Question

Lridium-192 is an isotope of iridium and has a half-life of 73.83 days. if a laboratory experiment begins with 100 grams of iridium-192, the number of grams, a, of iridium-192 present after t days would be a = 100 1 2       t 73.83 . which equation approximates the amount of iridium-192 present after t days?

Answer 1

Radioactive material undergoes first order dissociation kinetics.

For 1st order system,
k = 0.693 / t1/2
where, t 1/2 = half-life of the radioactive disintegration process.

Given that, t 1/2  = 73.83 days
Therefore, k = 0.009386 day-1

Also, for 1st order reaction,
k = $\frac{2.303}{t} log \frac{Co}{Ct}$

Given that, Co = initial concentration of Iridium-192 = 100 g

Therefore, 0.009386 = $\frac{2.303}{t} log \frac{100}{Ct}$

On rearranging we get, Ct = 100 $(0.990656)^{t}$

Answer: Ct = 100 $(0.990656)^{t}$ equation approximates the amount of Iridium-192 present after t days