Question

Cobalt 60 is used in many applications where gamma radiation is required. The half-life of cobalt 60 is 5.26 years.
a. Find the decay constant
b. If a sample has an initial activity of 2*10^(15) Bq, what will its activity be after 3 years?

Answer 1

(a) The decay constant

$$\lambda=\frac{\ln 2}{t_{1/2}}=\frac{\ln 2}{5.26\times 365.24\times 24\times 3600 \:\rm{s}}=$$

$$=4.18\times 10^{-9}\:\rm{s^{-1}}$$

(b) Activity

$$A=\lambda N=A_0 2^{-t/t_{1/2}}$$

$$=2\times 10^{15}\:\rm{Bq}\times 2^{-3/5.26}=1.35\times10^{15}\:\rm{Bq}$$