Question

A tumor is injected with 0.1 grams of Iodine-125, which has a decay rate of 1.15% per day. To the nearest day, how long will it take for half of the Iodine-125 to decay?

Answer 1

It will take 60 days for half of the Iodine-125 to decay

Explanation:

The amount of iodine-125 after t days is given by the following equation:

`Q(t) = Q(0)(1-r)^(t)`

In which Q(0) is the initial amount and r is decay rate, as a decimal.

Decay rate of 1.15% per day.

This means that
`r = 0.0115`

So

`Q(t) = Q(0)(1-r)^(t)`

`Q(t) = Q(0)(1-0.0115)^(t)`

`Q(t) = Q(0)(0.9885)^(t)`

To the nearest day, how long will it take for half of the Iodine-125 to decay?

This is t for which
`Q(t) = 0.5Q(0)`

So

`Q(t) = Q(0)(0.9885)^(t)`

`0.5Q(0) = Q(0)(0.9885)^(t)`

`(0.9885)^(t) = 0.5`

`\log{(0.9885)^(t)} = \log{0.5}`

`t\log{0.9885} = \log{0.5}`

`t = \frac{\log{0.5}}{\log{0.9885}}`

`t = 59.93`

To the nearest day

It will take 60 days for half of the Iodine-125 to decay