Calculation of true thickness of the bed

 

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} $