A) \[2.1\times {{10}^{-3}}{{s}^{-1}}\]
B) \[2.1\times {{10}^{-4}}{{s}^{-1}}\]
C) \[2.1\times {{10}^{-5}}{{s}^{-1}}\]
D) \[2.1\times {{10}^{-6}}{{s}^{-1}}\]
Correct Answer: D
Solution :
The percentage of radioactive substance left behind after time t, \[\frac{N}{{{N}_{0}}}=(100-18)%=82%\] Also \[\frac{N}{{{N}_{0}}}={{e}^{-\lambda t}}\] \[\therefore \] \[\frac{82}{100}={{e}^{-(\lambda \times 24\times 60\times 60)}}\] Or \[24\times 60\times 60\times \lambda =\log \left( \frac{100}{82} \right)\] \[\therefore \] \[\lambda =2.1\times {{10}^{-6}}{{s}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
