Question

A patient is given `0.050` `mg` of technetium-99m, a radioactive isotope with a half-life of about `6.0` hours. How long until the radioactive isotope decays to `6.3 * 10^-3` `mg`?

Answer 1

`"44 h"`

Explanation:

As you know, a radioactive isotope's nuclear half-life tells you exactly how much time must pass in order for an initial sample of this isotope to be halved.

The equation that establishes a relationship between the initial mass of a radioactive isotope, the mass that remains undecayed after a given period of time, and the isotope's half-life looks like this

`color(blue)(A = A_0 * 1/2^n)" "`, where

`A` - the initial mass of the sample
`A_0` - the mass remaining after a given time
`n` - the number of half-lives that pass in the given period of time

In your case, the initial amount of technetium-99 is said to be equal to `"0.050 mg"`. You are asked to find the amount of time needed to reduce this initial sample to `6.3 * 10^(-3)"mg"`.

Use the above equation to find the value of `n`

`6.3 * 10^(-3) color(red)(cancel(color(black)("mg"))) = 0.050 * 10^(-3)color(red)(cancel(color(black)("mg"))) * 1/2^n`

This is equivalent to

`(6.3 * 10^(-3))/0.050 = 1/2^n`

Take the natural log of both sides of the equation to get

`ln((6.3 * 10^(-3))/0.050) = ln( (1/2)^n)`

`ln((6.3 * 10^(-3))/0.050) = n * ln(1/2)`

`n = ln(1/2)/ln((6.3 * 10^(-3))/0.050) = 7.31`

Since `n` represens the number of half-lives that pass in a given period of time, you can say that

`n = t/t_"1/2" implies t = n * t_"1/2"`

In your case, `t_"1/2" = "6.0 h"`, which means that

`t = 7.31 * "6.0 h" = "43.86 h"`

Rounded to two sig figs, the answer will be

`t = color(green)("44 h")`