Part A Neutron activation analysis for a sample of a rock revealed the presence
ID: 785102 • Letter: P
Question
Part A Neutron activation analysis for a sample of a rock revealed the presence of 5926Fe ,which has a half-life of 46.3days . Assuming the isotope was freshly separated from its decay products, what is the mass of 5926Fe in a sample emitting 1.00 mCi of radiation?Part A Neutron activation analysis for a sample of a rock revealed the presence of 5926Fe ,which has a half-life of 46.3days . Assuming the isotope was freshly separated from its decay products, what is the mass of 5926Fe in a sample emitting 1.00 mCi of radiation?
Explanation / Answer
Radioactivity (Bq) R = (m/ma)*Na*ln(2)/t(1/2)
ma =59.26 grams/mol
Na=6.023*10^23 is the Avogadro Number
1 Ci = 37Giga Bq
Therefore
m = R*ma*t(1/2)/Na/ln(2) = (10^-3*37*10^9)*59.26*(46.3*24*3600)/(6.023*10^23)/ln(2) =2.1*10^-8 kg =0.0021 mg of (59)Fe isotope