# What ratio/percentage of a given object's molecules are absorbing/emitting photons at any given moment?

I've read a couple of times in the past that when the Sun is shining on the grass, even under bright sunlight, only a fraction of the grass's molecules are interacting with the Sun's photons at any given moment.....

Is this true? What is the fraction (for a brightly lit object)?

In order to get a feeling for the involved numbers let us consider a how many atoms are on a surface of $$1m^2$$ of water.
• The mass density of water is $$\rho_m = 1kg/dm^3 = 10^3 kg/m^3$$.
• A single water molecule has the mass $$m = 18 u$$. Thus, $$1mol$$ has the mass $$M=18g = 18\cdot 10^{-3}kg$$. Therefore, we have $$10^3/(18\cdot 10^{-3}) \approx 5 \cdot 10^{4} mol$$ in $$1m^3$$ of water. Since $$1mol$$ consists of $$N_A \approx 6.022 \cdot 10^{23}$$ particles we have $$N\approx 3\cdot 10^{28}$$ molecules in $$1m^3$$ of water. Hence, the number density is given by $$\rho_N = 3\cdot 10^{28}/m^3 = \frac{N}{V}$$. This yields $$V = N/\rho_N$$.
• Next, we have to find a formula for the volume $$V$$. In order to do so, we assume that the molecules are packed like little spheres. The volume per molecule would be $$V=\frac{4}{3}\pi R^3 \approx 4 R^3$$ -- we omit the packaging factor, which is approx 50%. Putting this into the upper formula we obtain the molecular radius $$R$$. It is approximately given by $$4 R^3 = V = \frac{1}{\rho_n}$$ which yields $$R \approx 0.2 nm$$.
• Now that we know the radius per molecule we can calculate the number of molecules on the surface. Taking only the area of $$A=1m^2$$ we have $$N_s=\frac{A}{\pi R^2} \approx 8\cdot 10^{18}$$ molecules building the surface. If we take a leaf and checks it's thickness, we will find that it is much thicker than $$1\mu m$$. Furthermore, if we would take a leaf and make it thinner, such that we can see some light through it, the thickness of the leaf would still be larger than $$1\mu m$$. Therefore, I suspect that more than 1000 layers of molecules are used to fully absorb the light. Hence, the number of molecules involved in the absorption is probably something like $$\tilde{N}_V \approx 10^{22}$$.
• Next, we ask how much power the sun provides per square meter. Wikipedia says it around $$1W/m^2$$. Let's further assume that there are only red photons with wavelength $$\lambda = 600nm$$. The energy per photon would be $$E = h \nu = h c/\lambda \approx 3.3 \cdot 10^{-19}J$$. Thus, the number of photons per square meter and second is given by $$1/E \approx 3 \cdot 10^{18}$$. Thus approximately $$40\%$$ of all "surface molecules" $$N_s$$ and only $$0.04\%$$ of $$\tilde{N}_V$$ molecules interact once with a photon per second.
• Finally, we have to know the "dead time" for each molecule. I mean, if a molecule absorbs a photon, how long does it take to be available for the next photon? I don't know this dead time. However, if it is much shorter than $$1s$$, the effective percentage of interacting molecules becomes small. In contrast, if it takes a "long" time (compared to $$1s$$), we certainly have to take $$\tilde{N}_V$$ instead of $$N_{s}$$ as a reference.