# Why doesn't hydrogen gas exist in Earth's atmosphere?

The root mean square velocity of hydrogen gas at room temperature is:
Gas constant: $$R=8.31\ \mathrm{J\ K^{-1}\ mol^{-1}}$$
Molar mass of hydrogen gas: $$M=2.02\times10^{-3}\ \mathrm{kg/mol}$$
\begin{align} v &= \left(\frac{3\times8.31\ \mathrm{J\ K^{-1}\ mol^{-1}}\times300}{2.02\times10^{-3}\ \mathrm{kg/mol}}\right)^{\frac12}\\ &= 3356.8377\ \mathrm{m/s}\\ &= 3.356\ \mathrm{km/s} \end{align}
The escape speed of Earth is $$11.2\ \mathrm{km/s}$$, which is larger than the root mean square velocity of hydrogen gas. But still, hydrogen gas doesn't exist in Earth's atmosphere. Why? Have I made any mistakes in my calculations?

• but there is... Jan 2 '17 at 21:57
• Wiki : Atmospheric escape
– J...
Jan 3 '17 at 14:20
• I think you really have two questions here. One is why the Earth isn't a gas giant, where any answer except "because" really needs to be asked on the astronomy site. The other is can you, starting from a primordial Earth that post giant-impact has an atmosphere that maybe (or maybe not: astrobio.net/geology/earths-early-atmosphere ) had lots of hydrogen in the form of methane, water vapor, ammonia, &c, work out where it went? Jan 3 '17 at 20:57
• The main reason belongs on chemistry.se, not physics.se: hydrogen reacts with things. Most of Earth's atmospheric hydrogen wound up in the oceans.
– Mark
Jan 3 '17 at 22:38
• The result of the calculation made in the question is mathematically incorrect. I get 1924.1817 rather than 3356.8377. I also don't understand where the factor 3 in the numerator comes from. Shouldn't it be 2 (hard to know really because no formula is presented or motivated)? Then the result would be 1571.0878 .
– jkej
Jan 4 '17 at 11:37

The answer to your question comes from Maxwell distribution of speed of the hydrogen molecules. If you take a look at this graph, about the speed of a particle $v$ and the probability of that speed $w$, you can see that there is a non-zero probability that the speed of a certain molecule is greater than the root mean square speed $v_{\mathrm{qm}}$ of that distribution.

In particular, you can calculate the probability that the speed of a certain molecule is greater than the escape velocity of Earth $v_{\mathrm{esc}} = 11000\,\mathrm{m/s}$. Under the hypothesis of ideal gas, this probability is: $$\mathcal{P} = \int_{v_{\mathrm{esc}}}^{\infty} w(v) dv$$ defining the probability density function: $$w(v) = 4 \pi \left( \frac{m}{2 \pi k_{\mathrm{B}} T}\right)^{3/2} e^{-\frac{mv^2}{2 k_{\mathrm{B}} T}} v^2$$ where $m$ is the mass of the hydrogen molecule, $k_{\mathrm{B}}$ is the Boltzman constant, $T$ is the absolute temperature (in Kelvin) and $v$ the speed.

By doing this calculation (please let me use other values I have already calculated, at this point it should be easy to apply that formula for every value) at a temperature $T=270\,\mathrm{K}$ and with a mass $m_{H_2} = 2 \cdot 1.67 \times 10^{-27}\,\mathrm{kg}$, we get that the root mean square speed is $v_{\mathrm{qm}}=1830\,\mathrm{m/s}$. On the other hand, the probability that a particle has a speed six times greater than this value (it is approximately the escape velocity of Earth) is $2 \times 10^{-9}$. This value is small, but not negligible; in a long enough time, every molecule of hydrogen will escape from Earth's atmosphere.

For a last example, you can consider the mass of the molecule of oxygen. Its mass is 16 times bigger than the hydrogen molecule and its root mean square speed is 4 times lower and 24 times lower than the escape speed. The probability to get enough speed to escape Earth's atmosphere is approximately $10^{-40}$: really, really small.

This is an intuitive, approximate explanation of why the molecular hydrogen concentration in Earth's atmosphere is really low, while the concentration of other, heavier molecules is higher. There is a difference in the probability, for a certain molecule, to have a speed greater or equal to the escape speed of the Earth. This influences the rate at which these kind of molecule escape the atmosphere and therefore will lead to a different equilibrium (a different concentration) for each molecule.

(source: F. Ciccacci, "Fondamenti di Fisica Atomica e Quantistica")

• While the $O_2$ probability is small, it is not zero, so the above reasoning would apply to $O_2$ as well, surely? I would imagine that there must be some rate equation involved rather than just a non zero escape speed probability? Jan 3 '17 at 19:36
• @copper.hat of course there should be some balance between the oxygen escaping from the atmosphere and mechanisms of replacing it, in order to keep its concentration constant. What I calculated is just the probability for a single molecule to have a certain velocity, then you have to consider how many molecules have enough speed to escape and how many of them are replaced. At equilibrium, the same number of molecules escaping will be somehow replaced, to maintain the concentration constant. Jan 3 '17 at 21:12
• What I tried to say is that we can estimate the number of molecules in the atmosphere in about 10^44 (as done here theweatherprediction.com/habyhints3/976) and consider that just the 21% is oxygen ( en.m.wikipedia.org/wiki/Atmosphere_of_Earth), therefore just a few oxygen molecules will escape from Earth in a certain interval of time, and each of them will be replaced if the concentrations in the atmosphere are at an equilibrium. Jan 3 '17 at 21:13
• This way understates the odds of escape--the outer layers of the atmosphere are hot. Jan 3 '17 at 21:55
• @copper.hat Yup, and indeed, oxygen does escape from the atmosphere. It's just that even if you ignored all the other effects and assumed an entirely inert atomsphere and only accounted for thermal escapes, there still wouldn't have been enough time for oxygen to escape yet. The amount of oxygen in the atmosphere changes all the time, it's no constant - though escape is still a negligible part of this over the existence of the Earth. Even geological time is too short for some things :) And then you realize that atmospheric oxygen is just a tiny fraction of Earth's oxygen, and being recycled. Jan 4 '17 at 11:18

The equilibrium concentration of hydrogen in the atmosphere is about 0.5 ppmv (parts per million per volume) according to Wolfram Alpha (with a tip of the hat to @AccidentalFourierTransform). This is a result of mechanisms of production, and destruction (chemical reactions, escape). You are right that the RMS velocity of hydrogen is less than the escape velocity - but that doesn't matter.

The thing to keep in mind is that not all molecules have the same velocity. The Maxwell-Boltzmann velocity distribution is of the form

$$p(v) = \sqrt{\left(\frac{m}{2\pi kT}\right)^3}4\pi v^2e^{-\frac{mv^2}{2kT}}$$

You can plot this for hydrogen and get the following:

This tells us that there is a small but finite probability of an individual molecule reaching escape velocity. Once that molecule is removed, it won't be coming back, and the velocity distribution will be re-established (because the atmosphere will remain at the same temperature). So there is a slow "leak" of hydrogen from the atmosphere. It is sufficient that that leak be faster than the rate of generation of new hydrogen, for the concentration to drop; eventually, equilibrium is reached.

Because Nitrogen and Oxygen have much heavier molecules, they represent a much larger fraction of the atmosphere. The probability that one of their molecules will reach escape velocity is many orders of magnitude smaller than the probability for hydrogen. Thus, over "geological time", almost all hydrogen disappears from the atmosphere.

Note - if you plot the above on a semilog scale you can see just how small the probability of the high velocities is - and then you remember that the upper atmosphere (above 100 km or so) is actually significantly hotter than the air closer to the surface - under certain conditions, the upper part of the thermosphere can reach temperatures over 2000 C during the day. At that temperature, the probability of hydrogen escaping increases very significantly. This is illustrated in this plot:

Those hot hydrogen molecules (and atoms) high up in the outermost layers of the atmosphere have a really good chance of escaping...

Final note - unless the mean free path of the molecule is very large, it will undergo another collision and most likely be sent back to earth. This is why only the temperature of the very outermost layers of the atmosphere matter for this calculation.

• There is an image on Wikimedia which gives the same $\mathrm{H}_2$ concentration (can click on larger PNG renditions if your browser will not let you zoom). However in extreme altitudes above the turbopause (100 km) you can find layers where the relative concentration of hydrogen is much closer to 100%. Of course the atmosphere is extremely thin up there. So the concentration 0.5 ppm is for the part below the turbopause where the gasses mix well. Jan 3 '17 at 15:43
• @JeppeStigNielsen - very interesting, thanks for sharing. Jan 3 '17 at 15:46
• @JeppeStigNielsen Much closer to 100%? Definitely not for molecular hydrogen. For atomic hydrogen on the other hand, we can start talking percentages in the exosphere (>500 km).
– jkej
Jan 4 '17 at 0:16
• @Floris The thermosphere is not where molecules typically escape from and even if some may escape from the upper part of the thermosphere (depending on where you draw the upper boundary) it is definitely much hotter than 360 K in this part of the thermosphere.
– jkej
Jan 4 '17 at 2:33
• @jkej - thanks, I had pulled my number from a plot that stopped at 100 km; looking a little deeper I saw temperatures could indeed be significantly hotter... updated answer accordingly. Jan 4 '17 at 4:22

The other answers are correct in terms of the principal reason that lighter molecules are much more likely to escape the atmosphere. However, it seems that the premise of the question (and perhaps also of some of the answers and comments) is based on an incorrect model of atmospheric escape.

Molecules from most parts of the atmosphere would never escape regardless of velocity.

The picture below is taken from this document (recommended reading) and illustrates that escape is basically only possible for molecules above 500 km (the exosphere). Molecules below that will (most likely) collide with other molecules before escaping.

So, what is the atmospheric composition like in the exosphere?

The figure below shows the mixing ratios for the main components between 0 and 1000 km based on data taken from NRLMSISE-00. The exosphere is shaded in grey. As you can see, the most abundant molecules in the exosphere are atomic oxygen, helium and hydrogen. Hence, it is kind of a moot point to discuss escape velocities of molecular hydrogen and oxygen. Of course, the mass ratio happens to be the same since they are both diatomic molecules (but note that a large part of the atomic hydrogen in the exosphere originated from water in the lower atmosphere rather than from molecular hydrogen).

It is also very hot in the exosphere (~1000 K) so making calculations for surface temperatures is also technically wrong (though again, it may happen to give the same qualitative result when comparing two species).

Yes, lighter molecules (or atoms) are much more likely to escape the atmosphere. The by far biggest reason for this is that they are much more likely to reach escape velocity (see the other answers). But another contributing factor is that they are much more likely to reach the upper layers of the atmosphere where they can escape because of gravitational stratification, which starts already in the heterosphere. However, most molecular species will be photolyzed into their atomic constituents by UV radiation before they reach altitudes high enough to escape, so it is really the mass of the atoms that matter. Another factor that is very important for which species reach the top of the atmosphere is of course chemical processes in the atmosphere, but that is a whole different chapter.

• Very nice complement to the existing answers, filling in important gaps. Jan 4 '17 at 4:24
• @Farcher Thank you! I have now edited so that the link works.
– jkej
Oct 21 '18 at 12:06