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A radioactive isotope having a half -life period of3 days was received after 12 days. If 3g of theisotope is left in the container, what wouldbe the initial mass of the isotope?[JEE(Main)-2013 (Online)](1) 36g(3) 24g(2) 48g(4) 12g |
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Answer» t½ = 3 days = ln2/k => k = ln2/3 now when t = 12 days , amount left = 3g so, 12 = (1/k)*ln[A°/3]=> 12 = (3/ln2)*ln(A/3)=> 4 = log2[A/3]=> A/3 = 2⁴ = 16gm=> A = 3*16 = 48g thankyou very much bhai |
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