Question

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 1

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%