Chemistry · Lesson

Excited states, photochemical pathways and energy transfer

Follow competing pathways after excitation, with event counts and system boundaries explicit. Distinguish first departures, final chemical yield and nonradiative energy transfer.

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By the end, you can

Calculate a lifetime and first-departure branches from supplied rate constants.
Use absorbed-photon counts for a stated product quantum yield.
Interpret a restricted Förster distance model without treating it as universal photon reabsorption.

Excitation opens competing routes

Absorption can prepare an excited state that relaxes radiatively, converts nonradiatively, transfers energy or participates in chemistry. Nuclear motion and coupling between electronic surfaces can help determine which routes occur. An absorbed photon does not by itself guarantee one final product molecule.

Spin multiplicity is 2S+1 for total electronic spin S. Internal conversion connects states of the same multiplicity, while intersystem crossing changes multiplicity. In common molecular usage, fluorescence is a multiplicity-preserving emission and phosphorescence involves different multiplicities. Fast versus slow emission alone is not the defining distinction, and spin mixing affects the idealized rules.

A first-departure model can be solved explicitly

Take one initially prepared state with constant independent departure rates k_f=2.00 × 10⁸ s⁻¹, k_IC=3.00 × 10⁸ s⁻¹ and k_ISC=5.00 × 10⁸ s⁻¹. Assume no return, re-excitation or additional departure channel. The total is k_total=1.00 × 10⁹ s⁻¹, lifetime τ=1/k_total=1.00 ns, and remaining population is exp(−k_total t).

The first-departure fractions are 0.200 fluorescence, 0.300 internal conversion and 0.500 intersystem crossing. By time t, the cumulative probability for channel j is (k_j/k_total)[1−exp(−k_total t)]. An ISC departure can undergo later processes in the receiving state; these fractions do not specify all ultimate chemical products.

Remaining population and cumulative departures share one probability budget

Fraction or probability

Time after preparation / ns

Remaining prepared state
Cumulative fluorescence departure
Cumulative internal-conversion departure
Cumulative intersystem-crossing departure
— Remaining population exp(−t/τ)

Underlying data

Time after preparation / ns	Fraction or probability
0	1 · Remaining prepared state
1	0.36787944117144233 · Remaining prepared state
2	0.1353352832366127 · Remaining prepared state
3	0.049787068367863944 · Remaining prepared state
5	0.006737946999085467 · Remaining prepared state
0	0 · Cumulative fluorescence departure
1	0.12642411176571153 · Cumulative fluorescence departure
2	0.17293294335267748 · Cumulative fluorescence departure
3	0.19004258632642723 · Cumulative fluorescence departure
5	0.1986524106001829 · Cumulative fluorescence departure
0	0 · Cumulative internal-conversion departure
1	0.1896361676485673 · Cumulative internal-conversion departure
2	0.2593994150290162 · Cumulative internal-conversion departure
3	0.2850638794896408 · Cumulative internal-conversion departure
5	0.29797861590027436 · Cumulative internal-conversion departure
0	0 · Cumulative intersystem-crossing departure
1	0.31606027941427883 · Cumulative intersystem-crossing departure
2	0.43233235838169365 · Cumulative intersystem-crossing departure
3	0.475106465816068 · Cumulative intersystem-crossing departure
5	0.49663102650045726 · Cumulative intersystem-crossing departure

Original single-state first-order model with total departure rate 1.00 × 10⁹ s⁻¹ and lifetime 1.00 ns. Initially one excitation is prepared; there is no return, re-excitation or other channel. Only the remaining excited-state population has a dashed curve; cumulative first-departure channels appear as selected markers. The four quantities sum to 1 at each time. Subsequent receiving-state chemistry and final product yields are outside this model.

State the event and count absorbed photons

A quantum yield is the number of a stated event or product divided by the number of absorbed photons, equivalently mol of product divided by mol of photons on a consistent amount basis. It differs from energy conversion efficiency and from a yield calculated using incident photons when not all incident light is absorbed.

For an illustrative 5.00 mW monochromatic beam at 4.00 × 10² nm vacuum wavelength, suppose exactly the stipulated fraction 0.500 is actually absorbed and product quantum yield is 0.100. With reflection and scattering excluded, stable absorption and no relevant depletion over 60.0 s, the absorbed photon rate is about 5.03 × 10¹⁵ s⁻¹ and the formed product amount is about 5.02 × 10⁻⁸ mol. Overall chain-reaction product yields can exceed one per absorbed photon; a probability for one elementary departure remains bounded by one.

A specified nonradiative pair model has a distance law

In the point-dipole weak-coupling Förster model, transfer rate k_T=(1/τ_D)(R₀/r)⁶ and efficiency E=1/[1+(r/R₀)⁶]. Here τ_D is donor lifetime without acceptor, r is pair separation and R₀ includes the applicable spectral-overlap, medium and orientation assumptions. The other donor decay rates are assumed unchanged.

For supplied τ_D=4.00 ns and R₀=5.00 nm, r=R₀ gives E=0.500. At r=10.0 nm, E=1/65≈ 0.0154. This is nonradiative coupling rather than an emitted photon travelling to the acceptor and being reabsorbed. Real-pair distances cannot be inferred from this formula without validating geometry, orientation and the model’s applicable regime.

Worked examples

Separate lifetime from branch probabilities

Use the stated k_f=2.00 × 10⁸, k_IC=3.00 × 10⁸ and k_ISC=5.00 × 10⁸ s⁻¹. Find lifetime, first-departure fractions and remaining population after one lifetime.

Add the rates to obtain 1.00 × 10⁹ s⁻¹.
Take its reciprocal: τ=1.00 × 10⁻⁹ s=1.00 ns.
Divide each channel rate by the total to obtain 0.200,0.300 and 0.500.
At t=τ, exp(−1)≈ 0.367879 remains in the initially prepared state.

τ=1.00 ns; first-departure fractions 0.200,0.300 and 0.500; remaining population after τ≈ 0.368. Later products are not determined.

Convert absorbed power into a product amount

Use the supplied 5.00 mW at 4.00 × 10² nm vacuum wavelength, actually absorbed fraction 0.500, product quantum yield 0.100 and exposure 60.0 s, with the stated constant-absorption/no-depletion assumptions.

The photon energy is hc/λ≈ 4.96611 × 10⁻¹⁹ J.
Absorbed power is 0.500 × 0.00500 W=0.00250 W.
Absorbed photon rate is 0.00250/E_photon≈ 5.03412 × 10¹⁵ s⁻¹.
Multiply by yield 0.100 and time 60.0 s, then divide by N_A to convert product entities into moles.

Product amount≈ 5.02 × 10⁻⁸ mol, or 50.2 nmol. This is an invented photon-budget calculation using absorbed light, not an empirical yield for a named compound.

Change only the distance in a stated pair model

Take τ_D=4.00 ns and R₀=5.00 nm, with the applicable isolated-pair assumptions already specified. Compare E at r=5.00 and 10.0 nm.

At r=R₀, (r/R₀)⁶=1, so E=1/2.
At r=2R₀, that ratio is 64.
E=1/(1+64)=1/65≈ 0.0153846.
The corresponding added transfer rate at 10.0 nm is(1/4.00 ns)/64≈ 3.91 × 10⁶ s⁻¹.

E=0.500 at 5.00 nm and approximately 0.0154 at 10.0 nm. The distance change is meaningful only within the fixed-overlap, orientation and decay assumptions.

Try it yourself

A chemical chain forms 3.00 × 10⁴ product entities from 1.00 × 10⁴ absorbed photons. What is its stated product quantum yield?
Show the answer and reasoning
3.00.
A chain product count can exceed one per absorbed photon without making a single departure probability exceed one.
For the supplied total departure rate 1.00 × 10⁹ s⁻¹, what is the lifetime?
Show the answer and reasoning
1.00 ns.
Take the reciprocal rate.
In the stated Förster model, what is E at r=R₀?
Show the answer and reasoning
0.500.
Both competing terms in the denominator then equal 1.
Does a first-departure ISC fraction alone give the ultimate chemical product yield?
Show the answer and reasoning
No.
The receiving state can undergo later competing processes.

What to carry forward

Name the prepared state and the competing channels.
A first-departure probability differs from a final chemical yield.
Quantum yields count stated events per absorbed photon.
Förster transfer is nonradiative and model-dependent.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.