Question

A sample of osmium contains 8.000 micrograms of Os-185. Determine the activity of the sample in millicuries? [Half-life and atomic mass data can be found in the reference table linked in Question-0.] millicuries Check

Answer 1

Radioactivity,R = K* N where, K is a constant defined by 0.693/half life and N is the number of nuclei available.

It‘s given that half life (Should be in seconds) =93.6 days = 93.6 days * 24 h* 3600sec = 8,087,040 seconds So, K = 0.693/8,087,040sec = 8.57 *10-8 sec-1

Number of nuclei = mass/atomic mass = 8*10-6g / 184.95g/mol   (Mass= 8µg = 8 * 10-6 g )

= 4.33 *10-8 mol = 4.33 *10-8 mol * 6.022*1023 mol-1

= 2.60 *1016   

So, Activity = K * N

= 8.57 *10-8 sec-1 * 2.60*1016  =2.23* 109 sec-1

To convert this value into curie, divide the value by 3.7 *1010

That is, 2.23* 109 /3.7 *1010 = 6.03 *1018 Curie = 6.03 *1021 millicurie.