Radiation seed therapy is a common method for treating some prostate cancers and brain tumors. Iodine-125, with a half-life of 60 days, is one isotope used for this treatment. How long does it take for the activity of iodine-125 to decrease to 1.0% of its initial value and essentially cease to be effective?

Question

Quizplus · Accepted Answer

The activity of a radioactive substance is given by the equation:

A = A0 e^(-λt)

Where A is the activity at a given time, A0 is the initial activity, λ is the decay constant, and t is the time.

To find the time it takes for the activity to decrease to 1.0% of its initial value, we can set A to 0.01A0:

0.01A0 = A0 e^(-λt)

Dividing both sides by A0 and taking the natural logarithm of both sides, we get:

ln(0.01) = -λt

Solving for t, we get:

t = ln(0.01)/(-λ)

The half-life of iodine-125 is 60 days, which means that λ = ln(2)/60.

Substituting this into the equation for t, we get:

t = ln(0.01)/(-λ) = ln(0.01)/(-ln(2)/60) = 13 months.

Therefore, the activity of iodine-125 decreases to 1.0% of its initial value and essentially ceases to be effective after 13 months. The answer is B.

Radiation Seed Therapy Is a Common Method for Treating Some