Answer:
The activity of 50mg of Sr is [tex]2.63x10^{11} atoms/second[/tex]
Explanation:
Every nuclear disintegration follows the first order rate, thus the half-life (t1/2) is [tex]t_{1/2} =\frac{Ln 2}{k}[/tex] where k is the constant of the reaction.
Considering 1 year equals 3.15x107 seconds
[tex]\frac{28.8year x.3.15x10^{7}s }{1 year} = 9.072x10^{8} s[/tex]
For Sr with t1/2 = [tex]9.072x10^{8} s[/tex] , the constant of the reaction is
[tex]k =\frac{Ln 2}{t_{1/2}}=\frac{Ln 2}{9.072x10^{8} s } = 7.64x10^{-10} s^{-1}[/tex]
The activity (A) of a nucleus is [tex]A = k.N[/tex] where N is the number of nucleus. Sr molecular mass is 87.6g/mol and every mol contains 6.023x1023 atoms, thus
[tex]\frac{50x10^{-3}g x 6.022x10^{23} atoms}{87.6g} = 3.44x10^{20} atoms[/tex]
Therefore, the activity of 50mg of Sr is
[tex]A = k.N = 7.64x10^{-10} s^{-1} x 3.44x10^{20}atoms = 2.63x10^{11} atoms/second[/tex]
