A veteran of the war who left Southeast Asia in 1971 was tested in 1991, and it

Question

A veteran of the war who left Southeast Asia in 1971 was tested in 1991, and it was found that her blood contained 12.5 picograms of dioxin per gram (pg/g) of blood fat. The half-life for dioxin excretion from the human body is 7.1 years. (The compound is excreted slowly but not metabolized.) Dioxin excretion follows first order kinetics.

a) What was her dioxin level in 1971 when she left Vietnam?
b) In what year will her dioxin level fall to the background level seen in the population of the industrialized world if that background level is 3.0 picograms per gram (pg/g) of blood fat?

Explanation / Answer

Half life T1/2= 7.1 yrs. That dioxin become 50% of its original concentration in every 7.1 yrs.

Concentration in the year 1991, NT = 12.5 pg/g. The total year passsed T= 20, N0 = ?

Then, for a first order decay, we can use the following equation.

NT = N02-T/T1/2

12.5 =N0 2-20/7.1

No =12.5/2-20/7.1 = 88.08 pg/g was the concentration at 1971.

New concentration or NT' = 3 pg/g, what will be the T' = ?

Now, NT' = N02-T/7.1

Now if we consider, N0 =12.5 pg/g, is the initial concentration then to become, NT'=3 pg/g.

We can use the following formula.

T1/2 = T'ln2(/lnN0-lnNT)

or T' =7.1 *(ln12.5 - ln3) /0.693 = 7.1*(2.5257-1.0986)/0.693 = 14.6210 yrs. That means another 14.621 yrs is required for the concentration to become down to 3 pg/g from the year 1991.

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