Calculate the mass of `C^(14)` (half life =5720 years) atoms give `3.7xx10^(7)` disintegrations per second.

Explanation

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)`