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GATE -2016 (47) the oxygen isotope ratio as

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47) The (\frac{18_{O}}{16_{O}}) of a quartz sample yields a value of 0.0019. The value of \delta^{18_{O}} of the quartz sample is_______(use the value of the ratio in Vsmow as 0.002005)

 

(Thanks to Chandrasekhar, ANU)

Solution:

 Given that (\frac{18_{O}}{16_{O}})_{sample} = 0.0019

 (\frac{18_{O}}{16_{O}})_{standard}=0.002005

 \delta^{18_{O}} =?

 

 We know that the oxygen isotope ratio as

 

\delta^{18_{O}} =(\frac{(\frac{18_{O}}{16_{O}})_{sample}-(\frac{18_{O}}{16_{O}})_{standard}}{(\frac{18_{O}}{16_{O}})_{standar}})\times 1000

 

\delta^{18_{O}} =(\frac{0.0019-0.002005}{0.002005})\times 1000

 

\delta^{18_{O}} =(\frac{-0.000105}{0.002005})\times 1000

 

\delta^{18_{O}} =(-0.05236)\times 1000

 

\delta^{18_{O}} =-52.369

 

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