43) A basaltic lava flow is found to have a $\frac{87_{sr}}{86_{sr}}$ ratio of 0.72 and $\frac{87_{Rb}}{86_{sr}}$ ratio of 0.750. if the initial $\frac{87_{sr}}{86_{sr}}$ value is determined to be 0.704 what was the age of the flow.(assume the decay constant = 1.42 *10-11 year-1)
(Thanks to Chandrasekhar, ANU)
Solution:
Given
that $\frac{87_{sr}}{86_{sr}}$ = 0.720
$\frac{87_{Rb}}{86_{sr}}$
= 0.750
$\frac{87_{sr}}{86_{sr}})_{o}$=
0.704
$\lambda$ = 1.42 *10-11 year-1
The Decay equation of the Rb-Sr method is as follows
$(\frac{87_{sr}}{86_{sr}})_{parent}=(\frac{87_{sr}}{86_{sr}})_{o}+(\frac{87_{Rb}}{86_{sr}})_{parent}(e^{\lambda
t}-1)$---------------------------(1)
By
substituting the given values in the equation 1 we get
$(\frac{87_{sr}}{86_{sr}})_{parent}=(\frac{87_{sr}}{86_{sr}})_{o}+(\frac{87_{Rb}}{86_{sr}})_{parent}(e^{\lambda
t}-1)$
$0.720=0.704+(0.750)(e^{1.42\times
10^{-11}\times t}-1)$
$(0.750)(e^{1.42\times
10^{-11}\times t}-1)=0.720-0.704$
$(0.750)(e^{1.42\times
10^{-11}\times t}-1)=0.016$
$(e^{1.42\times
10^{-11}\times t}-1)=\frac{0.016}{0.75}$
$(e^{1.42\times
10^{-11}\times t}-1)=0.02133$
$(e^{1.42\times
10^{-11}\times t})=0.02133+1$
$(e^{1.42\times
10^{-11}\times t})=1.02133$
$({1.42\times
10^{-11}\times t})=ln(1.02133)$
$({1.42\times
10^{-11}\times t})=0.021108$
$t
= \frac{0.021108}{1.42\times 10^{-11}}$
$t=
0.01486\times 10^{11} $
$t=
1.486\times 10^{9} $
nice explanation
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