Energy / entropy balance in radation process

In a photovoltaic process, energy from sunlight is absorbed, partially transformed into work, partially transformed into heat and partially re-emmited. I'm struggling to understand the derivation by Tom Markvart of the entropy created per absorbed photon throughout the process.

I'm working with the article "Solar cell as a heat engine: energy–entropy analysis of photovoltaic conversion" (DOI 10.1002/pssa.200880460).

The author performs the analysis in two steps.

1. Each photon with energy $u_{in}$ will produce a work $w$ and heat $q_w$, hence $$u_{in} = w + q_w$$ The entropy balance reads $$s_{in} = q_w / T_{0} -\sigma$$ where $\sigma$ is the entropy generated per photon. Considering $s_{in} = u_{in}/T_{in}$, this leads to $$w = (1-\frac{T_0}{T_{in}})u_{in} -T_0 \sigma$$

2. The author then performs a second balance for the absorption-emission process, now taking into account the emitted photons ($u_{out}$, $s_{out}$ per photon) - but not the work anymore $$u_{in} = u_{out} + q_{ph}$$ $$s_{in} = s_{out}/T_0 + q_{ph} / T_{0} -\sigma$$ This leads to $$T_{0} \sigma =(u_{in}- u_{out}) - T_0(s_{in}- s_{out})$$ and finally $$w = u_{out} - T_0 s_{out}$$

There is quite many things I don't understand here

• Why would the created entropy per photon be the same in the two processes ?
• As the number of absorbed and emitted photons are not the same, what does the second energy balance actually means ? What is this $q_{ph}$ heat representing ?