= 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> gives
$$
E=\frac{k_D(R_0/R)^6}{k_D[1+(R_0/R)^6]}
=\boxed{\frac1{1+(R/R_0)^6}}.
$$
The 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 $\tau_{DA}=1/(k_D+k_T)$. Therefore
$$
\frac{\tau_{DA}}{\tau_D}=\frac{k_D}{k_D+k_T}=1-E,
\qquad\boxed{E=1-\frac{\tau_{DA}}{\tau_D}.}
$$
Both 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.
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