Past exam of the biological physics course of the University of Oxford / 2016 / 6 / a / Solution

A Jablonski diagram distinguishes electronic excitation from subsequent relaxation. Photon absorption takes a molecule from the electronic ground state $S_0$ to a vibrationally excited level of $S_1$ or a higher singlet state. Rapid vibrational relaxation and internal conversion usually populate the lowest vibrational levels of $S_1$. Fluorescence then emits a photon on a transition to $S_0$, often ending in one of its excited vibrational levels. Nonradiative relaxation competes with emission; intersystem crossing can also populate triplet states. Energy lost to relaxation and solvent reorganization commonly makes the fluorescence maximum occur at lower photon energy, hence longer wavelength, than the photon absorption maximum: the Stokes shift.

The intrinsic fluorescence quantum yield is the number of photons emitted by fluorescence per absorbed excitation. For a simple excited state with radiative rate $k_r$ and other decay rate $k_{nr}$,A nearby acceptor adds a nonradiative energy-transfer channel, not an emitted photon flying from donor to acceptor. The electric dipole-dipole interaction transfers the donor excitation while returning the donor toward its ground state and exciting the acceptor. The acceptor then relaxes and may fluoresce. The donor yield becomesFor the acceptor, distinguish its intrinsic yield per acceptor excitation, $Q_A=k_{r,A}/(k_{r,A}+k_{nr,A})$, from the number of acceptor photons per donor excitation. Transfer increases the supply of acceptor excitations, giving sensitized emission $E{Q}_A$ per donor excitation in the ideal model. It does not by itself change $Q_A$ if the acceptor's own decay rates remain unchanged. Environmental changes, saturation or additional quenching can change those rates, but increased acceptor intensity alone does not establish an increased intrinsic fluorescence quantum yield.