Acerca de termodinámica biológica con algunas conversiones

Es un ejercicio del libro "termodinámica biológica" de Hayne. Entiendo el ejercicio pero en la resolución aparece un "6" dividiendo la energía (ya verán en la resolución) y no entiendo por qué aparece ese "6". Esta en ingles, es algo malo. Acá les dejo el ejercicio con su resolución. Gracias.

The energy of oxidation of glucose to H2O and CO2 is –2870 kJ
mol
1. At least 2870 kJ mol
1 are needed to synthesize glucose from H2O and CO2. ¿How many 700 nm photons must be absorbed to fix one mole of CO2? If the actual number needed is 3 to 4 times the minimum, what is the efficiency of the process?

Resolución:

At least 2870 kJ mol–1 are needed to synthesize glucose.

The energy of a 700 nm photon is E = hc/λ = 6.63 × 10–34 J s × 2.998 × 108m s–1 / (700 × 10–9 m) = 0.0284 × 10–17 J

One mole of such photons has an energy of 0.02839 × 10–17 J × 6.02 × 1023 = 171 kJ.

The energy required to fix one mole of CO2 is 2870 kJ / 6 = 478 kJ. --->(A este 6 me refiero)

The number of 700 nm photons required is 478 kJ mol–1 / 171 kJ mol–1 = 2.80.

Because 3–4 times that many photons are required, the efficiency of the process is
1/4–1/3, or 25–33%. In comparison with industrial chemical processes, the level of
Efficiency of biotic glucose synthesis is very good.