A) \[\lambda T=1\]
B) \[\lambda T=1/2\]
C) \[\lambda T={{\log }_{e}}2\]
D) \[\lambda =-{{\log }_{e}}2T\]
Correct Answer: C
Solution :
The number of radioactive atoms after time t\[N={{N}_{0}}{{e}^{-\lambda t}}\] where\[{{N}_{0}}=\]number of atoms initially if\[t=T,\]then \[N={{N}_{0}}/2\] \[\therefore \] \[\frac{{{N}_{0}}}{2}={{N}_{0}}{{e}^{-\lambda T}}\] \[{{e}^{\lambda T}}=2\] \[\lambda T={{\log }_{e}}2\]
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