Calculate the activity of nuclides:
Problem:
100 mg piece of a gold ornament containing 10% copper as impurity was irradiated with thermal neutrons at 2.1×1011 n/cm2/s for 2 days producing 198Au (t½ = 2.7 d) and 64Cu (t½ = 12.7 h). Calculate the activity of these nuclides in terms of counts/s at the end of irradiation and after a delay of 2 d and 4 d. How long will it take for the decay of activity due to 64Cu to 0.1% of its original activity? 197Au is monoisotopic and its thermal neutron absorption cross section (σ) is 99 b whereas 63Cu is 30.9% abundant and its σ value is 4.4 b.
Answer: 100 mg piece of gold ornament will contain 90 mg gold or 197Au and 10 mg of copper or 0.309×10 = 3.09 mg 64Cu
Using the Eq. (13.8) Activity due to 198Au at the end of irradiation
= 90×10-3 × 6.02×1023 ×2.1×1011(1- 0.5985)/197
= 22.96×108 cps
Activity after a delay of 2 d = 22.96×108×e(-0.693×2/2.7)
= 13.74×108 cps
Activity after a delay of 4 d = 22.96×108×e(-0.693×4/2.7)
= 8.22×108 cps
Activity due to 64Cu at the end of irradiation = 3.09×10-3×6.02×1023×4.4×10-24\(1-0.073)
= 2.53×107 cps
Activity after a delay of 2 d = 2.53×107×e(0.693×48/12.7)
= 18.4×105 cps
Activity after a delay of 4 d = 2.53×107×e (0.693×96/12.7)
= 13.4×104
Using the formula t = 1/λ lnNo/N
After substitution, time interval = 12.7/0.693 ln 2.53×107×0.1/100
= 185.8 h = 7.74 d