Question

Co-60 is used extensively in radiotherapy as a source of \(\gamma\) -rays. Its half-life is 5.27 years. (a) How many years will it take to decrease to \(30.0 \%\) of its original activity? (b) If a cobalt source is ten years old, what percent of its original activity remains?

Short answer

(Given half-life of Co-60 is 5.27 years) Answer: It will take approximately 8.38 years for the Co-60 source to decrease to 30.0% of its original activity, and after 10 years, about 26.9% of the original activity of the Co-60 source remains.

Step-by-step solution

Calculate the decay constant#for Co-60

To calculate the decay constant we can use the formula: \(\lambda = \frac{ln(2)}{T_{1/2}}\) Substitute the given half-life (\(T_{1/2} = 5.27 years\)): \(\lambda = \frac{ln(2)}{5.27} \approx 0.131\) year\(^{-1}\)

Calculate the time required for the Co-60 source to decrease to 30.0% of its original activity

To solve for the time, \(t\), we can re-arrange the radioactive decay equation as follows: \(t = \frac{ln(\frac{N(t)}{N_0})}{-\lambda}\) We are given that \(N(t) = 0.30 N_0\), so we have: \(t = \frac{ln(0.30)}{-0.131} \approx 8.38\) years It will take approximately 8.38 years for the Co-60 source to decrease to 30.0% of its original activity.

Calculate the percentage of activity remaining after 10 years

Given that the cobalt source is 10 years old, we need to find the percentage of its original activity remaining. We can plug this time into the decay equation: \(N(10) = N_0e^{-0.131(10)} = N_0e^{-1.31}\) Divide both sides by \(N_0\): \(\frac{N(10)}{N_0} = e^{-1.31}\) Convert this fraction into a percentage: \(\frac{N(10)}{N_0} * 100\% \approx 26.9\%\) After 10 years, approximately 26.9% of the original activity of the Co-60 source remains.