How long will it take a Co-60 source to decrease to 18% of its original activity?

To solve this problem, we need to use the formula for radioactive decay, which is:

N = N0 * e^(-λt)

where: N is the final quantity of the substance, N0 is the initial quantity of the substance, λ is the decay constant, t is the time.

We know that N/N0 = 0.18 (because we want to find out when the substance has decreased to 18% of its original activity). We also know that the half-life of Co-60 is 5.27 years. The decay constant λ can be found using the formula:

λ = ln(2) / half-life

Substituting the half-life of Co-60 into this formula gives:

λ = ln(2) / 5.27 ≈ 0.131 per year

We can now substitute N/N0 and λ into the decay formula and solve for t:

0.18 = e^(-0.131t)

Taking the natural logarithm of both sides gives:

ln(0.18) = -0.131t

Solving for t gives:

t = ln(0.18) / -0.131 ≈ 16.6 years

So it will take approximately 16.6 years for a Co-60 source to decrease to 18% of its original activity.