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In an ore containing uranium , the ratio of $\;U^{238}\;$ to $\;Pb^{206}\;$ nuclei is $3$ . Then calculate age of ore assuming that all lead present in ore is final stable products of $\;U^{238}\;$ . Take half life if $\;U^{238}\;$ to be $\;4.5 \times 10^{9}\;$years

$(a)\;2.12 \times 10^{13} years\qquad(b)\;1.26 \times 10^{7} years\qquad(c)\;1.87 \times 10^{9} years\qquad(d)\;3.2 \times 10^{10} years$

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Answer : (c) $\;1.87 \times 10^{9} years$
Explanation :
In this problem it is assumed that half life of intermediate products is very small in comparision to that of $\;U^{238}$
Let required age is 't' . If N be initial no. of $\;U^{238}\;$ nuclei then
$\large\frac{N_{0}e^{-\lambda t}}{N_{0}\; (1-e^{\lambda t})}=3$
$4 e^{-\lambda t}=3 \qquad \; \lambda t=ln (\large\frac{4}{3})$
$t=\large\frac{ln(\large\frac{4}{3})}{\lambda}=\large\frac{ln(\large\frac{4}{3})}{ln 2}\;T$
$=1.87 \times 10^{9} years \;.$
answered Feb 26, 2014 by yamini.v

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