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  3. Chemistry
  4. The half-life period of 53I125 is 60 days. The radioactivity after 180 days will be

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Q. The half-life period of $_{53}I^{125}$ is 60 days. The radioactivity after 180 days will be

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Solution:

100% $\ce{->[t_{1}/2]}$50% $\ce{->[t_{1}/2]}$ 25% $\ce{->[t_{1}/2]}$ 12.5
Total time =3t1/2 = 180 days