PID loop rate versus response time
\$T_0 \leq(\dfrac{1}{6}÷\dfrac{1}{15})T_{95}\$

\$T_{95}\$...95% of the step-response settling time

Author: V.Bobal et al., Digital Self-tuning Controllers

EDIT:

Let's say: \$y=(1-e^{-t/T})\$, where y is the output step response. Now, $$0.95=(1-e^{-T_{95}/T})$$ $$-\dfrac{T_{95}}{T}=ln(0.05)$$ $$\dfrac{T_{95}}{T}\approx3$$

This gives:
\$T_0 \leq(\dfrac{1}{2}÷\dfrac{1}{5})\cdot T\$