Supporting Information. Upconversion Luminescence with Tunable Lifetime. Size

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1 This journal is The Royal Society of Chemistry 212 -S1- Supporting Information Upconversion Luminescence with Tunable Lifetime in NaYF 4 :Yb,Er Nanocrystals: Role of Nanocrystal Size Jiangbo Zhao a, Zhenda Lu b, Yadong Yin b, Christopher McRae c, James A. Piper a, Judith M. Dawes a, Dayong Jin a*, and Ewa M. Goldys a* * To whom correspondence should be addressed. dayong.jin@mq.edu.au, ewa.goldys@mq.edu.au. a MQ BioFocus Centre, Faculty of Science, Macquarie University, NSW 219, Sydney, Australia b Department of Chemistry, University of California, Riverside, CA 92521, USA c Department of Chemistry and Biomolecular Sciences, Faculty of Science, Macquarie University, NSW 219, Sydney, Australia Contents: 1. Figure S1 (size histograms) 2. Figure S2 (upconversion power-dependence gradients) 3. Figure S3 (upconversion luminescence decay of ~31nm nanocrystals in dried state and organic solvent) 4. Rate Equations for Yb,Er-doped Nanocrystals and the Asymptotic Solutions

2 This journal is The Royal Society of Chemistry 212 -S2- Fig. S1. (a-f) The particle size distribution histograms corresponding to TEM images in Figure 1. Histograms of the particle sizes are drawn from analysis of > 15 particles for each sample. The mean and standard deviation for each nanocrystalline size are, respectively, 5.99±.38 nm in a; 9.2±.85 nm in b; 13.94±.97 nm in c; 19.89±.76nm in d; 3.9±1.11 nm in e; 44.81±1.54 nm in f only measuring top/bottom surface area.

3 -S3- een Power-Dependence Gradient Electronic Supplementary Material (ESI) for Nanoscale This journal is The Royal Society of Chemistry Green Red Intensity (arb.unit).1 1 Laser Power Density (kw/cm 2 ) Diameter (nm) Fig. S2. Gradients for power dependence of the green and red upconverted luminescence of NaYF 4 nanocrystals with various sizes from 6 nm to 45 nm. The log-log plots of upconversion intensities versus NIR excitation power were fitted to straight lines, and the gradients of these lines 1.7 have been plotted for green and red luminescence. One selected example plot is shown in the 1.6 corresponding inset, indicating the upconversion emission intensities of the green and red emission 1.5 Green 1.4 versus NIR excitation laser power density. 1.3 symbols represent silica-coated nanoparticles.

4 Intensity (a.u.) aintensity (a.u.) Electronic Supplementary Material (ESI) for Nanoscale This journal is The Royal Society of Chemistry 212 -S4-1. dried NPs on glassslide NPs in cyclohexane time (ms) 2. b 1. dried NPs on glassslide NPs in cyclohexane time (ms) Fig. S3. Normalized upconversion fluorescence decays for ~31 nm nanocrystals in dried state and organic solvent cyclohexane: green emission (a) and red emission (b), respectively. All the measurements were obtained at an excitation wavelength of 98 nm.

5 This journal is The Royal Society of Chemistry 212 -S5- RATE EQUATIONS FOR ER:YB UNPCONVERTING NANOPARTICLES AND ASYMPTOTIC BEHAVIOUR OF THE SOLUTIONS 1. Rate equations for an arbitrary excitation dn Er,2 dn Y b,2 = k F T N Er N Y b,2 k c2 N Er,2 N Y b,2 k c3 N Er,3 N Y b,2 = k c2 N Er,2 N Y b,2 W 2 N Er,2 + W 32 N Er,3 + W 52 N Er,5 + W 62 N Er,6 dn Er,3 = k F T N Er N Y b,2 k c3 N Er,3 N Y b,2 W 3 N Er,3 + W 53 N Er,5 + W 63 N Er,6 dn Er,5 = k c2 N Er,2 N Y b,2 W 5 N Er,5 + W 65 N Er,6 dn Er,6 = k c3 N Er,3 N Y b,2 W 6 N Er,6 The rate equations for the upconverting NaYF 4 doped with Er and Yb are: (1.1) (1.2) dn Er,2 (1.3) dn Er,3 dn Y b,2 = ρ p σ Y b N Y b k F T N Er,1 N Y b,2 k c2 N Er,2 N Y b,2 k c3 N Er,3 N Y b,2 = k c2 N Er,2 N Y b,2 ρ p σ ESA2 N Er,2 W 2 N Er,2 + W 32 N Er,3 + W 52 N Er,5 + W 62 N Er,6 = k F T N Er,1 N Y b,2 ρ p σ ESA3 N Er,3 k c3 N Er,3 N Y b,2 W 3 N Er,3 + W 53 N Er,5 + W 63 N Er,6 (1.4) dn Er,5 = k c2 N Er,2 N Y b,2 + ρ p σ ESA2 N Er,2 W 5 N Er,5 + W 65 N Er,6 dn Er,6 (1.5) = k c3 N Er,3 N Y b,2 + ρ p σ ESA3 N Er,3 W 6 N Er,6 Here, N Er(Y b),i denotes the population of level i of Er(Yb). W i denotes total radiative and nonradiative decay rate of the relevant population, while W ij is the decay rate of population of level i decaying into j. σ Y b is the absorption cross section of ground state Yb, while σ ESA2(3) refers to absorption cross section of Er in level 2(3) at 98 nm. K F T is the coefficient of forward energy transfer, while k c2(3) is the cooperative upconversion coefficient for the 2 5 and 3 6 upconversion. ρ p is the power constant defined as excitation power variable. See main body of the paper for justification of these equations and underlying assumptions. 1

6 This journal is The Royal Society of Chemistry 212 -S6-2. Rate equations for pulsed excitation, between the pulses In the case of pulsed excitation, after the excitation has ceased (at t = ), these equations assume a simpler form: (2.1) dn Y b,2 = k F T N Er N Y b,2 k c2 N Er,2 N Y b,2 k c3 N Er,3 N Y b,2 (2.2) dn Er,2 = k c2 N Er,2 N Y b,2 W 2 N Er,2 + W 32 N Er,3 + W 52 N Er,5 + W 62 N Er,6 (2.3) dn Er,3 = k F T N Er N Y b,2 k c3 N Er,3 N Y b,2 W 3 N Er,3 + W 53 N Er,5 + W 63 N Er,6 (2.4) dn Er,5 = k c2 N Er,2 N Y b,2 W 5 N Er,5 + W 65 N Er,6 dn Er,6 (2.5) = k c3 N Er,3 N Y b,2 W 6 N Er,6 Here, N Er,1 is approximated by N Er =const, the density of Er ions. 3. Asymptotic behaviour of the solutions We now establish the asymptotic behaviour of the solutions.this is done in several steps. 1. Positivity of the solutions It is easy to show that if the initial conditions are all positive then solutions stay positive for all times. 2. Upper bound of the solution for N Y b,2 First we find the upper bound for N Y b,2. It satisfies the equation: dn Y b,2 (3.1) Clearly, by 1. we have = k F T N Er N Y b,2 k c2 N Er,2 N Y b,2 k c3 N Er,3 N Y b,2 (3.2) Solving this inequality we obtain dn Y b,2 k F T N Er N Y b,2 (3.3) N Y b,2 (t) e k F T N Er t N Y b,2 () 3. Upper bounds of the solution for N Er,2 N Er,6

7 This journal is The Royal Society of Chemistry 212 -S7- We now consider equations for N Er,2, N Er,3, N Er,5 and N Er,6 (3.4) (3.5) dn Er,2 dn Er,3 = (k c2 N Y b,2 + W 2 ) N Er,2 + W 32 N Er,3 + W 52 N Er,5 + W 62 N Er,6, = (k c3 N Y b,2 + W 3 ) N Er,3 + W 53 N Er,5 + W 63 N Er,6 + k F T N Er N Y b,2, (3.6) dn Er,5 = W 5 N Er,5 + W 65 N Er,6 + k c2 N Er,2 N Y b,2, (3.7) Clearly dn Er,6 = W 6 N Er,6 + k c3 N Er,3 N Y b,2. (3.8) AND dn Er,2 dn Er,3 < W 2 N Er,2 + W 32 N Er,3 + W 52 N Er,5 + W 62 N Er,6 (3.9) < W 3 N Er,3 + W 53 N Er,5 + W 63 N Er,6 + k F T N Er N Y b,2 By point 2. for every ϵ > we can choose t = t (ϵ, N Y b,2 (), k F T N Er ), such that for all t t we have N Y b,2 (t) < ϵ k c2 AND N Y b,2 (t) < ϵ k c3. Then, for t t we have (3.1) dn Er,2 < W 2 N Er,2 + W 32 N Er,3 + W 52 N Er,5 + W 62 N Er,6 (3.11) dn Er,3 < W 3 N Er,3 + W 53 N Er,5 + W 63 N Er,6 + k F T N Er N Y b,2 (3.12) dn Er,5 < W 5 N Er,5 + W 65 N Er,6 + ϵn Er,2 dn Er,6 (3.13) Now, for a given A, B, C > let < W 6 N Er,6 + ϵn Er,3 V t = N Er,2 + AN Er,3 + BN Er,5 + CN Er,6.

8 This journal is The Royal Society of Chemistry 212 -S8- For simplicity we denote f(t) = k F T N Er N Y b,2. By taking the first derivative of V t and using the equations for N Er,2,..., N Er,6 we obtain: (3.14) dv = dn Er,2 + A dn Er,3 + B dn Er,5 + C dn Er,6 < W 2 N Er,2 + W 32 N Er,3 + W 52 N Er,5 + W 62 N Er,6 AW 3 N Er,3 + AW 53 N Er,5 + AW 63 N Er,6 + Af(t) BW 5 N Er,5 + BW 65 N Er,6 + BϵN Er,2 CW 6 N Er,6 + CϵN Er,3 = (W 2 Bϵ) N Er,2 (AW 3 W 32 Cϵ) N Er,3 (BW 5 W 52 AW 53 ) N Er,5 (CW 6 W 62 AW 63 BW 65 ) N Er,6 + Af(t) Now we need to choose A, B, C and ϵ in such a way that all the prefactors in brackets in front of N Er,2,3,5,6 are positive. So we need to ensure that, simultaneously (3.15) W 2 Bϵ > (3.16) AW 3 W 32 Cϵ > (3.17) BW 5 W 52 AW 53 > (3.18) CW 6 W 62 AW 63 BW 65 > This is easily achievable. We choose A = 1. Then we take Equation 3.17 and choose a sufficiently large B, so that the left hand side becomes positive, this is always possible. With this choice of B and the previously chosen A = 1 we take Equation 3.18 and choose C that is large enough to make the left hand side positive. Now we return to Equation We note that for our choice of A = 1 the value of AW 3 W 32 = W 3 W 32 is always positive, on the basis of physical argument that W 3 = W 31 + W 32. In order to satisfy the inequality 3.16 it is enough to choose a sufficiently small ϵ so that, for our previously chosen C the term Cϵ is still smaller than W 3 W 32. Finally we check whether the ϵ found in this way will satisfy Equation 3.15; if it does not we choose an even smaller ϵ that would satisfy 3.15, for example less than W 2 /B, and it would obviously also satisfy Equation Similar arguments are used to satisfy 3.17 and Now we reformulate Equation 3.14 so that we can see the form of V t on the right hand side. (3.19) dv < (W 2 Bϵ) N Er,2 (AW 3 W 32 Cϵ) N Er,3 (BW 5 W 52 AW 53 ) N Er,5 (CW 6 W 62 AW 63 BW 65 ) N Er,6 + Af(t) = (W 2 Bϵ) N Er,2 (W 3 W 32 /A Cϵ/A) AN Er,3 + Af(t) (W 5 W 52 /B AW 53 /B) BN Er,5 (W 6 W 62 /C AW 63 /C BW 65 /C) CN Er,6

9 This journal is The Royal Society of Chemistry 212 -S9- Now we choose α to be the smallest of these prefactors in brackets, so α = min (W 2 Bϵ, W 3 W 32 /A Cϵ/A, W 5 W 52 /B AW 53 /B, W 6 W 62 /C AW 63 /C BW 65 /C). We recall that A = 1, then, for t > t we have dv (3.2) αv t + f(t) Now we use the Gronwall Lemma which gives (3.21) V t e α(t t ) V t + t e α(t s) f(s)ds By point 2 above we have f(t) De k F T N Er t, therefore V t e α(t t) V t + De αt e αs k F T N Ers ds t (3.22) e α(t t) D ( V t + e k F T N Er t e α(t t ) k F T N Er t ) α k F T N Er We assumed here that the denominator α k F T N Er is nonzero, if it is then we should change α slightly. It follows that there exist positive constants M, γ >, γ = min(α, k F T N Er ) such that V t Me γt. We immediately get that N Er,2 (t) Me γt and N Er,3 (t) Me γt. Also N Er,5 M/Be γt and N Er,6 M/Ce γt. 4. Optimal decay rates for N Y b,2 and N Er,2 N Er,6 Now we need to find optimal rates of decay for all functions N Er,2,3,5,6. We start with N Y b,2. From Equation 3.1 we get [ ] (3.23) N Y b,2 (t) = N Y b,2 ()e (k F T N Er t) exp k c2 N Er,2 (s)ds k c3 N Er,3 (s)ds However we have just proved that k c2 N Er,2 (s)ds < and k c3 N Er,3 (s)ds <. Therefore again (3.24) < lim t [ e k F T N Er t N Y b,2 (t) ] < that is γ = k F T N Er is the optimal rate of decay for N Y b,2 For Er we use the following expression for the solution of the first order differential equation dy (3.25) = α(t)y(t) + f(t) It can be expressed using the function G(t, s) given by ( ) (3.26) G(t, s) = exp α(u)du s

10 This journal is The Royal Society of Chemistry 212 -S1- The solution can be expressed as (3.27) y(t) = G(t, )y() + For N Er,3 we obtain (3.28) N Er,3 (t) = G(t, )N Er,3 () + + G(t, s)k F T N Er N Y b,2 (s)ds where (3.29) G(t, s) = exp [ W 3 (t s) k c3 Since < N Y b,2 (u)du < we have G(t, s)f(s)ds (3.3) C 1 e W 3(t s) G(t, s) C 2 e W 3(t s) G(t, s) (W 53 N Er,5 (s) + W 63 N Er,6 (s)) ds s ] N Y b,2 (u)du These considerations indicate that N Er,3 decays exponentially. This is because G(t, )N Er,3 () decays exponentially and there are two other terms that can, potentially make convergence worse. But we have already shown in point 3 that N Er,3 has an exponential upper bound. Hence these two terms can not destroy the exponential decay rate, we will see this in more detail in a moment. We denote γ to be the decay rate for N Y b,2,γ i the decay rate for N Er,i, i = 2, 3, 5, 6. Now we calculate γ 3 the exact decay rate for N Er,3. By using 3.28 we obtain (3.31) e γ 3t N Er,3 (t) e (γ 3 W 3 )t N Er,3 () + e γ 3t + e γ 3t W 63 e W 3(t s) N Er,6 (s)ds + e γ 3t W 53 e W 3(t s) N Er,5 (s)ds k F T N Er e W 3(t s) N Y b,2 (s)ds

11 This journal is The Royal Society of Chemistry 212 -S11- Now we substitute the relevant exponential functions for N Y b,2, N Er,5 and N Er,6. We carry out the integrations to obtain e γ 3t N Er,3 (t) e (γ 3 W 3 )t N Er,3 () + e (γ 3 W 3 )t + e (γ 3 W 3 )t W 63 e (W 3 γ 6 )s ds + e (γ 3 W 3 )t W 53 e (W 3 γ 5 )s ds k F T N Er e (W 3 γ )s) ds = e (γ 3 W 3 )t N Er,3 () (3.32) + e (γ 3 W 3 )t 1 [ W 53 e (W 3 γ 5 )t 1 ] (W 3 γ 5 ) + e (γ 3 W 3 )t 1 [ W 63 e (W 3 γ 6 )t 1 ] (W 3 γ 6 ) + e (γ 3 W 3 )t 1 [ k F T N Er e (W 3 γ )t 1 ] (W 3 γ ) Now the terms on the right must not tend to infinity because γ 3 is optimal; this gives the conditions that γ 3 W 3 (3.33) γ 3 γ 5 γ 3 γ 6 γ 3 γ Here, γ 3 is, by definition, the largest number that satisfies this condition. In addition, because γ 3 is optimal, at least one of the terms on the right must not tend to zero. Hence at least one of the above inequalities is, in fact, an equality. By the same argument but using the rate equations for the N Er,2, N Er,5, N Er,6 we get: γ 2 W 2 (3.34) γ 2 γ 3 γ 2 γ 5 γ 2 γ 6 and at least one of the above inequalities is an equality. We also obtain that (3.35) γ 5 W 5 γ 5 W 6 γ 5 γ + γ 2 where at least one of the above inequalities is an equality, and γ 6 W 6 (3.36) γ 6 γ + γ 3 where at least one of the above inequalities is an equality.

12 This journal is The Royal Society of Chemistry 212 -S12- From the equations 3.33,..., 3.36 and noting that in each case above we have at least one equality we obtain that γ 3 = min (W 3, γ, γ 5, γ 6 ) (3.37) For W 6 > W 5 we obtain: γ 2 = min (W 2, γ 3, γ 5, γ 6 ) γ 5 = min (W 5, W 6, γ + γ 2 ) γ 6 = min (W 6, γ + γ 3 ) (3.38) γ 5 = min (W 5, γ + γ 2 ) Therefore, taking into account 3.33 we find that γ 3 = min (W 3, γ, W 5, γ + γ 2, W 6, γ + γ 3 ) (3.39) = min (W 3, γ ) Similarly, taking into account 3.34 we obtain γ 2 = min (W 2, W 3, γ, W 5, γ + γ 2, W 6, γ + γ 3 ) (3.4) = min (W 2, γ ) After one more substitution we get (3.41) γ 5 = min (W 5, W 2 + γ, 2γ ) (3.42) γ 6 = min (W 6, W 3 + γ, 2γ ) We recall that in γ is the decay rate for Yb ions which can be higher than each of the rates within Er, because it is controlled by (arbitrarily high) Er concentration and the forward energy transfer coefficient. In this case we get that γ 6 = W 6 and γ 5 = W 5. For low enough γ, the terms W 2 + γ, W 3 + γ and 2γ may become rate-limiting. In the case of W 6 < W 5 we get (3.43) γ 5 = min (W 6, W 2 + γ, 2γ ) (3.44) γ 6 = min (W 6, W 3 + γ, 2γ ) For high enough values of γ = k F T N Er we obtain that γ 6 = γ 5, contrary to the experimental observations. This indicates that W 6 > W 5.