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Q. 10 g of a radioactive isotope is reduced to 1.25 s in 12 yr. Therefore, half-life period of the isotope is

JamiaJamia 2013

Solution:

$ [R]=\frac{{{[R]}_{0}}}{{{2}^{n}}} $ $ \therefore $ $ 1.25=\frac{10}{{{2}^{n}}} $ or $ {{2}^{n}}=\frac{10}{1.25}=8={{2}^{3}} $ Hence, $ n=3 $ 3 half-lives $ =12\,yr $ or $ {{t}_{1/2}}=4\,yr $