Saved Bookmarks
| 1. |
The counting rate observed from the radioactive source at t= 0 second was 1600 counts/sec and t= 8 sec, it was 100 count/sec. The counting rate per sec at t =6 sec will be: |
|
Answer» 400 1ST CASE: when t=0, 1600/sec `=lambda N_(0)` 2nd case: when t=8sec. `(d N)/(dt)=lambda N_(0) e^(-8lambda)` `or 100 =lambda N_(0) e^(-8lambda) =1600 e^(-8 lambda)` `or 1/(16) =e^(-8lambda) or (1/2)^(4)=(e^(-2 lambda)^(4)` `e^(-2lambda)=1/2` 3rd case" t=6 sec `(d N)/(dt)=1600 e^(-6 lambda)=1600 (e^(-2lambda))^(3)` `=1600 xx (1/2)^(3)=1600 xx 1/8 =200//sec`. |
|