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A radioactive sample has a half life of `40` seconds. When its activity is measured `80` seconds after the beginning, it is found to be `6.932xx10^(18)` dps. During this time total energy released is `6xx10^(8)` joule `(ln2=0.6932)`:-A. The initial number of atoms in the simple is `1.6xx10^(20)`B. The initial number of atoms in the sample is `1.6xx10^(21)`C. Energy released per fission is `5xx10^(-13) J`D. Energy released per fission is `(5)/(3)xx10^(-13)J` |
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Answer» Correct Answer - B::C `A = A_(0) e^(-lambda t)` `6.932xx10^(18) = A_(0)e^(-(ln2)/(40)(80))` `= A_(0)e^(-ln4) = (A_(0))/(4)` `A_(0) = 4xx6.932xx10^(18)` `lambda N_(0)= 4xx6.932xx10^(18)` `(0.6932)/(40)N_(0) = 4xx6.932xx10^(18)` `N_(0)=160xx10^(19)` `N_(0) =1.6xx10^(21)` Also `(3)/(4) (1.6xx10^(21))E_(0)=6xx10^(8)` `E_(0)=5xx10^(-13)` |
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