A) 0.4 In 2
B) 0.2 In 2
C) 0.1 In 2
D) 0.8 In 2
Correct Answer: A
Solution :
Given\[{{N}_{0}}\lambda =5000,\,\,N\lambda =1250\] \[N={{N}_{0}}{{e}^{-\lambda t}}={{N}_{0}}{{e}^{-5\lambda }}\] where \[\lambda \] is decay constant per min. \[N\lambda ={{N}_{0}}\lambda {{e}^{-5\lambda }}\] \[1250={{N}_{0}}\lambda {{e}^{-5\lambda }}\] \[\therefore \] \[{{e}^{-5\lambda }}=\frac{5000}{1250}=4\] \[{{e}^{5\lambda }}=4\] \[5\lambda =2{{\log }_{e}}2\] \[\lambda =0.4\ln 2\]
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