Iodine-131, t_1/2 = 8.0 days, is used in diagnosis and treatment of thyroid gland diseases. If a laboratory sample of iodine-131 initially emits 9.95 × 10¹⁸ β particles per day, how long will it take for the activity to drop to 6.22 × 10¹⁷ β particles per day?

Question

Quizplus · Accepted Answer

Using the radioactive decay equation, we can set up the following equation:

A = A0 * (1/2)^(t/T)

where A is the final activity (6.22 × 10^17 β particles/day), A0 is the initial activity (9.95 × 10^18 β particles/day), t is the time elapsed, and T is the half-life of iodine-131 (8.0 days).

Plugging in the given values, we get:

6.22 × 10^17 = 9.95 × 10^18 * (1/2)^(t/8)

Taking the natural logarithm of both sides:

ln(6.22 × 10^17) = ln(9.95 × 10^18) - (t/8) * ln(1/2)

Solving for t:

t = 8 * (ln(9.95 × 10^18) - ln(6.22 × 10^17)) / ln(1/2) = 31.9 days

Therefore, the activity will drop to 6.22 × 10^17 β particles/day after about 32 days (choice C).

Iodine-131, T1/2 = 8

Iodine-131, T_1/2 = 8