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

`2.5523* 10^(11) atoms/seconds`is the activity, measured in of a 50 mg sample of 90-Sr.

Step-by-step explanation:

Half life of the 90-Sr ,
`t_(1/2)= 28.8 years`

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

`\lambda =(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 =
`(0.050 g)/(90 g/mol)=0.0005555 mol`

Number of atom in 0.0005555 moles of 90-Sr:

`0.0005555 mol* 6.022* 10^(23) mol^(-1)=3.3455* 10^(20) atoms`

`\lambda =0.0240625 year^(-1)=(0.0240625)/(3.154* 10^7 s)=7.6292* 10^(-10) s^(-1)`

1 year =
`3.154* 10^7 seconds`

Activity measured in atoms per seconds:

= Number of atoms ×
`\lambda`

`=3.3455* 10^(20) atoms* 7.6292* 10^(-10) s^(-1)`

`=2.5523* 10^(11) atoms/s`