GATE-2018 (24) :Wyllie – Time average equation

 

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 V_smow 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

 

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

 

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

 

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

 

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

 

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