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 =
`(2.303)/(t) log (Co)/(Ct)`

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

Therefore, 0.009386 =
`(2.303)/(t) log (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