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# Ruby Crystal for Demonstrating Time- and Frequency-Domain Methods of Fluorescence Lifetime Measurements

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## Abstract

We present experiments that are convenient and educational for measuring fluorescence lifetimes with both time- and frequency-domain methods. The sample is ruby crystal, which has a lifetime of about 3.5 milliseconds, and is easy to use as a class-room demonstration. The experiments and methods of data analysis are used in the lab section of a class on optical spectroscopy, where we go through the theory and applications of fluorescence. Because the fluorescence decay time of ruby is in the millisecond region, the instrumentation for this experiment can be constructed easily and inexpensively compared to the nanosecond-resolved instrumentation required for most fluorescent compounds, which have nanosecond fluorescence lifetimes. The methods are applicable to other luminescent compounds with decay constants from microseconds and longer, such as transition metal and lanthanide complexes and phosphorescent samples. The experiments, which clearly demonstrate the theory and methods of measuring temporally resolved fluorescence, are instructive and demonstrate what the students have learned in the lectures without the distraction of highly sophisticated instrumentation.

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1. 1.

Any repetitive function can be expanded in a Fourier series. The expression for this series expansion can be expressed either in real form $$y(t) = A_0 /2 + \sum\nolimits_{m = 1}^\infty {(A_m \,\cos \,m\omega _0 t + B_m \,\sin \,m\omega _0 t)}$$, or complex form: $$y(t) = \sum\nolimits_{m = - \infty }^\infty {C_m \,\exp (im\omega _0 t)}$$. ω0 is the fundamental frequency (the period is $$T = {{2\pi } \mathord{\left/ {\vphantom {{2\pi } {\omega _0 }}} \right. \kern-\nulldelimiterspace} {\omega _0 }}$$), $$m$$ is an integer. $$1/2A_0 ,A_m \;{\rm and}\;B_m$$ are the DC-offset and the mth sine and cosine component amplitudes, and $$C_m = (A_m^2 + B_m^2 )^{1/2}$$.

2. 2.

The Fourier expansion of a pure square wave of period T and minimum and maximum amplitude 0 and A is $$y(t)_{{\rm Sq}\,{\rm Wave}} = A/2 + 2A/\pi \sum\limits_{\quadm = {\rm odd} \hfill \atop {\rm positive}\;{\rm intergers} \hfill} {\frac{1}{m}\sin \left( {\frac{{m2\pi t}}{T}} \right)}$$

3. 3.

The frequency dependence of each component is described in complex notation by $$A_s (i\omega ) = A_{_{s,0} } /(1 + i\tau _s \omega )$$. If there are S different lifetimes, for every frequency ω1, the equations corresponding to Equations 5 and 6 are: $${\rm Modulation} = [ {( {\sum\limits_{s = 1}^{s = S} {\frac{{\alpha _s }}{{1 + (\omega _1 \tau _s )^2 }}} } )^2 + ( {\sum\limits_{s = 1}^{s = S} {\frac{{(\alpha _s \omega _1 \tau _s )}}{{1 + (\omega _1 \tau _s )^2 }}} } )^2 } ]^{1/2}$$, and $$\Phi _{{\rm F},\omega _1 } = \tan ^{ - 1} ( {{{\sum\limits_{s = 1}^{s = S} {\frac{{(\alpha _s \omega _1 \tau _s )}}{{1 + (\omega _1 \tau _s )^2 }}} } \mathord{/ {\vphantom {{\sum\limits_{s = 1}^{s = S} {\frac{{(\alpha _s \omega _1 \tau _s )}}{{1 + (\omega _1 \tau _s )^2 }}} } {\sum\limits_{s = 1}^{s = S} {\frac{{\alpha _s }}{{1 + (\omega _1 \tau _s )^2 }}} }}} \kern-\nulldelimiterspace} {\sum\limits_{s = 1}^{s = S} {\frac{{\alpha _s }}{{1 + (\omega _1 \tau _s )^2 }}} }}} )$$, where α i is the fractional amplitude of the ith frequency component at low frequencies, and $$\sum\nolimits_s {\alpha _s } = 1$$.

4. 4.

The digital transform of a repetitive function can be calculated easily from the data points in one period, if the points are equally spaced. Call the kth data point D k . Then form the following summations: $$F_{\sin } = \sum\nolimits_{k = 1}^K {D_k \sin (\theta _k )}$$, $$F_{\cos} = \sum\nolimits_{k = 1}^K D_k\cos(\theta_k), \sum\nolimits_{k=1}^n X^2_i$$, and $$F_0 = 1/K\sum\nolimits_{k = 1}^K {D_k}$$; $$\theta _k = 2\pi k/K$$, and K is the total number of points in one period. The modulation and phase of the fundamental frequency component is the easily calculated. Defining $$F_\omega = (F_{\sin }^2 + F_{\cos }^2 )^{1/2}$$, we have $$M_{F,\omega } = F_\omega /F_0$$ and $$\Phi _\omega = \tan ^{ - 1} (F_{\sin } /F_{\cos } )$$. This can be extended to higher harmonics.

5. 5.

The power spectrum of a repetitive signal is related to the squares of the Fourier coefficients. Let the Fourier decomposition of the time dependent signal be $$y(t) = A_0 /2 + \sum\nolimits_{m = 1}^\infty {(A_m \cos \,m\omega _0 t + B_m \sin \,m\omega _0 t)}$$ ; see Footnote 1. The total power in the mth component is then defined as $$P_y (m\omega _0 ) = 1/2(A_m^2 + B_m^2 )$$. The last expression plotted versus $$\log (m\omega _0 )$$ constitutes the power spectrum.

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## Acknowlgments

We thank the students of the Physics 590OS optical spectroscopy class for their enthusiastic participation, Ulai Noomnarm and Chittanon Buranachai for reading the manuscript and giving feedback, Jack Bopari, the director of the undergraduate teaching laboratories in the Physics department, for his generosity loaning equipment and the ruby sample, Eugene Colla for assistance assembling Setup 1 for partial fulfillment of the senior thesis project of DEC, and RMC thanks Gerard Marriott for his enjoyable participation years ago when the idea of listening to phosphorescence lifetimes was born.

## Author information

Correspondence to Robert M. Clegg.

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Chandler, D.E., Majumdar, Z.K., Heiss, G.J. et al. Ruby Crystal for Demonstrating Time- and Frequency-Domain Methods of Fluorescence Lifetime Measurements. J Fluoresc 16, 793–807 (2006). https://doi.org/10.1007/s10895-006-0123-7

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• Fluorescence
• Luminescence