Exam 2
Assigned: 2 March 2017
Due: 7 March 2017
2nd Due date: 10 March
2017
Lecture 4: Alpha Decay
Lecture 5: Beta Decay
Lecture 6: Gamma Decay
Lecture 7: Fission
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. Submission of the work is not required for
the 1st due date. Show your
work or references on a separate page and save electronically for the 2nd
due date.
There will be a skype meeting on the quiz on Tuesday 7 March
at 1000 in the HRC 4th floor conference room.
Please use 3 significant digits for your
answers.
For scientific notation please use X.XXEX (i.e, 1230
as 1.23E3). This can be read by EXCEL as
a number. DO NOT use y.yyx10^y or
y.yy*10^y.
For question 1, is the half-life of 14C 5730y (I'm not sure if "y" means years which is printed on the figure you provided) or is it 5715a as in the Chart of the Nuclides?
ReplyDeleteThanks!
Also, for logf beta- I am getting a negative number, which would not make sense given the formula for logft. So is the Q value provided on the figure (156.475) in MeV, because in the chart of the nuclides it shows it as 0.157MeV.
DeleteFrom page 12 of the table of the isotopes:
DeleteB. Energies:
All energies are given in keV. Level energies are shown in boldface type, and
transition energies in boldface italic type. Level energies are quoted relative to a
constant offset (x, y, z, ..) as x or 0+y, 1576.5+z, etc., when their relationship to the
ground state is unknown. Some g-ray energies are given as X or >0 when the
transition is known to exist but its energy is not known. Systematic level energies are
given in parentheses. Ground state energies are normally written as 0, not 0.0.
Years is commonly abbreviated with Y.
DeleteYou need to calculate the logft value, which includes the half-life. The logf value can be negative.
DeleteThanks!
DeleteFor Question 2.2, Q values for 258No and 260No can be determined. The alpha decay half-lives for these isotopes are likely to be much greater than the SF decay half-life listed for each isotope in the chart of the nuclides.
ReplyDeleteQuestion 1.2: how to determine Transition type from logft value, it is 1+ ?
ReplyDeleteQuestion 2.2: how to determine the alpha decay branching ratio for this isotope in percent?
review the beta decay lecture notes for the transitions
Deletefor the branching, this was covered in lecture 3, page 35. The slide is titled branching decay.
there is 1+ option for question 1.2
DeleteOn #7, for the 180Hf fission, if the 1st product is 99Ru, would the 2nd product be 81Ni? If so, how would we calculate the Q value? Both Q value calculators show no values.
ReplyDeleteYou can calculate this from the mass excess. You should know how to do this from slide 8 of lecture 2.
DeleteI also get an answer using http://nrv.jinr.ru/nrv/webnrv/qcalc/.
I am having some trouble with the first question. For logf I get -6.473 which seems reasonable, but for the logt I'm getting 25.92. Together, I get 19.45 which just seems very high (beyond third forbidden). Do either of these values give you a clue as to what I'm doing wrong?
ReplyDeleteAlso, is there a mistake on the second dropdown on 1.2? It asks for a transition type, but the provided answers seem to be spin and parities? Maybe I'm just misunderstanding the question.
Thank you!
I will correct the pull down in 1.2. that is for a later question. Sorry about the error.
DeleteFor the logt value, the half in seconds is
Delete5730 x 365.24* 24* 3600 = seconds
log of this value is around 11. For logf I get a value close to -2.
make sure your half life is in seconds!!!!
DeleteThank you for the reply! I must be doing something wrong, I will have to work through it again.
DeleteI do have another question though... On 8.3, it's asking which beta transition state is most likely. From what I understand, the 97Zr will most likely beta decay into a metastable 97Nb, which then gamma decays into a stable 97Nb. So for the question, is it referring to the spin and parity of the metastable or stable 97Nb?
Thank you so much for the help!
A change was made to a drop down menu in question 1.2
ReplyDeletethank you!!!!
DeleteI just submitted my quiz via email, thank you!
ReplyDeletethanks for the quiz and previous comments.
Deletehttp://radchem.nevada.edu/docs/course%20reading/TOI.PDF
ReplyDeleteTable of the isotopes link
submitted my quiz..thank you for all the clarification!!
ReplyDeletesubmitted my quiz, thanks
ReplyDeletequiz 2 submitted
ReplyDeleteTest 2 submitted. The comments in this thread really helped clarify many of my problems, thanks!
ReplyDeleteHappy to hear the blog was useful. In the future you could post a comment on a question and basically have all course members assist in answering.
DeleteSubmitted Quiz 2. I had trouble with question 2 because I didn't notice one of the choices for the constants was a negative number.
ReplyDeleteSent Quiz 2. The link helped a lot, as well as the other comments. Thanks.
ReplyDeleteI have submitted my exam
ReplyDeleteI've sent in my quiz 2 to your email. Thank you!
ReplyDeleteSent it quiz...thank you for helping us out this morning
ReplyDeleteJust submitted quiz number 2.
ReplyDeletefor 1.2 transition type for spin and parity...why is the answer 2nd forbidden??
ReplyDeleteFor question 2.1 you have two means to determine the transition type; the logft and change in spin and parity. The spin change from 14C to 14N is 1, with no parity change. Assume the decay is Fermi since A is low. A Fermi transition with a spin change of 1 and no parity change occurs over a 2nd order transition based on the data in the table. The log ft value, of 9, indicates 1st forbidden. The purpose of this question was to demonstrate different routes in evaluating data for beta decay.
DeleteThe actual result indicates a mixed system. From the literature, Study of the Beta-Spectra of C14 and S35, Phy Rev, 74, 1948 (http://journals.aps.org/pr/pdf/10.1103/PhysRev.74.548). The forbidden nature of 14C is discussed. The introduction includes the statement “The half lives are such that the 87 day S35 involves an allowed transition, whereas the 5100-year C14 is at least second forbidden.
thank you..I understand now.
DeleteSent my quiz 2 yesterday. Thanks!
ReplyDeletethanks for the comments and the exams
ReplyDeleteI have submitted my corrections via email, thank you.
ReplyDeleteSent my corrections via email. Thanks so much for answering all my questions!
ReplyDeleteSubmitted the quiz, thank you!
ReplyDeleteCorrected quiz submitted
ReplyDeleteSent corrected quiz 2 yesterday. Thanks
ReplyDeleteSent corrected quiz 2>
ReplyDeletecorrected quiz submitted
ReplyDeleteI wasn't certain if I was supposed to comment on these quizzes, so I figured I would go back and comment just in case. I submitted my quiz and quiz correction to you via email. Thanks!
ReplyDelete