Question

The isotope 90_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:

$2.5523\times 10^{11} atoms/seconds$is the activity, measured in of a 50 mg sample of 90-Sr.

Explanation:

Half life of the 90-Sr ,$t_{1/2}= 28.8 years$

Activity coefficient of the 90-Sr = $\lambda$

$\lambda =\frac{0.693}{28.8 years}=0.0240625 year^{-1}$

Mass of 90-Sr = 50 mg = 0.050 g

Molecular mass of 90-Sr = 90 g/mol

Moles of 90-Sr =$\frac{0.050 g}{90 g/mol}=0.0005555 mol$

Number of atom in 0.0005555 moles of 90-Sr:

$0.0005555 mol\times 6.022\times 10^{23} mol^{-1}=3.3455\times 10^{20} atoms$

$\lambda =0.0240625 year^{-1}=\frac{0.0240625}{3.154\times 10^7 s}=7.6292\times 10^{-10} s^{-1}$

1 year = $3.154\times 10^7 seconds$

Activity measured in atoms per seconds:

= Number of atoms × $\lambda$

$=3.3455\times 10^{20} atoms\times 7.6292\times 10^{-10} s^{-1}$

$=2.5523\times 10^{11} atoms/s$