Question

A 20 g sample of cobalt is irradiated in a power reactor with a flux of 10¹⁴ n/cm²/s for 6 years. Calculate (a) the activity of ^60mCo immediately upon removal from the reactor and (b) the activity of ⁶⁰Co 50 hours after removal. Assume no source depletion.

The answer for Part (a) = 11,600Ci ^60mCo

The answer for Part (b) = 11,154Ci ⁶⁰Co

Please show all steps and give a good explanation.

Answer 1

a) The activity of the sample after a time t of irradiation is given by

Where   is the decay constant for the product = 0.131449 per year

is the flux = 10¹⁴ neutrons /cm² s

N₀ is the number of target atoms = 6.022*10²³ * 20/60 = 2.00733*10²³

is the absorption cross section = 37.2 barns.

So,

b) Activity after a time t is given by

(divide by 8760 to convert to hours)

The answer given in the book is wrong. Even if we put the value of answer from part A to part B, it will not get the provided answer.