Find the decay constant of `""^(55)Co` radio nuclide if its activity is known to decrease 4% per hour. The decay product is non - radioactive.

Explanation

Initial activity of nuclei `A_(0)=lamdaN_0`
Activity of nuclei at time `t,A=lamdaN=lamdaN_(0)e^(-lamdat)`
`A=A_(0)e^(-lamdat)" "....(1)`
Since activity decreases at 4.0% per hour , activity of `""^(55)Co` radio nuclide at 0 = 1 hr.
`A=A_(0)-etaA_(0)=A_(0)(1-eta)`.
where `eta = 0.04 " "....(2)`
Taking log of
where `eta = 0.04 " "....(2)`
Taking log of Eq. (1) we get `lamda = -1/(t) ln (1- eta)`
On substituting values , we get `lamda~~ 1.1 xx10^(-5)s^(-1)`.