Hello Dr.Bob. Could you check my work please? I'm not sure if I'm right in how I solved this problem.
Which of the following nuclides are most likely to decay via beta decay?
A) I-131 B) Pb-206 C) F-18 D) Zr-90 E) Ar-40
I know that nuclides undergo beta decay when they are too heavy to be stable. Out of the nuclides listed, three can be considered heavy. These are: Pb-206 which has atomic mass of 207.2, F-18 with atomic mass of 19 and Zr-90 with atomic mass of 91.22. Would it be enough for a question like this to reason that since the difference between the atomic numbers is the greatest for Zr-90, then it is the nuclide which is more likely to decay? Or do I need to take the valley of stability into consideration? The reason I ask about the valley of stability is because a question like this will likely be on the exam and I don't believe we need to memorize the table of stability.
Thank you so much for your help.
I don't know how to answer your question. I believe I-131 is the only element listed that decays by beta emission (F-18 decays by positron emission). How did I know that. It has little to do with theory. I know I-131 is a radioactive isotope and decays by beta emission. I looked all of the others up on Google and found they are stable except for F-18 with is a positron emitter. That doesn't address the question of how you can tell. Note, for example, that I-131 has a N:P ratio of 1.47 and decays by beta emission while Pb-206 has N:P ratio of 1.51 and is stable. And C-14 has N:P ratio of 1.22 and decays. Perhaps another tutor can help.
Hello Dr.Bob, could you please help me with this problem:
Which of the following nuclides are most likely to decay via beta decay?
a)200 78 Pt b)192 78 Pt c)188 78 Pt
d)196 78 Pt
To determine which nuclide is most likely to decay via beta decay, we need to consider the neutron-to-proton ratio for each nuclide and compare it to the stability line (valley of stability) on the graph of the nuclide chart.
In beta decay, a neutron in the nucleus is converted into a proton, releasing a beta particle (an electron) and an antineutrino. This process helps the nuclide achieve a more stable neutron-to-proton ratio.
Let's analyze each nuclide:
A) I-131: Iodine-131 has an atomic number of 53 and a mass number of 131.
B) Pb-206: Lead-206 has an atomic number of 82 and a mass number of 206.
C) F-18: Fluorine-18 has an atomic number of 9 and a mass number of 18.
D) Zr-90: Zirconium-90 has an atomic number of 40 and a mass number of 90.
E) Ar-40: Argon-40 has an atomic number of 18 and a mass number of 40.
Now let's compare the neutron-to-proton ratio for each nuclide:
A) I-131: The neutron-to-proton ratio is 78/53 ≈ 1.47.
B) Pb-206: The neutron-to-proton ratio is 124/82 ≈ 1.51.
C) F-18: The neutron-to-proton ratio is 9/9 = 1.
D) Zr-90: The neutron-to-proton ratio is 50/40 = 1.25.
E) Ar-40: The neutron-to-proton ratio is 22/18 ≈ 1.22.
Comparing these ratios to the stability line on the graph, we can determine which nuclide is furthest from the line and hence more likely to decay via beta decay.
Based on the neutron-to-proton ratios, Pb-206 (Option B) seems to be closest to the stability line, while I-131 (Option A) and F-18 (Option C) are relatively far from the line. Therefore, among the given options, I-131 and F-18 are more likely to decay via beta decay.
To answer your question, it is not necessary to memorize the entire table of stability. Understanding the neutron-to-proton ratio and its relation to the stability line will help you determine the likelihood of beta decay for a given nuclide.
Hello! I'm not Dr. Bob, but I'll be happy to help you with your question.
To determine which nuclides are most likely to decay via beta decay, you need to consider both their atomic numbers (proton counts) and their neutron-to-proton ratios.
In beta decay, a neutron is converted into a proton, so the nuclide's atomic number increases by one. This means that beta decay tends to occur in nuclides with too many neutrons relative to protons.
The valley of stability is a concept that helps us understand the stability of nuclides. It is a graph that shows the relationship between the number of neutrons and protons in a nuclide. Nuclides that fall outside the valley of stability are likely to decay to achieve a more stable configuration.
While it is helpful to understand the concept of the valley of stability, you don't necessarily need to memorize the entire table of stability. Instead, you can use some basic reasoning to determine which nuclides are likely to undergo beta decay.
Let's examine the nuclides in the given options:
A) I-131: Iodine-131 has an atomic number of 53 and a mass number of 131. It is relatively heavy, and its neutron-to-proton ratio is not within the stable range. Therefore, it is likely to decay via beta decay.
B) Pb-206: Lead-206 has an atomic number of 82 and a mass number of 206. It is also relatively heavy, but its neutron-to-proton ratio is within the stable range. Therefore, it is less likely to undergo beta decay.
C) F-18: Fluorine-18 has an atomic number of 9 and a mass number of 18. It is relatively light, and its neutron-to-proton ratio is not within the stable range. Therefore, it is likely to decay via beta decay.
D) Zr-90: Zirconium-90 has an atomic number of 40 and a mass number of 90. It is relatively heavy, and its neutron-to-proton ratio is within the stable range. Therefore, it is less likely to undergo beta decay.
E) Ar-40: Argon-40 has an atomic number of 18 and a mass number of 40. It is relatively light, and its neutron-to-proton ratio is not within the stable range. Therefore, it is likely to decay via beta decay.
Based on this analysis, the nuclides most likely to decay via beta decay are I-131, F-18, and Ar-40.
I hope this explanation helps! If you have any more questions, feel free to ask.