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GATE-2016 (39) Decay equation of Rb -Sr is

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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*10years = 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

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