Question

The half-life of a dangerous radionuclide, strontium-90, ⁹⁰Sr, is 29 years. This isotope is produced during the nuclear fission reactions of atomic power reactors; if a receptor is compromised (at Chernobyl and Fukushima it happened) considerable amounts can be released. It has yet to be disclosed how much ⁹⁰Sr was released at Fukushima (the authorities won't say, though perhaps they don't know), so let's just suppose 9.0 g (possibly a gross underestimate).

a. Please find the rate constant for the decay of ⁹⁰Sr in units of s^-1.

b. Please determine the intial rate of decay (in nucleii/s) of 9.0 g ⁹⁰Sr.

c. Please determine how much of the original ⁹⁰Sr will remain after 44 years.

Answer 1

a)

rate constant is given via:

HL = ln(2) / k

k = ln(2) / Hl = ln(2) /(29) = 0.023901 1/y

change to 1/S --> 0.023901 1/y / 365*24*3600 s =0.023901 / ( 365*24*3600) = 7.57*10^-11 1/s

b)

the initial rate of decay fro m = 9 g

mol of Sr = mass/MW = 9/87.62 = 0.102716 mol

Rate = K(C)

Rate = ( 7.57*10^-11 )(0.102716 ) = 7.7756*10^-12 mol per second

c)

how much original will be left after t = 44 eyars

apply

Afinal = Ainitial *(1/2)^(t/HL)

Afinal /Ainitial = (0.5) ^(44/29) = 0.3493

Afinal = Ainitial * 0.3493 --> 34.93 % left