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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