WBJEE · Chemistry · Chemical Kinetics
The half-life period of \({ }_{53} I^{125}\) is 60 days. The radioactivity after 180 days will be
- A \(25 \%\)
- B \(12.5 \%\)
- C \(33.3 \%\)
- D \(3.0 \%\)
Answer & Solution
Correct Answer
(B) \(12.5 \%\)
Step-by-step Solution
Detailed explanation
Let, initial concentration \((a)=100\) Given, half-life \(\left(t_{1 / 2}\right)=60\) days To find, radioactivity, i.e. \((a-x)\) after time \(T\) ( 180 days) \[ \because \quad T=n \times t_{1 / 2} \] where, \(n=\) no. of half-lives…
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