 Solution

Let $k_D=k_r+k_{nr}$, so the lifetime without transfer is ${\tau }_D=1/k_D$. For a fixed isolated donor–acceptor pair, the specified transfer rate is $k_T=k_D(R_0/R{)}^6$. Inserting this rate into the definition of FRET efficiency givesThe donor population in the presence of the acceptor obeys ${\dot{N}}_D=-(k_D+k_T)N_D$, so its fluorescence decays exponentially with lifetime ${\tau }_{DA}=1/(k_D+k_T)$. ThereforeBoth relations assume the donor's nontransfer rates are unchanged by adding the acceptor, a single approximately fixed separation during each excited-state lifetime, and an available acceptor. The transfer law also presumes the usual weak-coupling dipole approximation and the spectral and orientation conditions contained in $R_0$. If the reference donor is differently quenched, the lifetime difference cannot all be attributed to transfer. Heterogeneous mixtures or a fluctuating separation can produce multiexponential decay, in which case blindly substituting an arbitrary fitted average lifetime into the single-pair formula need not give the ensemble's mean efficiency.