Q.
If one starts with 1 Curie (Ci) of radioactive substance (t½: 15 hr), the activity left after a period of two weeks will be about 0.02 x μCi. Find the value of x.
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answer is 9.
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Detailed Solution
k=0.69315hr=0.0462hr−1k=2.314×24hrlogc0ct0.0462hr−1=2.314×24hrlog1CictSolve for ct :∴ct = 1.82 × 10-7Ci ≈ 0.18 μCi = 0.02x μCi ∴ x = 9
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