Do atoms recoil when emitting a photon?

[sol47739]

TL;DR Summary

Do atoms recoil? If yes how is that consistent with all of the energy difference between the energy levels going to the photon?

Hello! I have some questions regarding the photon emission and whether the atom recoils or not.
When an electron in an atom emits a photon. One can calculate the energy of the photon by the difference between the energy levels from where it left to which it returned.

Let’s say it jumped from n=2 to n=1, the angular momentum are multiples of Plancks constant and when having calculated the energy difference E2-E1 one gets the energy of the emitted photon.

I understand all of this and it makes sense.
Now to my question:

I have read on a couple of websites and in some sources(not the best sources quora, physics stack exchange for example, Wikipedia) that the atom undergoes a recoil when emitting a photon. And that that is in order to conserve momentum like in a gun shot, the same amount of momentum that gets transferred to the bullet also gets transferred to the gun and hands of the person holding the gun. This means that the bullets momentum = the recoil momentum. I can’t understand how this could be the case in the atom and the photon problem since if this would be the case and the atom actually would recoil because of the emitted photon the emitted photon would only carry an energy/momentum half of the difference between the energy levels since half of it goes to the recoil of the atom. So if the atom actually would be recoiling the photon energy would be (E2-E1)/2 as would the recoil. As we know this is not the case. So from my own reasoning the atom do not recoil and this is because things work different on the quantum level. Am I correct or not by that assumption?

Or has it been proved that atoms recoil? In case it has how is it explained that the photon energy is exactly the same as the energy difference and not half of it amounting for the recoil? In the case they do recoil is it only the electron or the entire atom?
From my understanding I also feel that if the atom would recoil it would be extremely difficult to cook things down since although if an atom emitted a photon so that the atom is in the lowest energy level it would till have half of the momentum as recoil left, which can’t be “emitted” in the way a photon can. I also feel this would contradict the Blackbody spectrum. Since from what I know one an atom is in the lowest energy level it stands still. And when you heat something up the material absorbs photons and the atoms get their momentum from photons and start to move and when in equilibrium the exchange of these photons will result in the Blackbody spectrum. If the atoms only would emit half of the energy as a photon and the rest as recoil the Blackbody spectrum would be much more shifted towards the longer low energy wavelengths, as we know this is not the case.
So to summarize:

Do atoms recoil? If yes how is that consistent with all of the energy difference between the energy levels going to the photon?

If you know of some reliable material or scientific papers and experimental evidences about this subject, I am thankful too if writing me.

 

[PeterDonis]

I have read on a couple of websites and in some sources(not the best sources quora, physics stack exchange for example, Wikipedia)

The obvious response to this is that you should be looking at better sources; for example, have you tried looking at any textbooks that discuss this?

 

[sol47739]

The obvious response to this is that you should be looking at better sources; for example, have you tried looking at any textbooks that discuss this?

I haven’t found any good textbooks discussing exactly this topic, I also asked for recommendations in my question.

 

[PeterDonis]

the atom undergoes a recoil when emitting a photon.

Yes, that is correct; as you say, it's required by conservation of momentum.

if this would be the case and the atom actually would recoil because of the emitted photon the emitted photon would only carry an energy/momentum half of the difference between the energy levels

Not at all. Try doing the actual math instead of waving your hands.

from what I know one an atom is in the lowest energy level it stands still.

I have no idea where you are getting that from. Consider an atom in a gas at room temperature. The atom can be in its ground state and still be moving around.

 

[sol47739]

Yes, that is correct; as you say, it's required by conservation of momentum.Not at all. Try doing the actual math instead of waving your hands.I have no idea where you are getting that from. Consider an atom in a gas at room temperature. The atom can be in its ground state and still be moving around.

Here I found some calculations. As I see the momentum of the recoiled atom is very small. Are there any known experiments that has confirmed this?

 

[PeterDonis]

Here I found some calculations

Where? Please give a reference. You can't just post something and not tell us where it comes from.

As I see the momentum of the recoiled atom is very small

No, the energy of the recoiled atom is very small compared to the energy of the emitted photon.

The momentum of the recoiled atom is exactly equal in magnitude (and opposite in direction) to the momentum of the emitted photon; this must be true by conservation of momentum.

Are there any known experiments that has confirmed this?

There are many, many experiments which have confirmed the equations that govern photon emission from atoms, including the recoil of the atom.

 

[sol47739]

Here I found some calculations. As I see the momentum of the recoiled atom is very small. Are there any known experiments that has confirmed thi

Yes, that is correct; as you say, it's required by conservation of momentum.Not at all. Try doing the actual math instead of waving your hands.I have no idea where you are getting that from. Consider an atom in a gas at room temperature. The atom can be in its ground state and still be moving around.

How is that consistent with the Blackbody spectrum if an atom in its ground state can move around?

 

[sol47739]

How is that consistent with the Blackbody spectrum if an atom in its ground state can move around?

From where has it got it’s momentum from then?

 

[PeterDonis]

How is that consistent with the Blackbody spectrum if an atom in its ground state can move around?

Why do you think it wouldn't be?

 

[PeterDonis]

From where has it got it’s momentum from then?

I have no idea what you are asking. From where has what got its momentum? What are you talking about?

 

[DaveC426913]

If photons didn't have momentum that could be exchanged with atoms then solar sails wouldn't work.

 

[sol47739]

Why do you think it wouldn't be?

I have no idea what you are asking. From where has what got its momentum? What are you talking about?

If photons didn't have momentum that could be exchanged with atoms then solar sails wouldn't work.

Since what I have learned and understand is that you could think like this:
When Wien and Planck tried to figure out the interactions between light(electromagnetic waves and photons) the reasoning went like this:
First we have to remember the classical model of thermal equilibrium in the class oak theory of gases. That is the Maxwell Boltzmann distribution:
“The Maxwell–Boltzmann distribution applies fundamentally to particle velocities in three dimensions, but turns out to depend only on the speed (the magnitude of the velocity) of the particles.”https://en.m.wikipedia.org/wiki/Maxwell–Boltzmann_distribution
Rayleigh and Jeans tried to account for the Blackbody spectrum based on classical means that an accelerated charge emits electromagnetic radiation continuously, as we know this led to the ultraviolet catastrophe. (http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html)
Wien on the other hand came up with Wien’s law by relating the intensity of a particular wavelength at a temperature in the black body spectrum to the probability of velocity(as I quoted the Maxwell Boltzmann distribution depend only on the velocity in an ideal gas with all particles having the same mass) of the particles in the Maxwell Boltzmann distribution. This worked pretty well and although being an assumption back then it worked pretty well just because it is about photons, which carries momentum, whose energy thus momentum are directly proportional to the frequency or inversely proportional to the wavelength.
Later on Planck came to the conclusion that the energy exchange between the electromagnetic field and matter occurs in discrete steps and Plancks constant was born, which then give the right Blackbody spectrum function, which includes stimulated emission. In stimulated emission the emission gets stimulated by another photon, while in spunatnous emission it happens „without a cause“ because of the vacuum fluctuation, that is why it is random.
Thus from my understanding and what I have learned the momentum of atoms and molecules has it’s origin in photons that are „bound“ to the atom. And the atom slows down when emitting a photon.
All kinds of heat transfer (Heat transfer is classified into various mechanisms, such as thermal conduction, thermal convection, thermal radiation, and transfer of energy by phase changes.) are indeed caused by photons if you go to the fundamental level.

Solar sails would work since one can think of the photon momentum being bound to the atom.

Or has one seen an atom recoil experimentally?

 

[PeterDonis]

from my understanding and what I have learned the momentum of atoms and molecules has it’s origin in photons that are „bound“ to the atom.

Nothing in what you gave in your post up to this point says anything like this. So I still don't know where you are getting this from or what you think the issue with black body radiation is.

has one seen an atom recoil experimentally?

This question has already been answered.

 

[sol47739]

Nothing in what you gave in your post up to this point says anything like this. So I still don't know where you are getting this from or what you think the issue with black body radiation is.This question has already been answered.

While the photon that carries linear momentum is bound to the atom the atom goes faster as fast as the kinetic energy the photon that is bound to it carries. When the photon leaves the atom the momentum leaves the atom and it goes slower or stop moving in case it got emitted to the ground state.

 

[PeterDonis]

While the photon that carries linear momentum is bound to the atom the atom goes faster as fast as the kinetic energy the photon that is bound to it carries.

The photon is not carrying linear momentum while it's bound to the atom. That's why the atom recoils when it emits the photon: because the photon gains linear momentum when it's emitted and the atom has to gain an equal and opposite linear momentum to satisfy momentum conservation.

The energy that an emitted photon carries is not taken from the atom's kinetic energy. It's taken from the atom's potential energy.

 

[Vanadium 50]

How is that consistent with the Blackbody spectrum if an atom in its ground state can move around?

How does this have anything to do with your original question?

 

[Dale]

half of it goes to the recoil of the atom

If you actually work it out you will find that way less than half goes into the atom. It isn’t 0 but it is pretty small.

 

[sol47739]

If you actually work it out you will find that way less than half goes into the atom. It isn’t 0 but it is pretty small.

So it is true in a sense that the recoil alone can’t account for the velocities of atoms at a certain temperature? But photons are bound with momentum to the atom?

The photon is not carrying linear momentum while it's bound to the atom. That's why the atom recoils when it emits the photon: because the photon gains linear momentum when it's emitted and the atom has to gain an equal and opposite linear momentum to satisfy momentum conservation.

The energy that an emitted photon carries is not taken from the atom's kinetic energy. It's taken from the atom's potential energy.

Thanks for you explanation so far. According to you, what causes the momentum of the atoms when you heat up something with radiation? What is the original cause when you have let’s say a solid ice cube and heat it up with radiation from the stove? Isn’t it then that the photons in the radiation from the stove gives linear momentum to the atoms and if you were to be able to suddenly remove all photons from the system(which of course isn’t possible, but if) the the water would freeze again? Aren’t these atoms just being there and photons causing the momentum exchange? If it follows from recoil or from the photons being bound to the atom is the question? But photons are the origin of momentum.

 

[PeterDonis]

it is true in a sense that the recoil alone can’t account for the velocities of atoms at a certain temperature?

I have no idea why you would even think this. The velocities of atoms at a certain temperature are because of the temperature; that's what temperature means, that the atoms have kinetic energy. The kinetic energy comes from whatever heat source is putting the container full of atoms at that temperature.

But photons are bound with momentum to the atom?

Go read the first paragraph of my post #15 again.

 

[PeterDonis]

What is the original cause when you have let’s say a solid ice cube and heat it up with radiation from the stove?

How are you going to heat it up with radiation from a stove?

If the air is at normal room temperature, that will melt the ice by conduction and convection, probably much, much faster than radiation anyway.

A better example would be heating something in space, where it's vacuum so there is no conduction or convection.

Isn’t it then that the photons in the radiation from the stove gives linear momentum to the atoms

In a case where radiation is the only source of heat present, then yes, you could say photons are adding kinetic energy, and therefore also linear momentum, to the atoms of whatever is being heated up. But that's one particular case.

photons are the origin of momentum.

In the one particular case I described above, yes. But not in general, no.

 

[sol47739]

I have no idea why you would even think this. The velocities of atoms at a certain temperature are because of the temperature; that's what temperature means, that the atoms have kinetic energy. The kinetic energy comes from whatever heat source is putting the container full of atoms at that temperature.Go read the first paragraph of my post #15 again.

Yes, you are correct that it is because of the temperature. But I am trying to understand temperature on a deeper level. What is temperature? Internal energy and movements of atoms. And what is the original cause of this, if it is about neutral atoms and molecules, which you can’t accelerant electric fields. Are there any other sources than radiation/photons that can increase the temperature(movement of the atoms) in a ensemble of neutral atoms? If the ensemble collides with another ensemble of course but still that ensemble carry more internal energy that also had to come from somewhere, which are from photons. Since we know that all materials radiates that isn’t absolutely zero and that according to the Blackbody spectrum as long as it can be assumed to be a approximate black body.

According to you you can have an ensemble of atoms at a very high temperature and all atoms are in their lowest energy state and the ensemble won’t radiate at all? Am I understanding you correctly or am I failing again somewhere in the understanding?

 

[PeterDonis]

What is temperature? Internal energy and movements of atoms.

Yes.

what is the original cause of this

The fact that atoms at finite temperature have nonzero kinetic energy. There is no original answer to where that energy comes from; it's always been there, ever since the Big Bang. In fact the expansion of the universe has cooled things off tremendously.

According to you you can have an ensemble of atoms at a very high temperature and all atoms are in their lowest energy state and the ensemble won’t radiate at all?

I have said no such thing. Any object at finite temperature will emit radiation. Even if the individual atoms are all in their ground states, that just means the energy due to the finite temperature is in other degrees of freedom, such as the linear motions of atoms or the vibrations of molecules.

 

[sol47739]

How are you going to heat it up with radiation from a stove?

If the air is at normal room temperature, that will melt the ice by conduction and convection, probably much, much faster than radiation anyway.

A better example would be heating something in space, where it's vacuum so there is no conduction or convection.In a case where radiation is the only source of heat present, then yes, you could say photons are adding kinetic energy, and therefore also linear momentum, to the atoms of whatever is being heated up. But that's one particular case.In the one particular case I described above, yes. But not in general, no.

But in that particular case then, is it the recoil that causes the atoms to start moving or is it the photon being bound to the atom?
1.Let’s say a photon gets absorbed what happens with the momentum of the entire atom then?
2.While the photon is “bound” to the atom how is it affected by it?
3.When the photon leaves and gets emitted how does that affect the momentum of the atom?

 

[PeterDonis]

in that particular case then, is it the recoil that causes the atoms to start moving

Of course. What else could it be?

the photon being bound to the atom?

There is no photon "bound to the atom". The photon that gets emitted, gets created in the process of emission. It's not bouncing around inside the atom waiting to get out.

 

[sol47739]

Yes.The fact that atoms at finite temperature have nonzero kinetic energy. There is no original answer to where that energy comes from; it's always been there, ever since the Big Bang. In fact the expansion of the universe has cooled things off tremendously.I have said no such thing. Any object at finite temperature will emit radiation. Even if the individual atoms are all in their ground states, that just means the energy due to the finite temperature is in other degrees of freedom, such as the linear motions of atoms or the vibrations of molecules.

Yes but that still happens in quantized steps as photons. And vibration of molecules are also quantized, so if such a vibration is “activated” the molecule actually isn’t in its true ground state. Maybe in an electronic ground state but not in the ground state of all possible things that can get excitied in the molecule.
You said that the energy due to the finite temperature is in other degrees of freedom, I agree, but those degrees are still “converted” in to photons or were originally exicited by photons. So the amount of photons bound to a system is directly proportional to the temperature.
So if photons can be emitted from a system you can’t really say that it is in its true ground state.

 

[sol47739]

Of course. What else could it be?There is no photon "bound to the atom". The photon that gets emitted, gets created in the process of emission. It's not bouncing around inside the atom waiting to get out.

But as

If you actually work it out you will find that way less than half goes into the atom. It isn’t 0 but it is pretty small.

said. I wonder is that recoil enough since it is very small, is that enough to amount for the Maxwell Boltzmann distribution?

 

[PeterDonis]

that still happens in quantized steps as photons

Emitting radiation? Sort of, yes, but they won't all have the same energy or momentum. It will be a continuous spectrum.

vibration of molecules are also quantized, so if such a vibration is “activated” the molecule actually isn’t in its true ground state.

Nothing will be in a "true ground state" at any temperature above absolute zero.

those degrees are still “converted” in to photons or were originally exicited by photons

Not necessarily, no. I mentioned conduction and convection earlier. In most everyday situations, those are vastly larger sources of heat transfer than radiation.

the amount of photons bound to a system is directly proportional to the temperature

Nonsense. There aren't any "photons bound to a system". Go read the last paragraph of my post #24 again.

 

[PeterDonis]

I wonder is that recoil enough since it is very small, is that enough to amount for the Maxwell Boltzmann distribution?

No. The calculation you gave in post #5 makes that clear; the recoil energies are much smaller than thermal energies.

 

[sol47739]

No.

What then does give enough energy for it to correspond to Maxwell Boltzmann distribution. Was Dale Right that the recoil is negible?

And the Blackbody spectrum(the part amounting to spontaneous emission Wien’s distribution) is directly proportional to the Maxwell Boltzmann distribution. Thus there is a strong relationship between the energy/wavelength of a photon and the amount of atoms in the system haveing the probability of having a velocity thus momentum that corresponds to the amount of photons emitted. So I can’t understand how one doesn’t see that there is a strong relationship between the velocity of atom in the Maxwell Boltzmann distribution and the wavelengths of the photons in the Blackbody spectrum at that particular temperature.

Not necessarily, no. I mentioned conduction and convection earlier. In most everyday situations, those are vastly larger sources of heat transfer than radiation

But convection and conduction works because here two system interacts one with higher temperature(higher photon density exchange )and one with lower temperature (lower photon density).

 

[PeterDonis]

What then does give enough energy for it to correspond to Maxwell Boltzmann distribution.

Go read the second paragraph of my post #22 again.

You don't seem to be reading responses very carefully. I strongly suggest that you stop and take a step back and think for a while about what you have been told.

Was Dale Right that the recoil is negible?

Yes. Once again, the reference you yourself gave in post #5 clearly shows that. Did you actually read what it said?

there is a strong relationship between the velocity of atom in the Maxwell Boltzmann distribution and the wavelengths of the photons in the Blackbody spectrum at that particular temperature.

This is just another way of saying that temperature is the average kinetic energy of the atoms.

It does not, however, say that photons are "bound to the atoms", which is what I told you was nonsense.

But convection and conduction works because here two system interacts one with higher temperature(higher photon density exchange )and one with lower temperature (lower photon density).

No, convection and conduction are always present whenever systems are in contact. Whether any net heat transfer occurs by these modes will depend on the temperatures of the two systems. But the processes themselves are taking place (the atoms at the contact surface are still interacting and exchanging energy) even when the two systems are at the same temperature and aren't exchanging any net heat.

 

[sol47739]

Go read the second paragraph of my post #22 again.

You don't seem to be reading responses very carefully. I strongly suggest that you stop and take a step back and think for a while about what you have been told.Yes. Once again, the reference you yourself gave in post #5 clearly shows that. Did you actually read what it said?This is just another way of saying that temperature is the average kinetic energy of the atoms.

It does not, however, say that photons are "bound to the atoms", which is what I told you was nonsense.No, convection and conduction are always present whenever systems are in contact. Whether any net heat transfer occurs by these modes will depend on the temperatures of the two systems. But the processes themselves are taking place (the atoms at the contact surface are still interacting and exchanging energy) even when the two systems are at the same temperature and aren't exchanging any net heat.

I do read your responses. The problem I think is that I am too critical and I do not just accept something. And you are so to speak textbook correct. But since I doubt textbooks before accepting something, my questions might sound stupid.

Thanks for the exchange. I will think about it.

 

[PeterDonis]

you are so to speak textbook correct

My answers are not just repeating what textbooks say. I am using my understanding of the subject matter and ordinary common sense reasoning. You should be able to do the same to judge what I am saying. But you need to have a valid understanding of the subject matter to base your reasoning on.

 

[Vanadium 50]

I do read your responses.

About half of your messages are posted less than three minutes after the one you are responding to, and half of those are posted in a minute.

Is that enough time to read, think about and pose a thoughtful response?

 

[Dale]

those degrees are still “converted” in to photons or were originally exicited by photons.

This is not generally true. The degrees of freedom include atomic electron orbitals, molecular electron orbitals, linear kinetic energy, rotational kinetic energy, molecular vibrations, longer range vibrational modes, and probably others I missed. These modes can be excited by photons of the appropriate energy, but to assume that they always are, originally are, or even usually are is incorrect. They can also be excited through collisions, phonons, relaxation, and other similar processes. Matter simply doesn’t work as simplistically as that.

So the amount of photons bound to a system is directly proportional to the temperature.

I suspect that your concept of bound photons is probably wrong. When a photon excites some degree of freedom (regardless of what kind of degree of freedom it is) that photon ceases to exist. It is not bound, it is destroyed. If it were not, then its energy would be unavailable to excite the degree of freedom.

 

[hutchphd]

Also it is even incorrect to think that the "photon" interacts only with the electron. It interacts with the atom because the system has separate charges and therefore dipole moment that allows conservation of angular momentum and energy. The spectrum of transitions is complicated (as @Dale mentions) because the atom has other degrees of freedom. But the system atom+photon conserves energy,momentum,angular momentum, charge,..........etc