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.
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/t1/2) 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
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