Question

`._(84)^(218)Po` `(t_(1//2) = 183 sec)` decay to `._(82)^(214)Pb` (`t_(1//2) = 161`) sec by `alpha` emission, while `._(82)^(214)Pb` decay by `beta`-emission. In how much time the number of nuclei of `._(82)^(214)Pb` will reach to the maximum?
A. 182 sec
B. 247.5 sec
C. 308 sec
D. 194.8 sec

Answer 1

Correct Answer - b
`._(84)^(218)Po overset(lambda_(1) = (0.693)/(183) = 3.786 xx 10^(-3) sec^(-1))(to)`
`._(82)^(214)Pb overset(lambda_(2) = (0.693)/(161) = 4304 xx 10^(-3) sec^(-1))(to)`
`t_("max") = (2.303)/(lambda_(1) - lambda_(2)) log (lambda_(1))/(lambda_(2))`
`(2.303)/(3.786 xx 10^(-3) - 4.304xx 10^(-3)) log (3.786 xx 10^(-3))/(4.304 xx 10^(-3))`
`= - (2.303) xx 5.183 xx 10^(-4)) (-0.05569)`
`= 247.5 sec`