Question

A tumor is injected with 0.4 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

The Time period required for decay of Iodine-125 to half of its value is 60 days .

Explanation:

Given as :

The initial quantity of iodine-125 = 0.4 gram

The rate of decay = 1.15 %

Let The time period for decay = x day

The finial quantity after decay = half of initial quantity

I.e The finial quantity after decay = 0.2 gram

Now ,

The final quantity after decay = initial quantity ×
`(1-(\textrm rate)/(100))^(\textrm Time)`

Or, 0.2 gm = 0.4 gm ×
`(1-(\textrm 1.15)/(100))^(\textrm x)`

or,
`(0.2)/(0.4)` =
`(0.9885)^(x)`

Or, 0.5 =
`(0.9885)^(x)`

Or,
`(0.5)^{(1)/(x)}` = 0.9885

Taking log both side

log (
`(0.5)^{(1)/(x)}` ) = Log 0.9885

or,
`(1)/(x)` × log 0.5 = - 0.0050233

or,
`(1)/(x)` × ( - 0.301029 ) = - 0.0050233

or, x =
`(0.301029)/(0.0050233)`

∴ x = 59.92 ≈ 60 days

Hence The Time period required for decay of Iodine-125 to half of its value is 60 days . Answer