Question

The ridioisotope iodine - 131 (half-life = 8 days) is received from a medical supplier in a solution such that 0.5 cc of the solution has enough activity for injection for a certain organ scan at the time of supply. If it is impractical to use more than 8 cc of the solution per scan, what is the useful life of the solution?

Answer 1

T_1/2 = half life = 8 days

$\lambda$ = decay constant

decay constant is given as

$\lambda$ = 0.693/T_1/2

$\lambda$ = 0.693/8 = 0.087

N = 0.5

N_o = 8

t = useful life

using the equation

N = N_o e^{- $\lambda$ t}

0.5 = 8 e^-0.087t

t = 31.86 days