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3 November, 14:13

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.

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  1. 3 November, 14:25
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    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%
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