Quiz 2 is posted. The quiz covers:
Lecture 5: Beta Decay
Lecture 6: Gamma Decay
Lecture 7: Fission
Lecture 8: Nuclear Models
Use lecture notes, textbooks, Chart of the Nuclides, Table of the Isotopes, and web pages. Show your work or reference and send in as a separate document. Quiz 2 is assigned on 26-Jun-15. The 1st due date is 1-Jul-15. The 2nd submission date 6-Jul-15. Please post any questions to the blog.
Answers to Quiz 2 now posted. The 2nd submission is 7 July. If this date needs to be changed please comment on the blog.
ReplyDeleteOn question 8 it is a fairly simple calculation and I've double checked my work there's only a couple discrepancies on the chart. Is there any way to see a couple of your calculations? I don't understand what I'm doing wrong
ReplyDeleteExam for the 127I
ReplyDeleteusing Q value calculator http://www.nndc.bnl.gov/qcalc/
Target 127I
Ejectile 64Cr
10.0 MeV for Q value
From inspection the fission product A=63 (A total 127)
Z I = 53, Z Cr = 24
53-24 =29, which is Cu
Other fission product 63Cu
For Vc
(29*24)*1.44/(1.8((64^0.333)+(63^0.333)))
=69.8 MeV
Coulomb barrier greater than Q value, no fission
let me know if you have other questions or want a specific example.
For question 8 there were some discrepancies which are now corrected. For the key question 8 used http://nrv.jinr.ru/nrv/webnrv/qcalc/.
ReplyDeleteFor question 5, for Cs137, you have the correct answer as 9.6! I've done this calculation number times and I'm getting about 11. I even practiced the example problem given on slide 19 in lecture 5 and I got the correct answer so I know I'm doing the calculation right. Could you maybe show us your calculations for this?
ReplyDeleteI have no doubt your calculation is correct. However one does not need to do a calculation for this. Looking at the 137Cs decay scheme in question 5, the most likely decay, 94.4 %, goes to the 11/2- excited state in 137Ba. If you follow the line for this level you have 94.4 % (the transition %), 9.6 (log ft), 11/2- (spin and parity), the decay information, including mulitpole radiation, 661.660 (energy above the ground state in keV), and 2.552 m (excited state half life).
ReplyDeleteI just took the log ft value from the presented data.
please let me know if you have any additional questions.
should we not get the same answer though?
DeleteThe log ft calculation is an estimate. Also which states and energy did you use? You will need to account for the excited state. If you do this you should be close. I would not expect the values to be exact.
DeleteThe answers and comments to quiz 2 are posted.
ReplyDelete