How much energy does photosynthesis need per O2 molecule produced?

Question

I am not asking about chlorophyll or the process of photosynthesis. I understand 6H2O + 6CO2 + light energy -> C6H12O6+ 6O2 That info is available everywhere.

What I am asking is How Much light energy in watts or some other appropriate unit is required for production of each six molecules of O2?

Thank you

Jerry

Answer 1

The efficiency of the photosynthesis processes combined is roughly equivalent to a Diesel engine's, approx 35% of total of received energy, under ideal conditions.

Efficiency = energy in sugar / energy absorbed

So find out the heat of oxydation for glucose, you know the number of oxygen molecules generated, and the efficiency of the process.

Answer 2

Firstly a Watt is not a unit of energy. A Watt is the rate of energy usage. The appropriate unit is the Joule.

Since each molecule of CO2 processed requires about 8 to 12 photons. Assuming that our photons are 170nm then for 6 molecules and

6*10*h*c/170nm = 7 × 10-17 joules

But bear in mind that for real plants, 94% of photons are wasted.

Answer 3

This is not simple.  Fasten your seat belt.

It takes 4 electrons to convert one CO2 to carbohydrate (the final product of photosynthesis).  We know that at least 2 photons of visible light are used to boost each electron to an energy level capable of reducing CO2.  We also know (as you can see from your equation) that for each CO2 fixed the plant releases one O2.  Thus, we need to put in 6 x 4 x 2 = 48 photons of visible light to produce the 6 O2 molecules, theoretically.  Under ideal conditions some inevitable losses occur so that a more realistic estimate for the photon requirement might be 60 (this is in agreement with some of the most respected values in the photosynthesis literature).  So, now we need to calculate the energy content in those 60 photons of visible light.

A good median estimate for the wavelength of visible light (w) would be 550 nanometers (nm).  Energy content of a photon (E) is given by Planck's Law E = hc/w where h is Planck's constant (6.63 x 10^-34 Joule-sec) and c is the speed of light (3 x 10^17 nm/sec).  So, for a single photon of green (550 nm) light the energy content is

E = (6.63 x 10^-34) x (3 x 10^17)/(550) = 3.63 x 10^-19 Joule.

The energy content in 60 green photons is

60 x 3.63 x 10^-19 = 2.17 x 10^-17 Joule.