Saved Bookmarks
| 1. |
The radioisotope ""_(15)^(32)P is used in biochemical studies. A sample containing this isotope has an activity 1000 times the detectable limit. How long could an experiment be run with the sample before the radioactivity could no longer be detected ? Half-life of ""_(15)^(32)P is 14.2 days |
|
Answer» <P> Solution :The minimum NUMBER of P atoms, the radioactivity of which could be detected, is one, and so the sample will initially contain 1000 atoms of P.Thus, `(N^(0))/(N) = (1000)/(1)` Now, `lamda= (2.303)/(t) "log" (N^(0))/(N)` `(0.6932)/(14.2) = (2.303)/(t) "log" (1000)/(1)` or t=141.5 DAYS |
|