Question

Suppose you have an 80mg sample of iodine-125. If I-125 has a half-life of 60 days, how many mg are radioactive after one half-life?-- After 240 days?

Answer 1

Answer: The amount of I-125 after the given time is 5.002 mg.

Step-by-step explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

`t_(1/2)=(0.693)/(k)`

We are given:

`t_(1/2)=60days`

Putting values in above equation, we get:

`k=(0.693)/(60)=0.01155days^(-1)`

Rate law expression for first order kinetics is given by the equation:

`k=(2.303)/(t)\log([A_o])/([A])`

where,

k = rate constant =
`0.01155days^(-1)`

t = time taken for decay process = 240 days

`[A_o]` = initial amount of the reactant = 80 mg

[A] = amount left after decay process = ?

Putting values in above equation, we get:

`0.01155days^(-1)=(2.303)/(240days)\log(80)/([A])`

`[A]=5.002mg`

Hence, the amount of I-125 after the given time is 5.002 mg.