Dear Fuad Miskon,

I encountered the same problem just recently concerning the Eu-anomaly. I think, the one provided by Shields and Stille (2001) is missing a "divided by" and should instead be: Eu-anom = EuN / (SmN + GdN)0.5.  A calculation I find more appropriate, given the square root in that equation is: Eu-anom.= EuN / (SmN * GdN)0.5.

Lawrence and Kamber (2006) suggested to use this equation: Eu-anom = EuN / (SmN2 * TbN)1/3.

However, which one is best to use really depends on the nature of your sediments and their REE patterns. If either Sm, Gd or Tb seem enriched/depleted in respect to the overall pattern apply the equation that is not using the respective element. I.e. in a sediment that exhibits a "hat-shaped"-pattern, where Tb is relatively debleted in respect to Gd and Dy.

Also, you usually only use the anomaly for comparison within your own set of data, so you could also think of an equation that you feel is reasonable. Just make sure you use it consistently and if you compare it to other (peoples) data that you re-calculate the anomalies using your or the same equation.

Quick remark regarding the Ce-anomaly, be sure La is not enriched when you use the above mentioned calculations. I think Shields and Stille (2001) also refer to this.

Hope my answer is not too late and might still be helpful.

Regards,

Jess

Dear Fuad Miskon

From my concerned, in tropical or mostly in Australasia-Pacific regional in marginal area the best normalization Ce and Eu for sediment is PAAS and biota is using to  Chondrite. However  my idea when deal with river compartments the best equation for Ce anomaly conducted  through formula by Stein Jacobson from Harvard University and for Eu anomaly try searching formula that had been used by de Baar in his Phd thesis on Marine Geochemistry of REE (1983). And again i hardly inclined to you to introduced the first normalized REE standard specially in biota seen to hitherto no body concern on biota or the best could i say it has been over look by among REE scientists.

Regard's

Zolhizir bin Daud

Fuad,

I would do 2 things, first a spider plot of REE normalised to NASC or PAAS (a shale standard) to see how enriched they are in relation to a "normal' shale.  Secondly, in order to further quantify the anomaly i would calculate Ce*norm by comparing the Ce with the elements (La and Nd) Ce anom.=Ce/Ce*=3CeN/(2LaN+NdN) (N= normalised data.)

Good control as to your mineralogy should be of further help as many Ce enrichments may be related to detritus or phosphate, checking for other REE enrichments may help to identify this.   See Shields & Stille (2001) and MacLeod and Irving (1996) for more.

Hope this helps.

Sean

Thank you so much for the explaination and recommendation. I really appreciate it

Dear Fuad Miskon,

I encountered the same problem just recently concerning the Eu-anomaly. I think, the one provided by Shields and Stille (2001) is missing a "divided by" and should instead be: Eu-anom = EuN / (SmN + GdN)0.5.  A calculation I find more appropriate, given the square root in that equation is: Eu-anom.= EuN / (SmN * GdN)0.5.

Lawrence and Kamber (2006) suggested to use this equation: Eu-anom = EuN / (SmN2 * TbN)1/3.

However, which one is best to use really depends on the nature of your sediments and their REE patterns. If either Sm, Gd or Tb seem enriched/depleted in respect to the overall pattern apply the equation that is not using the respective element. I.e. in a sediment that exhibits a "hat-shaped"-pattern, where Tb is relatively debleted in respect to Gd and Dy.

Also, you usually only use the anomaly for comparison within your own set of data, so you could also think of an equation that you feel is reasonable. Just make sure you use it consistently and if you compare it to other (peoples) data that you re-calculate the anomalies using your or the same equation.

Quick remark regarding the Ce-anomaly, be sure La is not enriched when you use the above mentioned calculations. I think Shields and Stille (2001) also refer to this.

Hope my answer is not too late and might still be helpful.

Regards,

Jess

Different formula will give different answers and different opinions. Use any formula and reference it. I use that of Taylor and Mclennan,1985. Very simple; Eu/Eu*= 2x Eu*/ (Sm +Gd)N.

However, recent formula may be used.eg that by Lawrence and Kamber (2006).

I recommend Jessica's answer

Best Regards

Jessica is right here.