Small beads of iridium-192 are sealed in a plastic tube and inserted through a needle into breast tumors. If an Ir-192 sample has an initial activity of 560 dpm, how much time is required for the activity to drop to 35 dpm (t1/2 = 74 days) ?

296.1 day.

Explanation:

The decay of radioactive elements obeys first-order kinetics. For a first-order reaction: k = ln2 / (t1/2) = 0.693 / (t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2 / (t1/2) = 0.693 / (74.0 days) = 9.365 x 10⁻³ day⁻¹.

For first-order reaction: kt = lna / (a-x).

where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).

t is the time of the reaction (t = ? day).

a is the initial concentration of Ir-192 (a = 560.0 dpm).

(a-x) is the remaining concentration of Ir-192 (a - x = 35.0 dpm).

∴ kt = lna / (a-x)

(9.365 x 10⁻³ day⁻¹) (t) = ln (560.0 dpm) / (35.0 dpm).

(9.365 x 10⁻³ day⁻¹) (t) = 2.773.

∴ t = (2.773) / (9.365 x 10⁻³ day⁻¹) = 296.1 day.