A) \[t=T\log (1.3)\]
B) \[t=\frac{T}{\log (1.3)}\]
C) \[t=T\frac{\log 2}{\log 1.3}\]
D) \[t=\frac{\log 1.3}{\log 2}T\]
Correct Answer: D
Solution :
[d] Let initially there are total \[{{N}_{0}}\] number of nuclei At time \[t\,\frac{{{N}_{B}}}{{{N}_{A}}}=0.3\] (given) \[\Rightarrow \,\,\,{{N}_{B}}=0.3{{N}_{A}}\] \[{{N}_{0}}={{N}_{A}}+{{N}_{B}}={{N}_{A}}+0.3{{N}_{A}}\therefore \,\,\,{{N}_{A}}=\frac{{{N}_{0}}}{1.3}\] As we know \[{{N}_{t}}={{N}_{0}}{{e}^{-\lambda t}}\] or, \[\frac{{{N}_{0}}}{1.3}={{N}_{0}}{{e}^{-\lambda t}}\] \[\frac{1}{1.3}={{e}^{-\lambda t}}\Rightarrow \ln \,\,(1.3)=\lambda t\] or \[t=\frac{\ln (1.3)}{\lambda }\Rightarrow \,\,\,t=\frac{\ln (1.3)}{\frac{\ln (2)}{T}}=\frac{\ln (1.3)}{\ln (2)}T\]
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