# Photons and gases (pressure and gravity)

1. Nov 22, 2008

### Дьявол

I got one question for you. In the book that I learn from (Physics textbook), I read that "at very small dimensions of the particles (like the molecules of the gases) the force of the light pressuring at small particles can reach the force of gravity which acts on the particles upon."
If the light can "send" the gas particles out of the Earth I was wondering how gas particles exist on the Earth? Maybe, I misunderstood something, but how is possible that the pressure can be so "big"?
For example:
The Standard atomic weight of one atom of hydrogen is 1.00794(7) g·mol-1. To form the H2 gas it needs 2 atoms, so the mass of one molecule will be 2.00794(7) g·mol-1.
If the "ordinary" light (the light that we see) has 600 nm wavelength, 2 eV (energy), 3,6*10-36 and a impulse of 1,1*10-27, how "big" force acting on the hydrogen molecule it needs to "counter" the force of gravity?

P.S I am a total beginner so please use plain expressions and plain English as possible since I won't understand anything.

2. Nov 23, 2008

### buffordboy23

Interesting question. This is my thinking.

First, light particles carry momentum. When they collide with gas particles, they impart an impulse force on the gas molecules. Most gas molecules are diatomic, so this means a change in the rotational and kinetic energies of the gas molecule. For the molecule to leave Earth, I would expect that the resulting change in translational kinetic energy of the gas molecule must be large enough so that the molecule's velocity is greater than Earth's escape velocity.

The reason Earth can retain its atmosphere then is because there are a whole bunch of gas molecules. The gas molecule that once had sufficient velocity and direction to escape the Earth will likely collide with many other molecules, reducing its net velocity due to conservation of energy and momentum laws. At higher altitudes, where the atmosphere is less dense, your scenario would be more likely to happen.

I did some rough calculations for your quoted quantities. Assuming the hydrogen molecule has no initial translation kinetic energy, then about 1200 eV of translational kinetic energy must be transferred to the particle for it to escape. Consider the fact that a 600 nm photon has an energy of 2 eV. So, it appears that a lot of collisions for this scenario would be needed. Now, an x-ray photon (about 1 nm) can carry the required energy.

3. Nov 23, 2008

### Дьявол

Thanks for the reply.
Your way of thinking is great. Also I think that gas molecule should have constant speed above 9.8 m/s2 to get out of the Earth. I am wondering how did you calculated how "much" energy is required to leave the Earth after some constant collisions with other gas molecules?
Again, your right about the x-rays. An x-ray with 0.1 nm got 12,4*103 eV energy and mass for about 2,2*10-32kg, carrying a momentum of 6,6*10-24 kg*m/s.
Thanks again for the explanation.

4. Nov 23, 2008

### buffordboy23

9.8 meters-per-seconds-squared is the Earth's gravitational acceleration, not the gas molecule's velocity. During the photon/gas-molecule collision, the impulse force on the gas molecule is likely to be of very short duration, so the gas molecule likely experiences a large acceleration; this is likely many orders of magnitude larger than the Earth's gravitational acceleration. After the collision, the gas molecule's velocity is still affected by the Earth's gravity, so the gas molecule's velocity would decrease over time.

After the collision, the gas molecule had to be accelerated so that its velocity is about 11.3 km/s. With this speed the gravitational acceleration of the Earth is not strong enough to keep the gas molecule in the atmosphere.

Earth's escape velocity is about 11.3 km/s. So, the gas molecule needs about 1200 eV in translational kinetic energy K = (1/2)mv^2, where 1 J = 1.6 x 10^-19 eV. I pulled the mass from your quoted values; mass of diatomic hydrogen molecule = (mass per mole) / (# of molecules per mole).

EDIT: I found on the internet that particle collisions have maximum duration of about 50 microseconds. So, the average acceleration of the gas molecule during collision can be determined; a = (change in velocity) / (change in time). Assuming zero as the initial velocity of the gas molecule and using earth's escape velocity gives an acceleration of 2.24 x 10^8 m/s^2; this seems very large but remember that this acceleration is very brief, 50 microseconds.

Last edited: Nov 23, 2008
5. Nov 23, 2008

### Дьявол

I think you mean it receives short impulse but big acceleration, because of its lightness, since if we act with greater impulse force, we will do much bigger acceleration.

Can you please explain me what do you mean by that part?

$$K=\frac{m*v^2}{2}$$
In this case:
n=m/M
M=1.00794(7) g·mol-1
1mol got 6.02214179(30)×1023 atoms.
In this case we know that 6.02214179(30)×1023 atoms weight 1.00794(7)g.
We need two of the hydrogen atoms.
So two atoms of hydrogen would weight 2 x 0.1673723 x 10-23g
Now do you find the kinetic energy by putting the mass m=0.334744 x 10-23g and velocity v=11.3 km/s ?

6. Nov 23, 2008

### buffordboy23

Impulse has units equivalent to those for momentum. Its definition is impulse = (force) x (time of application). So, in a sense impulse can be thought of as a change in momentum:
http://en.wikipedia.org/wiki/Impulse

An incoming photon will have some initial momentum. During the collision it will exert a force on the gas molecule. This force is really small (about 5 x 10^-15 N), but since the gas-molecule has a very small mass, its subsequent acceleration is really large. A photon with greater momentum (smaller wavelength) would exert a larger force and acceleration on the gas-molecule.

Earth's gravitational acceleration is 9.8 m/s^2. However, during the 50 microseconds at which the impulse occurs, the acceleration of the gas-molecule is about 2 x 10^8 m/s^2; the ratio of this acceleration to earth's gravitational acceleration is about 2 x 10^7 (seven orders of magnitude larger).

Now, after photon and gas-molecule are no longer in contact, the only force acting on the gas-molecule is the gravitational force. If the gas-molecule does not have velocity equal to or larger than Earth's escape velocity at this point in the analysis, it can't escape the Earth.

Yes. The units will be in joules, so divide by 1.6 x 10^-19 J/eV to convert to eV.

It is possible for one photon, like an x-ray, to impart this kinetic energy upon the gas-molecule. Also, many photons of lower energy, such as those from the visible part of the spectrum, could accomplish the same result as long their overall effect on the gas-molecule's velocity tended to be in the same direction (the photon collisions would push the gas-molecule in the same direction).

With the random motion and collisions of gas molecules in the atmosphere, I can't determine at this moment how likely such a scenario would be. My initial thoughts are that it is very unlikely, unless we are talking about the upper atmosphere. We can probably use the density of earth's atmosphere to determine the spacing between adjacent molecules, and ultimately, the probability of such a scenario happening, if we make the assumption that once accelerated the gas-molecule cannot collide with any other molecules in the atmosphere.

Last edited: Nov 23, 2008
7. Nov 23, 2008

### Дьявол

The mass turned into kilos is 3.34744 x 10-27kg. The speed turned into m/s is 11200 m/s. So the kinetic energy now is: Ek=209951436.8 x 10-27 J=2.099514368 x 10-19 J = 1.31219648 eV
So it needs 1.31219648 eV to escape the earth, if hypothetically there aren't any molecules on the way. When impulse force is applied does the whole energy of the photon pass away to the gas molecule?

Now for the acceleration.

a= (change in velocity) / (change in time)

a= 11200 m/s /50*10-6s

a=224*106=2.24*108m/s2

The density of air at sea level is about 1.2 kg/m3(1.2g/L)

Could you please confirm my calculations?

Now for your post. I absolutely agree with you. I think that the mean role here have the lightness of the gas molecules and the impulse of the photon (the wave length, since f=c/λ and pф=hf/c).

Thanks again for the great posts.

8. Nov 23, 2008

### buffordboy23

Your calculations look good. I actually made a mistake with one of my earlier calculations. I forgot that your mass value for hydrogen was in grams, and I never converted to kilograms. So, the required kinetic energy is on the order of 1.2 eV rather than 1200 eV like I said before. So, this is within the range of photons in the visible spectrum, assuming they transfer a large portion of their momentum/energy to the gas-molecule.

It makes sense to assume that the collision is completely elastic. However, I can't even offer any advice on what amount of energy/momentum is transferred. If all of the energy of the photon was transferred, then this would imply that the photon now has a wavelength of infinity after the collision:

$$E = \frac{hc}{\lambda}$$

Last edited: Nov 23, 2008
9. Nov 24, 2008

### Дьявол

You are right. If the energy of the photon after the collision would be zero, then one of h or c should be zero, which is impossible, or even to be close to zero λ should be very big.