Calculation of true thickness of the bed

 

39) In a rock sample the values of $(\frac{87_{sr}}{86_{sr}})_{parent}$ and $(\frac{87_{Rb}}{86_{sr}})_{parent}$ are 0.7125 and 0.2 respectively. The decay constant (ʎ) $87_{Rb}$ is 1.42 * 10^-11 year ^-1 and the time before present (t) is 1000 million years. The value of the initial ratio of $(\frac{87_{sr}}{86_{sr}})_{o}$ is ______

 (Thanks to Chandrasekhar, ANU)

 

Soltuion:

 Given that of $(\frac{87_{sr}}{86_{sr}})_{parent}$ = 0.7125

                        $(\frac{87_{Rb}}{86_{sr}})_{parent}$=0.2

                          ʎ = 1.42 * 10^-11 year ^-1

                          t = 1000 million years  

                             = 1000*10⁶ years = 10⁹ years

 We know that decay equation of Rb -Sr is

$(\frac{87_{sr}}{86_{sr}})_{parent}=(\frac{87_{sr}}{86_{sr}})_{o}+(\frac{87_{Rb}}{86_{sr}})_{parent}(e^{\lambda t}-1)$---------------------------(1)

 

By substituting the given values in the above equation

$(\frac{87_{sr}}{86_{sr}})_{parent}=(\frac{87_{sr}}{86_{sr}})_{o}+(\frac{87_{Rb}}{86_{sr}})_{parent}(e^{\lambda t}-1)$

 

$0.7125=(\frac{87_{sr}}{86_{sr}})_{o}+0.2(e^{1.42\times 10^{-11}\times 10^{9} }-1)$

 

$0.7125=(\frac{87_{sr}}{86_{sr}})_{o}+0.2(e^{1.42\times 10^{-2} }-1)$

 

$0.7125=(\frac{87_{sr}}{86_{sr}})_{o}+0.2(e^{0.0142}-1)$

 

$0.7125=(\frac{87_{sr}}{86_{sr}})_{o}+0.2(1.0143-1)$

 

$0.7125=(\frac{87_{sr}}{86_{sr}})_{o}+0.2(0.0143)$

 

$0.7125=(\frac{87_{sr}}{86_{sr}})_{o}+0.00286$

 

$(\frac{87_{sr}}{86_{sr}})_{o}=0.7125-0.00286$

 

$(\frac{87_{sr}}{86_{sr}})_{o}=0.7096$