Monday, January 18, 2016
CHEM 418 Nuclear Chemistry, Winter 2016: Lecture 5 Beta Decay
Beta decay is presented in this lecture. The neutrino hypothesis and its relationship with beta decay is discussed. A review of Q value calculations for beta decay is provided. The importance of spin and parity, and how it can be used to assess beta decay, is discussed. Modeling beta decay through the weak force is provided. The impact of Coulomb interactions on positron and electron spectral shape is presented. The use of Kurie plots in understanding beta decay is introduced. Selection rules in beta decay and beta transitions are explained. Calculating logft and its relation to spin and parity are presented. Double beta decay is discussed.
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ReplyDeleteCalculating log(ft) values... I'm having some troubles.
ReplyDeleteSo for beta decay we may use the following formula.
log(f) = 4*log(Qrxn) + 0.78 +(0.02)*Zdaughter- 0.005*(Zdaughter - 1)*log(Qrxn)
Qrxn = delta(241-Pu) - delta(241-Am)
Qrxn > 0
Qrxn = 0.0208 MeV
Let Zdaughter = 95
and Zdaughter -1 = 94
then falling from the calculation of log(ft)
log(ft) = -10.9910... [MeV units?]
This value must be wrong. Can someone point me in the right direction?
Thanks!
-Faruq
So if I push through on the calculation of log(ft)
Deletelog(ft) = log(f) + log(t1/2)
t1/2 = 14.290*(3.157*10^7) [seconds]
log(f) = -10.9910 [? probably time]
log(ft) = -10.9910 + + log(14.290*(3.157*10^7))
log(ft) = 8.93629
6 < log(ft) < 10
and is therefor the first Forbidden Transition.
How wrong is this answer?
A great way to check your answer for the log ft data is the table of the isotope. For 241Pu, go to A=241. The beta decay of 241Pu goes to 241Am. If you look at this level you see 241Pu goes to the 241Am ground state. The logft is 5.8. You number should be closer to this.
Deletelog f (B-) = 4logE+0.78+0.02Z-0.005(Z-1)logE
DeleteE=0.020, Z=95 (the Z for Am), you have 94. I would try this again. The half life in seconds is 4.53E8, very close to yours.
Thanks Ken, I'll swing it again
Deleteexcellent. let me know if you need any further assistance.
DeleteSo I was able to get all the questions except for the log(ft) value for Ti's beta decay. I have trouble making calculations on this beta decay reaction because Q is quite negative...
Delete45-Ti --> 45-V + e + v + Q
Q < 0
45-Ti = -39.009121 MeV
45-V = -31.880549 MeV
Mass Difference: Q = (45-Ti) - (45-V)
Q = -7.12857 MeV
Use the chart of the nuclides to determine the decay route of 45Ti. From the chart of the nuclides 45Ti decays by positron emission to 45Sc. Find the Q value for this reaction, use the Z of Sc, and use the log f equation for positron decay. You should get a logft around 5.
DeleteUse the chart of the nuclides to determine the decay route of 45Ti. From the chart of the nuclides 45Ti decays by positron emission to 45Sc. Find the Q value for this reaction, use the Z of Sc, and use the log f equation for positron decay. You should get a logft around 5.
Deleteemailed quiz 5, sorry im late
DeleteSubmitted the quiz via email.
ReplyDeleteI was a little confused on the 4 problems the neutrino fixed for beta decay. Antiparticle, energy conservation, and spin conservation are clear, but is the zero net charge supposed to be the 4th? That doesn't seem like it "fixes" a problem, it's more that the neutrino doesn't create any additional problems, if my logic makes sense. I listed it anyway, since I'm likely overthinking it.
Thanks!
-Taylor
thanks for the comment, definitely thoughtful.
DeleteFor the neutrino the lack of a charge is one of the reasons it is difficult to observe. Furthermore, a charge, or lack of one, needs to be assigned to the particle. The charge balance of the other particle means tne neutrino mush have no charge.
Lecture viewed and quiz submitted.
ReplyDeleteI was confused when working on question 1, more specifically with the derivation of log ft. Why is it that we use the half life of the parent for log t, but the Z of the daughter for solving logf?
the f value can be thought of as half life corrected for differences in Z and W of the decay. W is the total kinetic energy, which is Q. Z is based on the perturbation from the Fermi function. This is the transition from the parent to the daughter,so the daughter Z is needed. For positron decay the Z is one less than the parent. For beta minus decay the daughter Z is greater than the parent. It is for this reason the Z of the daughter needs to be described.
DeleteI sent the lecture 5 pdf quiz as an attachment through email.
ReplyDeleteI have a question about calculating the logft value. When I looked the Q value for the Ti45, I found out the Q value of Ti45 for EC is twice the Q value of poistron decay. In addition, when I looked the table isotope, I found the Q value of EC right below the nulcide. Why we find the log ft value of positron decay instead of EC?
Thank you very much
This is described on page 2 of the beta decay lecture. The positron decay needs to include 2 electron masses. This due to the positron and the extra electron in the daughter. Since the daughter is one less proton than the parent, it will have an extra electron that needs to be considered.
DeleteThis is also explained in detail on page 6 of lecture 2.
DeleteSo higher Q value does't mean it will be the primary decay mode? The primary decay has to be followed the chart of nulcide?
Deletethat is correct, especially comparing Q values for different decay modes.
DeleteLecture watched. Quiz submitted through email. Thank you!
ReplyDeleteQuiz submitted through the Canvas email!
ReplyDeleteQuiz has been submitted via Canvas.
ReplyDeletepdf quiz received, thanks!
Deletequiz was submitted via email
ReplyDeleteHave the PDF quiz, thanks!
DeleteI have emailed you PDF quiz 5.
ReplyDeleteThanks for the pdf quiz!
DeleteLecture 5 viewed and pdf quiz emailed! (....late....)
ReplyDelete