Since this is rare-earth ions, it means that they are in some environment. What kind of environment in your case? Between which the energy levels there is a transfer of energy? Your question needs to specify. What is the basis for your assertions regarding the transfer of energy? First of all it is necessary to determine the mechanism of energy transfer between different RE ions.

@Yuri Gnatenko: I hope , there may be energy transfer from 4G5/2 of Sm3+ to 5D0 of Eu3+ (already reported).

It has been reported that the successive J components of lanthanides are very well separated from each other (enegy difference is much more than kT) .But in Sm (III) with 4f5 configuration and 6H5/2 ground term and Eu (III) with 4f6 configuration and 7F0 ground term , the energy difference is ≈ kT among the J components obtained from their respective ground terms. This, property, in fact, serves as a deriving force for their mutual energy transfer. This is all the more facilitated if both these ions are doped in some suitable foreign ions which may change their surface or even in molecular structures.

The following given relation is a well known relation for calculating the energy of any J component in lanthanides in ground terms:

Energy of a ground J component=1/2 λ [J (J+1)-L (L+1)-S(S+1)] --------- (a)

[I] Sm (III) to Eu (III)

Sm (III) is 4f5 with Ground term=6H5/2 has

L=5, S=5/2, J=5/2, λ= Zeta/2S=1157/5=231.4 cm-1

Eu (III) is 4f6 with Ground term=7F0 has

L=3 ,S=3,J=0,λ= Zeta/2S=1326/6=221.0 cm-1.

Put values of L,J, S for 6H5/2 of Sm (III) to obtain energy= - 1012.4.

Put values of L, J, S for 7F0 of Eu (III) to obtain energy= - 1326.0 

As energy of 6H5/2 of Sm (III) is higher than that of 7F0 of Eu (III), so it

flows from Sm(III) to Eu(III). But the reverse cannot happen as energy can not

flow from lower energetic Eu (III) to energetically higher Sm (III) unless excited by giving energy from some outer source

[II] In case of Eu (II) is 4f7 with Ground term=8 S 7/2 has

L=0, S=7/2, J=7/2 (take modulus), λ= Zeta/2S=1326/7=189.4 cm-1.

Put these values of L, J, S for 8S7/2 for Eu (II) to obtain energy= 0.0.

So energy cannot flow from Sm (III) to Eu(II) of its own unless excited by giving energy from some outer source. But the reverse can happen as the energy can flow of its own from energetically higher Eu (II) to energetically lower Sm (III).

Different relation are needed for non ground term J components .So energy relation between 4G5/2 of Sm (III) and 5D0 of Eu (III) cannot be calculated by relation(a).

what is zeta

 It is single- electron spin orbit coupling parameter (ζ ) which measures the strength of the interaction between the spin and orbital angular momenta of a single electron of the configuration. The operator that corresponds to the perturbation by spin- orbit  coupling is ζ  times the sclar  product between spin and orbital momenta vectors of the electron

The ζ   is a positive quantity but λ  can be positive( when the sub shell is less than half full) and  negative( when the sub shell is more than half full.

The ζ value are reported in the literature for 1st , 2nd and 3rd transitin and lathalide ions

 See In troduction to Ligand Fields By B. N. Figgis – Chapter-3[Atomic Spectroscopy]

Table:3.4.

Thank you Manohar. Please help me. if any one know zeta values of Sm3+ and Dy3+ ions let me know. is zeta values of Re3+ is same in any host? is it possible the energy transfer from Dy3+ to Sm3+. 

1.Plz. note Zeta for Sm [ not Sm^3] is 1157 Cm-1. Now you have lambda for Sm^3

=Zeta divided  by  twice of its total spin. This can be calculated  as follows

 Outermost conf  of Sm^3= 4f^5. So it has total spin=5/2; multiply by 2

. It becomes 5. So its 2S=5.

.Now divide 1157 by 5to get=1157/5=231.4 which is lambda for Sm^3

2. Now from the table( attached),  note zeta for Dy which is 2074.

 Its outermost electronic configuration is= 4f^106s^2. Remove three electrons

 From Dy to obtain Dy^3= 4f9. It will have five unpaired electrons with total S=5/2

 So its 2S= 5. Now divide 2074 by 5to obtain=415(appxo)

3. Again from the  Table 3.4, Zeta for Re=2500. The outer most electronic configuration of Re=5d^5 6s^2 and that of Re^3( after loosing 3 electons ) will be= 5d^4 with total spin=2 and thus 2S=4. So divide 2500/4=625 Cm-1.

Lastly:

4. As calculated in my first answer,

Sm (III) is 4f5 with Ground term=6H5/2 has

L=5, S=5/2, J=5/2, λ= Zeta/2S=1157/5=231.4 cm-1

Put values of L,J, S for 6H5/2 of Sm (III) to obtain energy= - 1012.4.

 Now come Dy^3

Dy (III) is 4f9 with Ground term=6H15/2[ because 4f^9 is more than half full, J=L+S] has

L=5, S=5/2, J=15/2, λ= Zeta/2S=1157/5=231.4 cm-1

Energy of a ground J component=1/2λ [J (J+1)-L (L+1)-S(S+1)] --------- (a)

 15/2( 15/2+1) -5(5+1)-5/2(5/2+1)].1.λ/2

      [63.75-30-8.75].415/2

     25.207.5 = 5187.5 

 As energy Dy^3 is higher than Sm^3, it can flow from Dy^3 to Sm^3.

Thank you and you are soooo knowledged person. sorry for disturbing you i have problem i.e i am using excitation at 405 nm in Dy3+ and Sm3+ co-doped materials. if i am using 405 nm, mostly Sm3+ will be excited because at 405 nm  absorption is high for Sm3+. but i am getting blue, yellow (Dy3+) and organge-red (Sm3+) emissions. Thats why i am asking is there any for energy transfer from Sm3+ to Dy3+ .

Plz. check the last line of my above given answer:

As energy Dy^3 is higher than Sm^3, it can flow from Dy^3 to Sm^3;meaning ,thereby , that the energy cannot flow from Sm^3 to Dy^3.

Thank you sir

Actually i am getting blue and yellow emissions under 405 nm excitation thats why i asked again. may be it occurs by other reason

Dear sir (Manohar Sehgal ),

The zeta values of Sm and Dy are in which host. Actually i dont have Figgis text book sir. thats why i am asking.

It depends on the energy of the excited state, from which the transition takes place. For eg: the excited energy level of Sm3+ is higher than that of Eu3+, therefore energy transfer takes place from Sm3+ to Eu3+. So you need to find the excited energy level of Eu2+ and Sm3+ to have a clear idea.