Assigned 8-Feb-17
1st Due Date: 13-Feb-17
2nd Due Date: 16-Feb-17
Lecture 6: Gamma Decay
Lecture 7: Fission
Lecture 8: Nuclear Force and Nuclear Structure
Lecture 9: Nuclear Reactions
Lecture 10: Radiation Interactions
The first iteration of the exam is due 13-Feb-17. The answers will be posted on 14-Feb-17. Any incorrect answers can be resubmitted by 16-Feb-17. Changed answers will be worth 50 % of the original grade. For the 2nd resubmission the work related to the changes must be shown. Please post all questions to the blog.
Use lecture notes, textbooks, Chart of the Nuclides, Table of the Isotopes, and web pages. Use the chart of the nuclides as your primary dataset for isotope half-life. Show your work or references on a separate page and save electronically. Please use 3 significant digits for your answers
For 4.1, would I consider the a,pn curve at 35eV? I think the a,2n curve definitely shows a possible isotope at 35eV but I was wondering you were looking for two isotopes for this question.
ReplyDeleteFor 4.3, would the effect of the cross section affect the number of isotopes produced more than the decay constant (half-lives)?
For 4.1, the question is: What is the primary isotope produced at 35 MeV alpha energy?. The answer should be the isotope that is made in the greatest amount.
ReplyDeleteFor 4.3 the question is: For a 10 minute irradiation at 20 MeV alpha, which isotope is produced with the largest number of atoms?
The 10 minute irradiation time is the key factor in answering your question. If a produced has a half-life in the minutes ranges, say 2 minutes to 30 minutes, then there would be an effect on the overall production. If the produced isotopes have half-lives greater than a few hours, then the amount of isotope that decays in 10 minutes can be ignored.
thanks for the question.
I can't figure out the 2nd fission product for the Hf-180 isotope on question 6. I arrived at the answer Ni-81, but there is no Ni-81 in the chart of the nuclides and so I'm wondering if this is incorrect (and thus my other answers for this question would be incorrect as well)
ReplyDeleteWouldn't it be 78Ni since fission spits out 3 neutrons?
DeleteAh, thank you!
DeleteFollowing up, should I expect the Q-values to be a lot less than Vc?
DeleteWould that be the case for the whole table? Because in this example from the lecture notes the # of neutrons doesn't change:
Delete238U --> 91Br + 147La + Q
I'll have to redo some of my calculations, but I got Q << Vc for some, and Q > Vc for others.
DeleteOh, actually, I think you were right. I think that example and the question on the exam are concerned with spontaneous fission, not neutron-induced fission. So the process just produces two fission products.
DeleteI was looking at the first lecture slides that lists types of decay and it lists spontaneous fission as having 4 neutrons as products.
Deletegreat discussion everyone. I would use the idea from Kathryn listed in the discussion.
Delete238U --> 91Br + 147La + Q
If one uses http://nrv.jinr.ru/nrv/webnrv/qcalc/, set for decay of 180Hf and have 1 isotope as 99Ru the answer comes out.
As for 81Ni, there is no data in the chart of the nuclides. However the calculated Q value is available. This question also demonstrates that one can get answers regarding isotopes that have not been identified.
DeleteAs for Vc compared to Q, this defines if an isotope can fission. If the Q is greater than the Vc, then there is enough energy for fission. As an example, the isotope 144Ce does not fission, but 252Cf does undergo spontaneous fission. You should see difference between Vc and Q for this isotopes.
DeleteThank you for the clarification. I was also wondering if you could confirm one more thing. For number 6, the coulomb barrier is calculated using the fission products fission but for number 9, we use the target and the projectile?
DeleteAlso, I was wondering if there are two separate coulomb barriers, center-of-mass and lab. frame. So in Lecture 9, when you multiplied (Atarget+Aprojectile)/(Atarget) with the center-of-mass lab. frame, is that a different coulomb barrier for the lab. frame, or is it simply the necessary reaction energy that should be different from the one and only coulomb barrier that we calculate from using the formula?
DeleteThe normal Coulomb barrier calculation is for center of mass. The (Atarget+Aprojectile)/(Atarget) is the normalization from the center of mass to the laboratory frame. This is the last few bullets on page 7 of lecture 9.
DeleteCenter of mass, need to bring to laboratory frame
Consider kinetic energy carried by projectile
3.36x ((14+4)/14) = 4.32 MeV alpha needed for reaction
Regarding the comment
Delete"Thank you for the clarification. I was also wondering if you could confirm one more thing. For number 6, the coulomb barrier is calculated using the fission products fission but for number 9, we use the target and the projectile?"
Different Coulomb barriers are being evaluated. For fission the Coulomb barrier that results from the fission products just touching is being evaluated. The fission process must have more energy than this Vc, or else the two fission products could not form.
For the reaction, the target and the projectile must touch to form the compound nucleus. If the reaction energy cannot force the two nuclei together then the reaction will not occur.
So the last problem on the exam is asking for just the center of mass Coulomb barrier, and the Coulomb barrier shouldn't be the same as the normalization, or the energy required for reaction?
DeleteThe question states "Provide the Coulomb barrier and threshold energy in the laboratory frame." The equation given is for center of mass. You can review page 7 of lecture 9
DeleteThis is the last few bullets on page 7 of lecture 9.
Center of mass, need to bring to laboratory frame
Consider kinetic energy carried by projectile
3.36x ((14+4)/14) = 4.32 MeV alpha needed for reaction
I see, thank you.
DeleteExam 2 has been submitted via email.
ReplyDeleteexam 2 submitted
ReplyDeleteExam 2 has been submitted via email.
ReplyDeleteExam 2 submitted on Canvas, thank you for the help!
ReplyDeleteExam 2 has been submitted
ReplyDeleteThanks for the quizzes and comments.
ReplyDeleteThe is an update to the answer key. For question 6 I solved for 232Th instead of 233Th. Also the Vc of the 180Hf reaction is updated.
ReplyDeleteExam 2 resubmission has been sent via email (with attachments)
ReplyDelete