= Solution
The stationary occupation ratio is $p_1/p_2=2$. Balance of the two <Probability flux>[probability currents] gives $p_1k_{12}=p_2k_{21}$, and <Boltzmann distribution>[Boltzmann occupation] of the trapped macrostates gives $p_2/p_1=e^{-(G_2-G_1)/(k_BT)}$. Therefore
$$
\boxed{\frac{k_{12}}{k_{21}}=\frac12,\qquad G_2-G_1=k_BT\log2.}
$$
The sketch has two wells separated by 1 micrometre, with the second free-energy level higher by $k_BT\log2$ and an intervening escape barrier. <Free energy> includes the volume of each trapped basin; equal minimum potential heights alone would not ensure equal occupations if basin shapes differed.
\Image[/past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2017-trap-tilt-and-carbon-response.png]
{title=Two-state free-energy tilt and the distinct fast and slow timescales of a carbon-pulse response}
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