The half life of `.^(212)Pb` is 10.6 hour. It undergoes decay to its daughter (unstable) element `.^(212)Bi` of half life 60.5 minute. Calculate the time at which the daughter element will have maximum activity.

Explanation

`lambda_(p) = (0.693)/(10.6 xx 60) = 0.001089 "min"^(-1)`
`lambda_(d) = (0.693)/(60.5) = 0.01145 "min"^(-1)`
`t_("max") = (2.303)/((lambda_(d) - lambda_(p))) log_(10) (lambda_(d))/(lambda_(p))`
`= (2.303)/(0.01145 - 0.001089) log_(10) [(0.01145)/(0.001089)]`
`= 222.2758 log_(10) (0.01145)/(0.001089)`
= 227.1 min
= 3.785 hours