InterviewSolution
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InterviewSolution
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Calculate the mass of `C^(14)` (half life =5720 years) atoms give `3.7xx10^(7)` disintegrations per second. |
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Answer» Let the mass of `.^(14)C` atoms be m.g Number of atoms in mg of `.^(14)C = (m)/(14) xx 6.02 xx 10^(23)` `lambda = (0.693)/("half life") = (0.693)/(5720 xx 65 xx 24 xx 60 xx 60) = 3.84 xx 10^(-12) sec^(-1)` We know that, `- (dN)/(dt) = lambda .N` i.e., Rate of disintegration `= lambda xx` no. of atoms `3.7 xx 10^(7) = (0.693)/(5720 xx 365 xx 24 xx 60 xx 60) xx (m)/(14) xx 6.02 xx 10^(23)` `3.7 xx 10^(7) = (3.84 xx 10^(-12) xx m xx 6.02 xx 10^(23))/(14)` |
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