A) \[2.89\times {{10}^{-2}}{{T}^{-1}}\]
B) \[3.89\times {{10}^{-3}}{{T}^{-1}}\]
C) \[4.89\times {{10}^{-2}}{{T}^{-1}}\]
D) none of these
Correct Answer: A
Solution :
Key Idea: \[k=\frac{2.303}{t}\log \,\frac{{{N}_{0}}}{{{N}_{t}}}\] where t = time = 48 days \[{{N}_{0}}=\] Initial amount of radioactive substance = 100 \[{{N}_{t}}=\] Initial amount of radioactive substance left after time t = 25 \[\therefore \] \[k=\frac{2.303}{48}\log \frac{100}{25}\] \[=0.0479\log 4=0.0479\times 0.6020\] \[=2.89\times {{10}^{-2}}T{{}^{-1}}\]
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