1.

._(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?

Answer»

182 SEC
247.5 sec
308 sec
194.8 sec

Solution :`._(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`


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