A) 1500 years
B) 300 years
C) 449 years
D) 810 years
Correct Answer: C
Solution :
\[{{\lambda }_{\alpha }}=\frac{1}{1620}\] per year and \[{{\lambda }_{\beta }}=\frac{1}{405}\] per year and it is given that the fraction of the remained activity \[\frac{A}{{{A}_{0}}}=\frac{1}{4}\] Total decay constant \[\lambda ={{\lambda }_{\alpha }}+{{\lambda }_{\beta }}=\frac{1}{1620}+\frac{1}{405}=\frac{1}{324}per\,year\] We know that \[A={{A}_{0}}{{e}^{-\lambda t}}\]Þ \[t=\frac{1}{\lambda }{{\log }_{e}}\frac{{{A}_{0}}}{A}\] Þ\[t=\frac{1}{\lambda }{{\log }_{e}}4=\frac{2}{\lambda }{{\log }_{e}}2\]=324 ´ 2 ´ 0.693 = 449 years.
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