Question

3H decays with a rate constant of 0.0563/yr. What is the activity of a 0.543 g sample in curies?

Answer 1

rate constant k = 0.0563 yr-1 = 0.0563 / ( 365.25 x 24 x 3600) s-1 = 1.784 x 10^ -9 s-1

Moles fo tritium ( 3H) = mass/ molar mass of tritium = 0.543/3.016 = 0.18

number of atoms = molex avagadro number = 0.18 x 6.023 x 10^23 = 1.08438 x 10^23

now actvity = rate constant in s-1 x number of atoms

            = 1.784 x 10^ -9 x 1.08438 x 10^23 = 1.9345x 10^14

activity in Curie = 1.9345 x 10^14 / ( 3.7x10^10) = 5228.5 Curies