Assigned: 28-March-2017
1st Answers: 2-Apr-2017
2nd Answers: 5-Apr-2017
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. Submission of the work is not required for the 1st due date. Please use 3 significant digits for your answers. For scientific notation please use X.XXEX (i.e, 1230 as 1.23E3).
Lecture 8: Nuclear Structure and Models
Lecture 9: Nuclear Reactions
Lecture 10: Radiation Interaction
Lecture 11: Speciation
Lecture 12: Uranium Chemistry
Lecture 13: Neptunium Chemistry
1st Answers: 2-Apr-2017
2nd Answers: 5-Apr-2017
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. Submission of the work is not required for the 1st due date. Please use 3 significant digits for your answers. For scientific notation please use X.XXEX (i.e, 1230 as 1.23E3).
Lecture 8: Nuclear Structure and Models
Lecture 9: Nuclear Reactions
Lecture 10: Radiation Interaction
Lecture 11: Speciation
Lecture 12: Uranium Chemistry
Lecture 13: Neptunium Chemistry
for question 6, how to calculate the beta and k(sp).
ReplyDeletefor question 1, can you give some ideal for me .
Thanks
There is an example on: https://en.wikipedia.org/wiki/Stability_constants_of_complexes
Deletelook for the section on hydrolysis constants.
submitted
ReplyDeletethanks for the submission
DeletePlease see below some comments on question 6 that may be helpful.
ReplyDeleteou have K from the reaction. logK = -5.64
H2O + PuO2 2+ <----> PuO2OH+ + H+
For simplicity PuO2 2+ = M2+ and PuO2OH+ = MOH+
K = [MOH+][H+]/[M2+], ignore water
Kw = [OH-][H+] =1.01E-14
One needs to solve for the reaction
M2+ + OH- <----> MOH+
B= [MOH+]/[M2+][OH-]
So B and K have a common term, [MOH+]/[M2+]
Substitute K into the equation for B
K/[H+] = [MOH+]/[M2+]
B = [MOH+]/[M2+][OH-] =K/[H+][OH-]
with Kw = [H+][OH-]
B = K/Kw
One now has the value for B
At a selected pH, [OH-] can be found from pH + pOH =pKw
At any given pH one can calculate
[MOH+]/[M2+] = B [OH-]
From the question you have the sum of the relative concentrations is 1, so
[MOH+] + [M2+] = 1
One can use these two equations to solve for the terms [MOH+] and [M2+]
submitted
ReplyDeletethanks for the submission
DeleteJust completed and submitted my exam via email, thank you!
ReplyDeletethanks for the submission
DeleteI just sent in the exam, thanks!
ReplyDeleteQuiz Submitted
ReplyDeleteI submitted quiz 3. Thank you!
ReplyDeletethanks for the exams!
ReplyDeleteHi everyone, I'm having difficulties interpreting question 4. Lecture 9 part 1 mentions the graph but I still can't interpret what is going on. Any help would be very appreciated.
ReplyDeleteThe graph shows cross sections for reaction. The alpha,n means the target has an alpha particle added to it, the it loses a neutron. For the energies select the reaction with the largest cross section.
DeleteFor example, 45 MeV the largest cross section is for alpha, 2pn. The means 55Fe is produced.
Deletecomment on question 1 is below
ReplyDeleteThis questions is described on page 8 in lecture 10.
The equation is:
Eloss = Stopping Power (SP) *thickness (t) *density (r)
The energy loss is (35-5) MeV. Stopping power is provided. The density is given for graphite, you will need to find it for other materials. Solve for thickness.
Done and submitted
ReplyDeleteI complete the exam and submitted it as an attachment via email. Thank you!!
ReplyDeleteQuiz 3 has been submitted!
ReplyDeleteSubmitted!
ReplyDeleteSubmitted
ReplyDeleteHi Ken! Sent in Quiz 3 via email.
ReplyDeleteI submitted my quiz..is there any way you can do a sample calc for question 5..please
ReplyDeleteAn example for question 5 is provided with the reaction of a proton beam on a 208Pb target to produce stable Au
DeleteReaction notation
208Pb(1H, 12Be)197Au
Q value is found from the Q-value calculator from BNL: -8.40 MeV
For threshold energy (see lecture 9 page 5)
T≥ (8.40) ((1+208)/208) = 8.44 MeV. This value is also provided by the Q value calculator.
An example for the Coulomb barrier is provided on lecture 9 page 7. Using the provided equation the center of mass frame Coulomb barrier is: 9.47 MeV.
Converting from center of mass to laboratory frame is done by the ratio (Aprojectile + Atarget)/Atarget
9.47 ((208+1)/208) = 9.52 MeV
The larger value is the laboratory frame Coulomb barrier, 9.52 MeV. This is the necessary reaction energy.
I've sent in my completed quiz, thank you!
ReplyDeleteSubmitted quiz 3. thank you.
ReplyDeleteSubmitted! This is off-topic, but I was wondering if you received my email regarding quiz 2? I haven't heard back yet and didn't know if I should resend it?
ReplyDeleteThanks!
an e-mail was sent to you regarding Exam 2.
DeleteSubmitted
ReplyDeleteHi, can you clarify #5 a bit? After looking at the solutions, my understanding is that the Vc calculation with the conversion to the laboratory frame is the Coulomb barrier. And then the necessary reaction energy is either the Coulomb barrier or threshold energy, just depending on which one is higher. But I'm not sure if that's the correct way to look at it. Thanks!
ReplyDeletethat is the exact way to look at it. The threshold energy may not be sufficient to overcome the coulomb barrier. In this case the coulomb barrier becomes the minimum energy.
DeleteThis comment has been removed by the author.
ReplyDeleteGood afternoon!
ReplyDeleteI'm having a little trouble understanding why the concentrations for question 6 are assigned to the species that you assigned them to in your answer key. Based on what I saw from the sample calculations you showed on this blog, I was under the impression that they would be reversed. Can you please clarify why the concentrations were assigned to the species that you assigned them to?
Thanks,
Halli
Sorry, my error. I will correct and repost. The plutonyl species should decrease with increasing pH.
Deleteupdated exam 3 answers are available.
DeleteCorrected quiz submitted.
ReplyDeletemay I ask just how the coulomb barrier is being calculated? nothing that I try is giving me the correct answer.
ReplyDeleteI resubmitted my test..Vc calculation was a bit confusing please explain that formula on the lecture..9p6,7 was missing info that I did not take account when I calculated the coulomb barrier when I first submitted my quiz.
ReplyDeleteTest 3 re-submitted!
ReplyDeleteI have submitted correction to exam 3. I would like you to explain how the coulomb barrier is different between 5.1 and 5.2 as the equation suggests, they should be the same, regardless of the target. Also, I could not get the answer provided in the answer sheet. I do not understand how you are getting the answers given.
ReplyDeleteExam 3 has ben resubmitted, thanks!
ReplyDeleteI have resubmitted Exam 3 however I also could not reproduce the answers for question five. Will you be explaining this afterwards?
ReplyDeleteQuiz Resubmitted.
ReplyDeleteCorrections have been sent via email. Thanks!
ReplyDeleteI sent in my corrected quiz. Thanks!
ReplyDelete