Iiodine -131 is a radioactive isdotpe. If 1.0 mg of `""^(131)I` has an activity of `4.6xx10""^(12)`Bq. What is the half-life of `""^(131)I` (in days)

Explanation

1. We can relate the activity R to the disintegration constant `lamda` with Eq. 40 -18 but lets it as `R=lamda N_(40)`, where `N_(40)` is the number of `""^(40)K` nuclei (and thus atoms) in the banana. 2. We can relate the disintegration constant to the known half life `T_(1//2)` with Eq, 40 -19 `(T_(1//2)=(In 2)//lamda)`.
Calcutaions: Combining Eqs 40 -19 and 40 -18 yields
`R=(N_(40)"In"2)/(T_(1//2))` (40-21)
We know that `N_(40)` is 0.0117% of the total number N of potassium atoms in the banana. We can find an expression for N by combining two equations that give the number of moles n of potassium in the banana. From Eq. 20-2, `n=N//N_(A)`, where `N_(A)` is Avogadro number `(60.2x10^(23)"mol"^(-1))`. From Eq. 20-3, `n=M_("sam")//M` where `M_("sam")` is the sdample mass (here the given 600 mg of potassium) and M is the molar mass of potassium. Combining those two equations to eliminate n, we can write
`N_(40)=(1.17xx10^(-4))(M_("sam")N_(A))/M` 40-22
From Appendix F, we see that the molar mass of potassium is 39.102 g/mol. Equation 40-22 then yields
`N_(40)=(1.17xx10^(-4))((600xx10^(-3)g)(6.02xx10^(23)"mol"^(-1)))/(39.102g//"mol")`
`=1.081xx10^(18)`
Substituting this value of `N_(40)` and the given half life of `1.25xx10^(9)Y` for `T_(1//2)` into Eq. 40-21 leads to
`R=((1.081xx10^(18))xx("In"2))/((1.25xx10^(9)y)(3.16xx10^(7)s//y))`
`=18.96Bq~~19.0Bq`
This is a bout 0.51 n Ci. Your body always has about 160 of potassium. If you repeat our calculation here, you will find that the`""^(40)K` component of that everday amount has an activity of `5.06xx10^(3)` Bq (or `0.14 muCi`). so, eating a banana adds less than 1% to the radiation your body receives daily from radioactive potassium.