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:

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.

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.

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