Question

A patient is adminsitered 20mg of iodine-131. How much of this isotope will remain in the body after 40 days if the half-life for iodine is 8 days? Please show all work in a neat order Thank you.

Answer 1

Radioactive decay is a first order reaction. So, for a first order reaction:

k = (2.303/t) log a/(a-x)

Where, k = rate constant, t = time, a = initial amount and (a-x) = amount remained after time t

k = (2.303/t_1/2) log1/0.5

k = (2.303/8) log 2 ............................ (i)

After 40 days:

k = (2.303/40) log 20/(a-x) .......................(ii)

From Equation (i) and (ii)

(2.303/8) log 2 = (2.303/40) log 20/(a-x)

log 20/(a-x) = (0.3010 x 40)/8

log 20/(a-x) = 1.505

20/(a-x) = 31.99

(a-x) = 0.625 mg