A) 4
B) 2
C) 0.25
D) 0.75
Correct Answer: A
Solution :
Activity, \[A=\lambda N=\frac{0.693}{{{T}_{1/2}}}N\] where \[{{T}_{1/2}}\] is the half-life of a radioactive sample. \[\therefore \] \[\frac{{{T}_{1}}}{{{A}_{2}}}=\frac{{{N}_{1}}}{{{T}_{1}}}\times \frac{{{T}_{2}}}{{{N}_{2}}}\] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{{{A}_{1}}}{{{A}_{1}}}\times \frac{{{N}_{1}}}{{{N}_{2}}}\] \[=\frac{2{{A}_{1}}}{{{A}_{1}}}\times \frac{2{{N}_{2}}}{{{N}_{2}}}=\frac{4}{1}\]
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