Trace Element Analysis Trace Element Analysis of Geological, Biological & of Geological, Biological & Environmental Materials Environmental Materials By Neutron Activation Analysis: By Neutron Activation Analysis:An Exposure An ExposureILA PILLALAMARRI ILA PILLALAMARRIEarth Atmospheric & Planetary Sciences Earth Atmospheric & Planetary SciencesNeutron Activation Analysis Laboratory Neutron Activation Analysis LaboratoryMassachusetts Institute of Technology Massachusetts Institute of TechnologyCambridge, MA 02139 Cambridge, MA 02139IAP 12.091 Session 5, January 19, 2005 IAP 12.091 Session 5, January 19, 2005Session 5  12.091 Course – Students’ individual presentations  12.091 Assignment Review  Neutron Activation Analysis – Trace Element Geochemistry – Examples of applications Mineralogy Petrology Examples of studies done using MIT-EAPS INAA Laboratory  Conclusion – Advantages of Neutron Activation Analysis IAP 12.091 Session 5, January 19, 2005 212.091 Course Student Presentations Wen-Fai Fong: Delayed Neutron Activation Analysis Martin D. Lyttle: Epithermal Neutron Activation Analysis Ian Garrick-Bethell: Prompt Gamma Neutron Activation Analysis IAP 12.091 Session 5, January 19, 2005 3 (see Projects section)Activity Equation Review A = number of decays per second (Activity) dps N = number of atoms of the target isotope = mx θ x 6.023 x 1023 tW m = mass of the element in the irradiated sample g θ = isotopic abundance w = Atomic weight of the element λ = decay constant = 0.693/t1/2 1/2 = Half-life of the isotope φ = neutron flux n.cm-2 .sec-1 σ = activation cross-section 10-24 cm2 IAP 12.091 Session 5, January 19, 2005 4Activation Equation Review … tirr = irradiation time A = N σ φ [ 1 - exp(-λtirr) ] After a delay of time td A = N σ φ [ 1 - exp(-λtirr) ]exp(-λtd) For a counting time of tc A = N σ φ [1 - exp(-λtirr)]exp(-λtd) [1 - exp(-λtc)] IAP 12.091 Session 5, January 19, 2005 512.091 Assignment 1)In a house in Cambridge, the water from the faucet suddenly started showing some particulate matter, which is suspected to be copper from a pipe. It was brought to the MIT Reactor for analysis. You are asked to calculate the activity that would be produced by thermal neutron activation, if 1 gram of copper is irradiated in the reactor flux of 4 x 1012 n.cm-2.sec 1 for 2 hours. i)On removal what is the activity of each copper isotope? ii)What will be the activity of each isotope 1 hour after the removal from the reactor? The answer should contain: the activity equation, the parameters and values used, and the activity calculated. Suggested Use: 1) The Chart of Nuclide Handout, 2) Table of Nuclides Appendix D, p 606 -650, Gamma-ray sources Appendix E 651-660, Nuclear and Radiochemistry by G. Friedlander, J. Kennedy, E. S. Macias, J. M. Miller Answer: The stable isotopes of copper are 63Cu and 65Cu. Atomic weight of Cu = 63.546, flux = 4E12 n/cm2.sec 63Cu (n,γ) 64Cu and 65Cu (n,γ) 66Cu are the thermal neutron activation reactions. 65Cu (n,γ) 66Cu: σ = 2.17E-24 cm2 , θ = 30.83%, 66Cu half life = 5.10 m, The activity of 66Cu upon removal = 0.69 curies and after 1 hour 0.197 mCi 63Cu (n,γ) 64Cu: σ = 4.4E-24 cm2 , θ = 69.17%, 64Cu half life = 12.75 h, The activity of 66Cu upon removal = 0.321 curies and after 1 hour 0.304 mCi IAP 12.091 Session 5, January 19, 2005 6 -12.091 Assignment 2)An entrepreneur wants to know whether a particular area of interest has Molybdenum and Antimony. So what are the radioisotopes that can be used for the thermal neutron activation analysis. Provide all the relevant information of the X (n,γ)Y reaction, identify the parent and daughter nuclei, the activation cross section, the half-life of the daughter product, and the predominant gamma-ray energy for identification. Suggested Use: Table of Nuclides Appendix D, p 606 -650, Gamma-ray sources Appendix E 651-660, Nuclear and Radiochemistry by G. Friedlander, J. Kennedy, E. S. Macias, J. M. Miller Answer: The stable isotopes of Molybdenum are 92Mo, 94Mo through 98Mo, and 100Mo. The thermal neutron activation reactions are: 92Mo (n,γ) 93Mo, σ = 0.006E-24 cm2 , θ = 14.8%, 93mMo half life = 6.9 h 98Mo (n,γ) 99Mo, σ = 0.131E-24 cm2 , θ = 24.13%, 99Mo half life = 65.94 h 100Mo (n,γ) 101Mo, σ = 0.200E-24 cm2 , θ = 9.63%, 101Mo half life = 14.6 m 98Mo (n,γ) 99Mo can be selected and the identifying gamma-ray energy is 739.4 keV IAP 12.091 Session 5, January 19, 2005 712.091 Assignment Answer 2 continued. The stable isotopes of Antimony are 121Sb and 123Sb The thermal neutron activation reactions are: 121Sb (n,γ) 122Sb, σ = 6.33E-24 cm2 , θ = 57.3%, 122Sb half-life = 2.72 d 121Sb (n,γ) 122mSb, σ = 0.05E-24 cm2 , θ = 57.3%, 122mSb half-life = 4.21 m 123Sb (n,γ) 124Sb, σ = 4.08E-24 cm2 , θ = 42.7%, 124Sb half-life = 60.2 d 123Sb (n,γ) 124mSb, σ = 0.03E-24 cm2 , θ = 42.7%, 124mSb half-life = 93 s 121Sb (n,γ) 122Sb can be chosen for short irradiation time, the identifying gamma-ray energy is 564 keV. 123Sb (n,γ) 124Sb can be chosen for longer irradiation time, the identifying gamma-ray energy is 603 keV. Note: The meta stable isotopes are short lived. IAP 12.091 Session 5, January 19, 2005 812.091 Assignment 3)An unknown sample powder was found in an envelope. It was brought to the reactor for analysis. The gamma spectrum revealed significant gamma-ray peaks of energy 320 KeV, 1368 keV and 2754 keV. Identify the content of the powder. Suggested use: Appendix 5, Table 2 Neutron Activation Analysis By D. De Soete, R. Gijbels, J. Hoste Answer: 50Cr (n,γ) 51Cr; 320 kev gamma-ray is emitted by 51Cr 23Na (n,γ) 24Na; 1368 kev and 2754 keV gamma-rays are emitted by 24Na. So the powder contains Sodium and Chromium. IAP 12.091 Session 5, January 19, 2005 912.091 Assignment 4)The weights of empty vial, empty vial + sample powder were taken 6 times . Write the formula for the propagation of errors, calculate the error in the weight of the sample powder. Interpret the results. Weights (in grams) of the empty vial, weighed separately for 6 times: 1.14470, 1.14475,1.14472, 1.14476, 1.14478, 1.14475 Weights (in grams) of the vial + sample powder, weighed separately for 6 times: 1.35041, 1.35040, 1.35029, 1.35018, 1.35026, 1.35035 Answer: Weights (in grams) of the empty vial, weighed separately for 6 times: 1.14470, 1.14475,1.14472, 1.14476, 1.14478, 1.14475 Average weight = 1.14474 Standard Deviation = 0.000026 Precision =