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}
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