EXAM 1 is now available
Assigned: 21-Jan-16
1st Due Date: 26-Jan-16
2nd Due Date: 29-Jan-16
The first iteration of the exam is due 26-Jan-16. The answers will be posted on 27-Jan-16. Any incorrect answers can be resubmitted by 29-Jan-16. 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.
There will be office hours on Friday 22 January and Tuesday 26 January. This quiz covers:
Lecture 1: Introduction, Chart of the Nuclides
Lecture 2: Nuclear Properties
Lecture 3: Decay Kinetics
Lecture 4: Alpha Decay
Lecture 5: Beta Decay
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.
I am having a problem with downloading the ERG Calculator needed for question 3. I went on their website and downloaded their Nuclide Decay program, but the program won't install correctly.
ReplyDeleteIs there a similar program or calculation I can use for this problem?
You can use the Bateman equation, but it would be better to use the program. I will send you a copy. If the error persists let me know. We can also review this question in office hours on Tuesday.
DeleteProfessor, I am having similar issues. Can you send me a copy as well?
ReplyDeleteFor ERG program the file is places at:
Deletec/program files (x86)
the file name is ERG
then make a short cut or link
For Lecture 3, decay kinetics: On slide 41, why do you divide by the molecular mass of iron if the isotope you have is iron-58? Isn't the molecular mass of iron based on iron-56?
ReplyDeleteYou start with 1 gram of natural Fe. I use the Fe mass to find the total amount of Fe atoms. From there I know that 0.282 % of those atoms are 58Fe. If I had 1 g of 58Fe to start with, then I would use the 58Fe mass to find the number of atoms.
DeleteI understand. So, we can't infer that what we start with is pure 58Fe. Thanks.
DeleteCorrect. We start with 1 g of natural Fe.
DeleteI sent exam 1 as an attachment through email.
ReplyDeleteThank you for having a office hour for discussing questions in exam.
thanks for your exam!
DeleteExam #1 is sent through email. Thank you!
ReplyDeletethanks for the exam!
DeleteExam 1 sent through the Canvas email!
ReplyDeleteexam received
DeleteEmailed Exam/Quiz1
ReplyDelete-Faruq
Resubmitted Exam #1 via email. Most of my values are in solid agreement with your. Not identical but close, very close figures.
DeleteThanks for the opportunity to correct our exams and resubmit.
-Faruq
exam 1 send via email -Raghav
ReplyDeletethanks for exam 1
DeleteI have emailed you Exam#1.
ReplyDeletethe exam was received, thanks!
DeleteExam 1 sent through Canvas.
ReplyDeletethanks for the exam!
DeleteExam 1 sent via email!
ReplyDeletethanks for the exam
DeleteLecture 4, Slide 19, Proton decay: You say the proton is barely held by the nucleus and the atom is proton rich and neutron poor. Later on, it says the barriers are high for the reaction to happen. Can you explain why barriers would be high if we already are proton rich?
ReplyDeleteGreat question. I should state that this is in relative terms. For the protoN emission the barrier is relatively low for the proton in neutron deficient nuclei compared to other nuclide. However the proton emission generally must be competitive with positron decay or electron capture. In this case the relative probability for proton emission is low. The proton needs to penetrate the nuclear well. Positron or electron capture have a different decay route, which is often more favorable that the probability of a proton penerating the barrier. I will clarify this point in future lectures. Thanks!
DeleteGreat question. I should state that this is in relative terms. For the protoN emission the barrier is relatively low for the proton in neutron deficient nuclei compared to other nuclide. However the proton emission generally must be competitive with positron decay or electron capture. In this case the relative probability for proton emission is low. The proton needs to penetrate the nuclear well. Positron or electron capture have a different decay route, which is often more favorable that the probability of a proton penerating the barrier. I will clarify this point in future lectures. Thanks!
DeleteMakes sense, thank you.
DeleteProfessor,
ReplyDeleteFor number 2.1, 2.2, and 2.3, I am not sure where those values came from. When I did that problem, I was looking at Co60 and saw the sigma subscript gamma values of 2.0, 4 and so I chose 2.0 barns for question 2.1, 60 barns for question 2.2, and 230 barns for question 2.3.
This is explained in the 2nd part of lecture 1 in the overview of the chart of the nuclides.
DeleteThe key component here is to understand the relevant isotope and the data. The cross section date is for a reaction on that isotope. If one asks for the thermal neutron capture cross section for 60Co, the relevant reaction is:
59Co + n -->60Co
The cross section data for this reaction is in the 59Co box. Now if one looks at the 59Co cross section data, you have 4 choices; thermal neutron to isomeric state, thermal neutron to ground state, resonance neutron to isomeric state, and resonance neutron to ground state. Page 44 of the chart of the nuclides has the examples.
The thermal neutron cross section data are in the 1st parentheses, with the ground state the 2nd value. This is 16 barns.
For the thermal neutron reaction
n + 59Co --> 60mCo
The relevant cross section is 21 barns.
Please let me know if that helps.
Oversight on my part, I apologize. Yes, that does help, thank you.
DeleteExcellent, glad this helps
DeleteA general comment on question 10.
ReplyDeleteYou need to equate 2 rate equations. One is
A=lambda * N
The other is
R=cross section * N * flux
The decay constant is from the half life. The cross section is from the chart of the nuclides. Remember that the question asks for 10 times the rate.
Dr. Czerwinski,
ReplyDeleteFor number 4, are we supposed to take the Hatsukawa equation, find C(Z,N) and X, then solve for A(Z) and plug that into the Geiger-Nuttall equation for A and then solve for B?
You have the data for log t1/2 and 1/sqrtQ.
Deleteplot the as y= mx +b, where y is log t1/2 and x is 1/sqrtQ. This provides the slope and intercept, the values you need for A and B. Once you have this you have the equation you need.
Let me know if this helps and if you need more time
You have the data for log t1/2 and 1/sqrtQ.
Deleteplot the as y= mx +b, where y is log t1/2 and x is 1/sqrtQ. This provides the slope and intercept, the values you need for A and B. Once you have this you have the equation you need.
Let me know if this helps and if you need more time
Thank you very much! That clears up a lot! It's also much easier than what I was trying to do.
DeleteGreat to hear.
DeleteGreat to hear.
DeleteSubmitted the Exam!
DeleteOn Canvas, I sent you an email containing my exam re-submission and work. Thank you!
ReplyDeleteI have emailed you my re submission with a calculation sheet, and excel graph for problem 4.
ReplyDeleteI sent my correction of exam 1 as an attachment through email. Thank you very much.
ReplyDeleteSo Sorry for the late post. I submitted my correction for exam #1 on Jan 30th but I forgot to post on here. I have sent you an email with my exam #1 resubmission, work, and excel file. Thank you!
ReplyDelete