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 :
Undisintegrated part \[\frac{N}{{{N}_{0}}}=(100-18)%=82%\] Using relation \[N={{N}_{0}}({{e}^{-\lambda \,t}})\] \[\frac{82}{100}={{e}^{-(24\times 60\,\lambda )}}\] \[\therefore \] \[24\times 60\times 60\times \lambda =\log \,\left( \frac{100}{82} \right)\] or \[\lambda =2.1\times {{10}^{-6}}{{s}^{-1}}\]
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