A) \[t=T\log (1.3)\]
B) \[t=\frac{T}{\log (1.3)}\]
C) \[t=\frac{T\log 2}{2\log 1.3}\]
D) \[t=T\frac{\log 1.3}{\log 2}\]
Correct Answer: D
Solution :
A time t \[\frac{{{N}_{B}}}{{{N}_{A}}}=.3\Rightarrow {{N}_{B}}=.3{{N}_{A}}\] also let initially there are total N0 number of nuclei \[{{N}_{A}}+{{N}_{B}}={{N}_{0}}\] \[{{N}_{A}}=\frac{{{N}_{0}}}{1.3}\] Also as we know \[{{N}_{A}}={{N}_{0}}{{e}^{-\lambda t}}\] \[\frac{{{N}_{0}}}{1.3}={{N}_{0}}{{e}^{-\lambda t}}\] \[\frac{1}{1.3}={{e}^{-\lambda t}}\ell n(1.3)=\lambda t\]or \[t=\frac{\ell n(1.3)}{\lambda }\] \[t=\frac{\ell n(1.3)}{\frac{\ell n(2)}{T}}=\frac{\ell n(1.3)}{\ell n(2)}T\]
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