One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at an annual rate of approximately 2.7% per year.

a. Write the percent of strontium-90 remaining, P, as a function of years, t, since the nuclear accident.

b. Estimate the half-life of strontium-90.

c. After the Chernobyl disaster, it was predicted that the region would not be safe for human habitation for 105 years. Estimate the percent of original strontium-90 remaining at this time.

Answer: a) % (C/Co) = (e^ (-0.027t)) * 100

b) t1/2 = 25.67years

c) 5.872%

Explanation:

a) Radioactive reactions always follow a first order reaction dynamic

Let the initial concentration of Strontium-90 be Co and the concentration at any time be C

The rate of decay will be given as:

(dC/dt) = - KC (Minus sign because it's a rate of reduction)

The question provides K = 2.7% per year = 0.027/year

(dC/dt) = - 0.027C

(dC/C) = - 0.027dt

∫ (dC/C) = - 0.027 ∫ dt

Solving the two sides as definite integrals by integrating the left hand side from Co to C and the Right hand side from 0 to t.

We obtain

In (C/Co) = - 0.027t

(C/Co) = (e^ (-0.027t))

In percentage, % (C/Co) = (e^ (-0.027t)) * 100

(Solved)

b) Half life of a first order reaction (t1/2) = (In 2) / K

K = 0.027/year

t1/2 = (In 2) / 0.027 = 25.67 years

c) percentage that remains after 105years,

% (C/Co) = (e^ (-0.027t)) * 100

t = 105

% (C/Co) = (e^ (-0.027 * 105)) * 100 = 5.87%