Question

How long will it take for 3/4 of the sample of 131 iodine that has half-life of 8.1 days?

Answer 1

Since you didn't specify whether 3/4 of the sample remains or undergoes decay, I'll show you both cases.

Here's the equation for exponential decay used in nuclear half-life calculations

`A(t) = A_0 * (1/2)^(t/t_("1/2"))`, where

`A(t)` - the amount left after t years;
`A_0` - the initial quantity of the substance that will undergo decay;
`t_("1/2")` - the half-life of the decaying quantity.

First case - 3/4 of the sample undergoes radioactive decay.

If 3/4 of the sample undergoes radioactive decay, you wil be left with 1/4 of the original sample. This means that `A(t)` will be equal to `A_0(t) * 1/4`. Plug this into the above equation and you'll get

`A_0(t) * 1/4 = A_0(t) * (1/2)^(t/t_("1/2"))`, or `1/4 = (1/2)^(t/t_("1/2"))`

This implies that `t/t_("1/2") = 2`, since `1/4 = (1/2)^2`.

Therefore, `t = 2 * t_("1/2") = 2 * 8.1 = "16.2 days"`

It will take 16.2 days for 3/4 of your sample to undergo radioactive decay.

Second case - 3/4 of the sample remains, i.e. does not undergo radioactive decay.

The same principle applies in this case as well, only this time 1/4 of the sample will decay and 3/4 will remain. This means that `A(t) = A_0(t) * 3/4`. So,

`A_0(t) * 3/4 = A_0(t) * (1/2)^(t/t_("1/2"))`, or `3/4 = (1/2)^(t/t_("1/2"))`

As a result,

`t/t_("1/2") = log_(("1/2"))(3/4) = 0.415`

`t = 0.415 * t_("1/2")`

`t = 0.415 * 8.1 = "3.36 days"`

It will take 3.36 days for 1/4 of the sample to undergo radioactive decay.