fact that elements with a total number of protons and neutrons that is a multiple of two have a higher relative abundance is due to the nature of nuclear decay processes. Among the four nuclear decay processes (α-decay, β−-decay, β+-decay, and electron capture), only α-decay changes the baryon number and does so in increments of four. The baryon number is the total number of protons and neutrons in the nucleus. Elements with a total number of protons and neutrons that is a multiple of two have a higher likelihood of undergoing α-decay, resulting in a higher relative abundance.
To answer these questions, we need to understand the concept of radioactivity and radioactive decay. Radioactivity is the process by which certain atomic nuclei emit particles or rays, resulting in the transformation of one element into another. These transformations are known as radioactive decays.
1. To identify which transformation is an example of β−-decay, we need to understand that β−-decay involves the emission of a beta particle, which is an electron. In this case, the transformation from 31/14Si to 31/15P indicates the conversion of a neutron into a proton. Since a neutron consists of one up quark and two down quarks, while a proton consists of two up quarks and one down quark, this transformation means one of the down quarks transformed into an up quark, resulting in the emission of a beta particle (electron). Therefore, the answer is that 31/14Si becomes 31/15P because a neutron is converted to a proton.
2. The question mentions the unstable uranium nuclide 238/92U, which has a very long half-life. The term half-life refers to the time it takes for half of the radioactive sample to decay. In this case, the long half-life of 238/92U means that it decays very slowly, resulting in a small rate of decay. Therefore, the answer is that the very long half-life of 238/92U means a very small rate of decay.
3. The question describes a beta minus decay process in which 13/7N decays into 13/6C, emitting a beta particle and an antineutrino. The process is proposed to involve the conversion of nitrogen's proton into a neutron. However, the question states that this process is not possible. The reason for this is the principle of charge conservation. In any nuclear decay process, the total charge before and after the decay must remain the same. In this case, nitrogen (13/7N) has 7 protons, while carbon (13/6C) has 6 protons. If the process were to convert a proton in nitrogen to a neutron in carbon, the total charge would no longer be conserved. Therefore, the answer is that the process is not possible because charge conservation is not satisfied.
4. The question discusses the relative abundance of elements in which the total number of protons and neutrons is a multiple of two. This observation can be explained by considering the different types of radioactive decay processes. There are four main types of nuclear decay: alpha (α)-decay, beta (β)-decay, gamma (γ)-decay, and electron capture. Out of these four processes, only α-decay changes the baryon number and does so in increments of four. The baryon number represents the total number of protons and neutrons in an atomic nucleus. Therefore, when α-decay occurs, the total number of protons and neutrons in the nucleus changes by four, leading to a relative abundance that is greater for elements with a total number of protons and neutrons that is a multiple of two.
5. True or False: Radioactive decay always results in the formation of a stable nucleus.
False. Radioactive decay can result in the formation of a stable nucleus, but it can also lead to the formation of another unstable nucleus, which can then decay further.