Saved Bookmarks
| 1. |
The rate of disintegration of a certain radioactive sample at any instant is `8100` dpm. Fifty minutes later the rate becomes `2700` dpm. The half of the radioactive sample will be (Take `log2=0.3, log3=0.5, log5=0.7`)A. 30 minutesB. 61 minutesC. `47.8` minutesD. `33.3` minutes |
|
Answer» Correct Answer - A `r_(1)=lamdaN_(1)` `8100=lamdaN_(@)` `2700=lamdaN_(t)` `lamda=(2.303)/50 "log" (N_(@))/(N_(t))=(0.693)/(t_(1//2))` |
|