Question

The isotope 'Sr has a half-life of 28.8 years. What is the activity, measured in atoms/second, of a 50 mg sample of "Sr? (10 pts.)

Answer 1

Answer:

The activity of 50mg of Sr is $2.63x10^{11} atoms/second$

Explanation:

Every nuclear disintegration follows the first order rate, thus the half-life (t1/2) is $t_{1/2} =\frac{Ln 2}{k}$ where k is the constant of the reaction.

Considering 1 year equals 3.15x107 seconds

$\frac{28.8year x.3.15x10^{7}s }{1 year} = 9.072x10^{8} s$

For Sr with t1/2 = $9.072x10^{8} s$ , the constant of the reaction is

$k =\frac{Ln 2}{t_{1/2}}=\frac{Ln 2}{9.072x10^{8} s } = 7.64x10^{-10} s^{-1}$

The activity (A) of a nucleus is $A = k.N$ where N is the number of nucleus. Sr molecular mass is 87.6g/mol and every mol contains 6.023x1023 atoms, thus

$\frac{50x10^{-3}g x 6.022x10^{23} atoms}{87.6g} = 3.44x10^{20} atoms$

Therefore, the activity of 50mg of Sr is

$A = k.N = 7.64x10^{-10} s^{-1} x 3.44x10^{20}atoms = 2.63x10^{11} atoms/second$