1 / 19

Particle Emission Probabilities

Particle Emission Probabilities Edgardo Browne Decay Data Evaluation Project Workshop May 12 – 14, 2008 Bucharest, Romania. Particle Emission Probabilities. Directly measured Deduced from g -ray intensity balance. 240 Pu Alpha Spectrum I 2 I 1.

ludwig
Télécharger la présentation

Particle Emission Probabilities

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Particle Emission ProbabilitiesEdgardo BrowneDecay Data Evaluation Project WorkshopMay 12 – 14, 2008Bucharest, Romania

  2. Particle Emission Probabilities • Directly measured • Deduced from g-ray intensity balance

  3. 240Pu Alpha SpectrumI2 I1

  4. The a-particle emission probabilities are: p1(%)= I1 x 100/(I1 + I2), and p2(%)= I2 x 100/(I1 + I2) I1 and I2 are spectral areas

  5. Uncertainties assuming uncorrelated spectral areas (I1, I2) and small dI1 and dI2 values. p1(%)= I1 x 100/(I1 + I2) p2(%)= I2 x 100/(I1 + I2) dp12=(dp1/dI1)2 dI12 + (dp1/dI2)2 dI22 dp22=(dp2/dI1)2 dI12 + (dp2/dI2)2 dI22 = 1002 x [(I12 dI22 + I22 dI12 )] 2 (I1 + I2 ) Notice that if I1 = I2 = I, and dI1 = dI2, then dp1/p1 = dp2/p2 = (2)1/2/2 dI/I Surprising??

  6. Of course NOT! p1 and p2 are correlated! p1= I1 x 100/(I1 + I2) = 100 x 1/(1 + I2/ I1) p2= I2 x 100/(I1 + I2) = 100 x 1/(1 + I1/ I2)

  7. Back to 240Pu

  8. dp12 = dp22 = 1002 x [I12 dI22 + I22 dI12], so (I1+I2)4 dp12 = dp22 = [73.512 x 0.212 + 26.392 + 0.362] = 328.56 1002 1002 dp1 = dp2= (328.56)1/2 = 0.18 100 Finally, p1 = 73.51 + 0.18 % p2 = 26.39 + 0.18 %

  9. General CaseNucl. Instr. Meth. In Phys. Res. A265, 541 (1988)

  10. Particle Emission Probabilities Deduced from Decay Scheme

  11. The uncertainties become

  12. If p1(%)=0, then T1 = T2 + T5

  13. 133Ba EC Decay to 133Cs

  14. Normalization factor N=100/(T81 + T161 + T384) = 1.0044 The uncertainties become: de384 = 0.74, de437 = 2.0 (Notice that they are not equal). e437 = 86 + 2 % and e384 = 14.00 + 0.74 % 133Ba EC Decay to 133Cs

  15. 133Ba EC Decay to 133Cs If we assume e81 = e162 = 0%, then N = 100/(T356+T276+T384+T223) = 1.00090, and e437 = 86.00 + 0.67 %, and e384 = 14.00 + 0.67 % For the general case see Nucl. Instr. Meth. In Phys. Res. A265, 541 (1988)

  16. 145Ba Total Absorption g-Ray Spectrum (1997Gr09)

  17. 145Ba Decay Scheme Based on Total Absorption g-Ray Spectrum1997Gr09 145Ba b- 3900 keV 68% 3000 keV 2566 keV 25% 7.0+2.2% 0 keV 145La

  18. Thank you!

More Related