Question

8. A fresh spent nuclear fuel was removed from a reactor core and placed into a spent fuel pool. One of the activation products in the nuclear fuel was Cobalt (Co-60). After 8 year of storage in the pools, the activity of the Co-60 was reduce to 0.5x105 Ci. Please determines: i)The initial activity of the Co-60 at the day when the spent nuclear fuel was stored in the pool. [Given that half-life, t1/2 of the Co-60 is 5.26 yr] ii) After 20 years of storage in the pool, the spent nuclear fuel is plan to be transferred into an interim dry cask storage. Assuming that the dose rate generated from the Co-60 without shielding is 10 Sv/hr, what is the thickness of the lead shielding needed to reduce the dose rate to 1HSv/hr? [Given that attenuation coefficient, u for lead is 1.23 cm]

Answer 1

(i) A = A_0 e^-lambda t lambda = 0.693/t_1/2 0.5 times 10^5 = A_0 e^- 0.693/5.26 times 8 A_0 = e^- 0.693/5.26 times 8 A_0 = 0.5 times 10^5 times 2.8690 A_0 = 1.434 times 10^5 ci (ii) D = D_0 e^-mu t D = 10^-6 sv/h and D = 10 sv/h 10 = e^+ 1.23 t times 10^-6 10^7 = e^-1.23 t implies ln 10^7 = 1.23 times t t = 13.10 cm