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GATE-2007 (43) A basaltic lava flow is found to have

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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}

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