Emission, luminescence and energy pathways

Absorbed energy can emerge as light

Luminescence involves light emission following excitation, through electronic or other excited states, rather than ordinary thermal radiation from a hot body. Some energy can instead become heat or enter other nonradiative pathways. Emitted wavelength, lifetime and efficiency describe different features: lifetime concerns how quickly an excited population decays, while efficiency concerns the fraction taking a specified output pathway.

A sample absorbs 100 photons and emits 60. The simple worked case gives a photon yield of 0.60, but two questions remain: how much energy did those 60 photons carry, and how long did emission take? A photograph alone answers neither. We will extend that count account with explicitly supplied photon energies, then use a separate pulse model to follow the timing.

Photoluminescence uses absorbed light to prepare excitation. Electrical injection or a chemical reaction can also supply an excited population, but their complete input accounts are different. An absorbed-optical-photon denominator is useful for the photoluminescence examples here; it is not automatically the electrical efficiency of a light-emitting device. All new numerical inputs below are fictional teaching records, not specifications for a named phosphor, dye or semiconductor.

Excitation changes a state; relaxation chooses a pathway

An excitation puts a molecule, ion or solid-state electronic system into a state with more energy than the relevant lower state. Allowed energies and transition probabilities depend on bonding and structure. A radiative transition emits a photon; nonradiative relaxation transfers energy through another channel, such as coupling to molecular vibrations or lattice vibrations. A lattice vibration is collective atomic motion, not an emitted light photon.

In a common photoluminescence sequence, absorption first reaches a higher excited level. Some energy is transferred to the surrounding material before the system reaches the state from which it emits. The later photon can then carry less energy than the absorbed photon. The first figure models exactly this sequence; it is not a universal level diagram or a complete band structure.

A defect can introduce local electronic states and pathways. In a semiconductor, capture at a defect can transfer electronic energy into several lattice-vibration excitations instead of a photon. Some defects can also participate in radiative transitions. Their chemical identity, geometry and coupling matter; counting defects or observing a darker sample does not alone determine which pathway changed.

Transition constraints also matter. In molecular photochemistry, phosphorescence involves a change of spin multiplicity, meaning the quantum classification of the combined electronic spin. Such a radiative return is often less probable than a prompt fluorescence return. A slow radiative channel still competes with nonradiative loss; a long-lived signal is not automatically a high photon yield. The simple one-population model below does not identify fluorescence versus phosphorescence or resolve several interacting states.

Efficiency needs a denominator

A photon yield compares emitted photons with a specified absorbed or incident photon count. Photoluminescence quantum yield conventionally uses absorbed photons; an incident-photon basis is a different efficiency measure. Energy efficiency additionally accounts for photon energies and other inputs. A bright image cannot establish these values without excitation and collection information.

Define N_inc as incident excitation-photon count, N_abs as the count absorbed by the specified sample and N_out as emitted luminescence photons leaving it over the stated observation. The photoluminescence quantum yield is Φ = N_out/N_abs. The incident-photon yield is N_out/N_inc = (N_abs/N_inc)Φ. These counts need excitation light separated from emission, the same sample boundary and an adequate observation interval.

An energy ratio weights photons by their energies. For monochromatic excitation and a single supplied emission energy, the absorbed-optical-energy efficiency is η_abs = N_out E_em/(N_abs E_ex) = Φ(E_em/E_ex). Its incident-energy counterpart includes the absorbed fraction as well. These two η quantities are defined for the optical account here, not a universal device efficiency.

If emission spans a band, add the energy in its wavelength intervals rather than multiplying every photon by the energy at the brightest peak. A detector signal must also be corrected for its wavelength-dependent response. A photon-number spectrum and a radiant-power spectrum use different weights; the peak of one need not represent the mean photon energy. Reflection, absorption and transmission develops collection of the excitation light; the present account follows energy after absorption.

Close the energy account as well as the photon account

Extend the 100-to-60 count example with an original repeat-averaged record, normalised to 100 absorbed excitation photons. There are 250 incident photons at vacuum wavelength 400 nm, 100 are absorbed and 60 luminescence photons leave at 620 nm. The other 150 excitation photons leave without absorption; their directions are unspecified. Counts here are supplied averages, not a claim that each small trial has exactly the same branching outcome.

Photon energy is E_γ = hf = hc/λ_vac, where h is Planck’s constant, f is frequency, c is vacuum light speed and λ_vac is vacuum wavelength. Use the deliberately rounded product hc = 1240 eV·nm for this arithmetic: E_ex = 3.10 eV and E_em = 2.00 eV. One electronvolt is an energy unit; use the rounded conversion 1 eV = 1.60 × 10⁻¹⁹ J if joules are needed. This photon relation does not require a refractive-index calculation.

In the stipulated route, every absorbed photon first prepares one excitation at 3.10 eV. Fast relaxation transfers 1.10 eV to the material and leaves one 2.00 eV excitation. Sixty return radiatively and emit 60 × 2.00 = 120 eV. Forty instead return nonradiatively and transfer 40 × 2.00 = 80 eV. The initial relaxation transferred another 100 × 1.10 = 110 eV. Thus absorbed energy is 310 eV = 120 eV of emitted light + 190 eV transferred nonradiatively.

For this completed account, stipulate no remaining stored excitation, chemical change, other output or extra energy input; the nonradiative transfer ultimately becomes thermal energy. Even the radiative events transferred 1.10 eV before emission. The photon yield is 60/100 = 0.60, but η_abs = 120/310 = 12/31 ≈ 38.7%. The missing 61.3% of absorbed energy is not merely the 40% of events that failed to emit.

Relative to the incident beam, photon yield is 60/250 = 0.24 and optical-energy efficiency is 120/(250 × 3.10) ≈ 15.5%. The full energy check is 775 = 465 + 120 + 190 eV: unabsorbed excitation light, emitted luminescence and nonradiative transfer. These energies would all acquire the same factor when converted to joules.

The longer emission wavelength in this example follows from its supplied energy loss. It is not a rule that all luminescence has these two wavelengths or these losses. Real materials may have several emitting levels, energy transfer, reabsorption or additional energy inputs. Name the applicable account before using a one-photon picture.

A decay time describes all exits from the emitting population

Now use a separate ideal pulse experiment. Immediately after the fast preparation stage, N₀ = 1000 independent excitations occupy one emitting population at time t = 0. Excitation then stops. There is no later feeding, re-excitation, interaction between excitations, photon reabsorption or change of state. Every radiative event produces one escaping photon; the only alternative is an irreversible nonradiative return. This at-most-one-photon model is deliberately limited.

Let k_r and k_nr be positive constant transition rates per occupied excitation, in s⁻¹ or ns⁻¹. For a sufficiently short interval dt, the expected radiative count is k_r N dt and the nonradiative count is k_nr N dt. Adding the losses gives dN/dt = −(k_r + k_nr)N. Constant-rate loss gives N(t) = N₀ exp(−t/τ), where τ = 1/(k_r + k_nr). The exponential exp describes repeated fractional loss, rather than removal of a fixed number each nanosecond.

τ is the population-decay time constant and mean residence time in this model. At t = τ, N/N₀ = 1/e ≈ 0.368; it is not the time at which every excitation emits or disappears. An individual event can happen earlier or later. One nanosecond is 10⁻⁹ s; for example, 0.0600 ns⁻¹ = 6.00 × 10⁷ s⁻¹.

The radiative share of every loss interval is k_r/(k_r + k_nr). Therefore the eventual photon yield is Φ = k_r/(k_r + k_nr) = k_rτ, provided each absorbed photon prepared one such excitation and every radiative photon escapes. The expected cumulative emitted count is N_γ(t) = Φ[N₀ − N(t)]; its long-time limit is ΦN₀. A fixed detector factor makes a corrected intensity trace proportional to k_rN(t), so a normalised single-exponential intensity decay can identify the same τ under these assumptions.

Equal lifetimes can hide different light yields

The table supplies three original rate pairs for separate copies of the pulse model. A has total rate 0.100 ns⁻¹, τ = 10.0 ns and Φ = 0.60. B has both rates halved, so τ = 20.0 ns but Φ is still 0.60. Slowing all exits changes timing without changing the branch fraction. C has the same total rate and lifetime as A but a smaller radiative share, Φ = 0.20.

For N₀ = 1000, the initial emitted-photon rates k_rN₀ are 60.0, 30.0 and 20.0 photons/ns for A, B and C. Their eventual emitted counts are 600, 600 and 200. The two-panel figure shows the distinction: A and C have identical normalised population curves, while their cumulative photon outputs differ. B produces the same eventual number as A more slowly.

At 10.0 ns, A has about 367.879 excitations remaining and has emitted 379.272 photons. Another 252.848 excitations have returned nonradiatively. The three expected counts sum to 1000. A detector gate ending then has not collected A’s eventual 600 photons. B has emitted only 236.082 by the same time, so a short gate can make its equal eventual yield look smaller.

If a single-population model is independently justified and Φ and τ are measured on compatible states/boundaries, the two rates follow: k_r = Φ/τ and k_nr = (1 − Φ)/τ. Lifetime alone supplies only their sum. Increasing only k_nr would shorten τ and reduce Φ, but observing that pattern does not uniquely prove a particular defect or quencher; the preparation, radiative coupling and optical collection may also have changed.

Separate emitted light from what the instrument records

Let D be background-corrected detected photon count and ε the independently supplied effective collection-and-detection fraction for the emission under study. In the simple monochromatic, linear detector model, D = εN_out. Collection angle, filters, emission direction and detection probability can all enter ε. It is not automatically one because the apparatus includes an integrating sphere.

For the separate averaged observations in the next table, corrected counts give Φ_A = 6/(0.100 × 100) = 0.60 and Φ_C = 10/(0.250 × 200) = 0.20. C gives more detector counts, although its supplied quantum yield is smaller. It absorbs more excitation and is collected more efficiently. The table reuses the A/C branch fractions but not the pulse figure’s common N₀ = 1000.

An emission spectrum locates outgoing photon energies; a corrected decay constrains timing; an excitation/absorption measurement supplies the input denominator. Together these constrain more than an image does. To distinguish a changed decay mechanism from a changed optical path, compare the absorbed excitation, calibrated spectral collection and corrected time response at a matched state. A reference material or an apparatus check addresses a different question from chemical or structural identification.

A real decay may contain several populations, delayed feeding or a detector response comparable to the decay time. A fitted exponential is then not automatically a unique intrinsic τ, and its normalised shape need not reveal the eventual yield. Evaluate excitation pulse width, detector timing, background, observation window, reabsorption and sample state before making that inference. Calibration, uncertainty and repeatability supplies the measurement framework; Charge carriers and electronic energy bands and Doping, defects and interfaces develop relevant solid-state mechanisms.

Luminescence can coexist with ordinary thermal radiation from a warm sample. The distinction concerns how the emitting population is prepared and whether it is in thermal equilibrium, not a claim that a luminescent sample has no temperature. A wavelength, lifetime or brightness comparison alone does not establish composition or stability in service.

Worked examples

Try the reasoning

1. Can heating compete with light emission?

   Show the answer and reasoning

   Yes.

   Nonradiative relaxation can dissipate excitation energy.

2. Does the longest emission lifetime mean the highest yield?

   Show the answer and reasoning

   No.

   Rates of radiative and nonradiative pathways both matter.

3. A new completed one-photon account absorbs 100 photons at 500 nm and emits 80 at 620 nm. Use the same rounded hc. Find absorbed-photon yield, absorbed-energy efficiency and energy transferred nonradiatively if no other output or stored energy remains.

   Show the answer and reasoning

   Φ = 0.80; η_abs = 20/31 ≈ 64.5%; nonradiative transfer is 88.0 eV.

   Excitation energy is 2.48 eV per photon, so input is 248 eV. Light carries 80 × 2.00 = 160 eV. The difference is 88 eV. A higher photon fraction still needs energy weighting; these stipulated wavelengths do not identify a real material.

4. A justified single-population model has τ = 4.00 ns and Φ = 0.50 at the same state and photon boundary. Infer k_r and k_nr in ns⁻¹ and s⁻¹. Could τ alone have supplied each rate?

   Show the answer and reasoning

   Each rate is 0.125 ns⁻¹ = 1.25 × 10⁸ s⁻¹. The lifetime alone fixes only their sum, 0.250 ns⁻¹.

   Use k_r = Φ/τ and k_nr = (1 − Φ)/τ. Many other positive pairs share that total. The inference also requires that absorbed photons create the one measured population and that radiative photons are counted at the stated boundary.

5. For the original separate pulse models A and B at common N₀ = 1000, a detector gate ends at 10.0 ns. Calculate emitted counts up to that time. Has B’s smaller gated output established a smaller eventual quantum yield?

   Show the answer and reasoning

   A gives about 379.272 photons; B about 236.082. Both still have eventual yield 0.60 in the supplied model.

   Use N_γ(t) = 600[1 − exp(−t/τ)] with τ = 10 and 20 ns. The gate samples different fractions of their decays. Correct finite-window loss using a justified time response, rather than silently treating the gated count as the complete output.

6. A separate record absorbs 200 photons and detects 12. Its effective ε is only known to lie from 0.100 to 0.200, an admissible range rather than a probability interval. Under the same one-population model with τ = 10.0 ns, bound Φ and k_r. Does the record prove Φ > 0.50?

   Show the answer and reasoning

   Φ lies from 0.30 to 0.60; k_r lies from 0.0300 to 0.0600 ns⁻¹. The stated information does not establish Φ > 0.50.

   Φ = 12/(200ε) decreases when ε increases. Divide its endpoint range by τ for k_r. The complementary k_nr changes to preserve the same 0.100 ns⁻¹ total; these are correlated model inferences, not independently measured rates or confidence bounds.

7. A report declares an at-most-one escaping photon per absorbed photon model, no extra excitation input, N_abs = 80, detected count 50 and independently established ε = 0.50. What corrected yield follows, and may it be clipped to one?

   Show the answer and reasoning

   N_out = 100 and Φ = 1.25. This conflicts with the declared model; clipping would conceal the mismatch.

   Check absorption/background/count calibration and whether the stated one-photon mechanism actually applies. The bound of one belongs to this model. The calculation alone does not prove that every possible luminescent process must obey that same photon-count bound.

8. After pulse excitation a corrected signal initially rises and then decays in two clearly different stages. May its late fitted τ be inserted into Φ = k_rτ as though it describes the entire initial population?

   Show the answer and reasoning

   Not from that observation alone. The single prepared population with constant rates and no later feeding has not been established.

   Later feeding, several populations or an unresolved instrument/pulse response can change the curve. Check those possibilities and the integrated output. One late decay constant does not supply every population, branch or photon count.

What to carry forward

• Trace excitation and relaxation pathways.
• State photon or energy denominators.
• Photon-number yield and optical-energy efficiency weight different quantities; retain excitation and emission energies.
• In the one-population constant-rate model, τ = 1/(k_r + k_nr) and Φ = k_rτ. Equal τ does not fix the radiative fraction.
• A finite detector window or a different collection fraction can change a brightness comparison without changing the eventual yield.
• Use spectral, absorption and time evidence together before assigning a physical pathway.