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

 24. Given that $\left( \frac{18_{O}}{16_{O}}\right)_{V-SMOW}=2005.2*10^{-6}$,the $\frac{18_{O}}{16_{O}}$  of a sample whose $(\delta^{18}O)_{V-SMOW}$=+25% is__________      $*10^{-6} $ (round off to 1 decimal place).

     

Thanks to Saroja

Solution:

 

$\left( \frac{18_{o}}{16_{o}}\right)_{sample}=?$

 

oxygen isotope ratio $(\delta^{18}0)$

 

 $(\delta^{18}O) =\left(\frac{\left(\frac{18_{O}}{16_{O}}\right)_{sample}}{\left(\frac{18_{O}}{16_{O}}\right)_{standard}}-1\right)1000$

 

$25=\left(\frac{\left(\frac{18_{O}}{16_{O}}\right)_{sample}}{\left(\frac{18_{O}}{16_{O}}\right)_{standard}}-1\right)1000$

 

$\frac{25}{1000}=\left(\frac{\left(\frac{18_{O}}{16_{O}}\right)_{sample}}{\left(\frac{18_{O}}{16_{O}}\right)_{standard}}-1\right)$

 

$\frac{\left( \frac{18_{O}}{16_{O}}\right)_{sample}}{2005.2*10^{-6}}=0.025+1 =1.025$

$\left( \frac{18_{O}}{16_{O}}\right)_{sample}=2055.33*10^{-6}$