A) \[\frac{In\,2}{4}\]
B) \[4\ln 2\]
C) 2ln2
D) \[\frac{In\,2}{2}\]
Correct Answer: A
Solution :
\[\,\because \,\,\,\,\,A\,\,=\,\,\lambda N\] \[\frac{{{R}_{B}}}{{{R}_{A}}}=\frac{{{A}_{0}}{{e}^{-{{\lambda }_{B}}t}}}{{{A}_{0}}{{e}^{-{{\lambda }_{A}}t}}}={{e}^{-3t}}\] \[{{e}^{-({{\lambda }_{B}}-{{\lambda }_{A}})t}}={{e}^{-3t}}\] \[\Rightarrow {{\lambda }_{B}}-{{\lambda }_{A}}=3\] \[\because {{\lambda }_{A}}=1\] \[\Rightarrow {{\lambda }_{B}}=4\] \[{{({{T}_{1/2}})}_{A}}\,=\,\,\ln 2\,\,\,\,=\,\,\,\frac{\ln 2}{{{\lambda }_{A}}}\] \[\Rightarrow {{({{T}_{1/2}})}_{B}}=\frac{\ln 2}{{{\lambda }_{B}}}=\frac{\ln 2}{4}\]
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