Photon Momentum/Energy: Is It Equal?

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In summary, the conversation discusses the relationship between the energy and momentum of a photon, and how this applies to photon absorption by atoms. It is noted that the energy of a photon depends on the reference frame, and that the change in momentum of an atom due to photon absorption is negligible due to the atom's much larger mass. The concept of "atomic momentum" is defined as Mv, where M is the atomic mass, and it is mentioned that the angular momentum of a photon is often referred to as "spin" rather than "orbital" momentum.
  • #1
jamie.j1989
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Hi,

Is the energy a photon carries with respect to its frequency the same as that of its momentum energy? My understanding is that it is by the energy relations,

$$hf=E$$
$$E^2=p^2c^2+m^2c^4=p^2c^2$$ for a photon with ##m=0##, frequency ##f## and plank constant ##h## . So we have from both,

$$p=\frac{hf}{c}$$ Where ##p## and ##c## are the momentum of the photon and speed of light respectively.

The issue I have is with photon absorption, as an atom absorbs a photon of frequency ##f## and excites an electron to the next highest state, that state being ##hf## higher than the previous, then how does the photon also transfer momentum to the atom if all its energy is taken by the excitation of the electron?
 
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  • #2
jamie.j1989 said:
The issue I have is with photon absorption, as an atom absorbs a photon of frequency ##f## and excites an electron to the next highest state, that state being ##hf## higher than the previous, then how does the photon also transfer momentum to the atom if all its energy is taken by the excitation of the electron?
A photon with energy exactly equal to the required delta will fail to excite the atom. The kinetic energy of the recoiling atom needs to be provided.

However, if one has a body with atoms in random thermal motion, some of them will be moving toward the oncoming photons and some will be moving away. This has the effect of smearing out the absorption band -- some atoms will be able to absorb low energy photons and some will be able to absorb high energy photons.
 
  • #3
But if the atom is moving towards the photon, it see's a blue shift in the photon's frequency, if this blue shifted frequency matches that of delta then surely it can be absorbed even though the energy of the photon is actually less, the extra energy coming from the recoil of the atom in the opposite direction?
 
  • #4
jamie.j1989 said:
But if the atom is moving towards the photon, it see's a blue shift in the photon's frequency, if this blue shifted frequency matches that of delta then surely it can be absorbed even though the energy of the photon is actually less, the extra energy coming from the recoil of the atom in the opposite direction?
Yes.

Note that the notion of the photon's "actual" energy is not well founded. The energy of a photon depends on your choice of reference frame. There is no one particular energy that is more actual than the others. If one chooses a center-of-momentum reference frame in which the combined momentum of the atom plus photon is zero then the photon's energy can all be used up in increasing the energy level of the electron.
 
  • #5
jamie.j1989 said:
The issue I have is with photon absorption, as an atom absorbs a photon of frequency ##f## and excites an electron to the next highest state, that state being ##hf## higher than the previous, then how does the photon also transfer momentum to the atom if all its energy is taken by the excitation of the electron?

The entire atom takes up the momentum of the photon. However, since the atom (with its nucleus) is so much more massive, and the momentum of a photon is extremely small, the change in momentum of the atom is negligible.

Zz.
 
  • #6
ZapperZ said:
The entire atom takes up the momentum of the photon. However, since the atom (with its nucleus) is so much more massive, and the momentum of a photon is extremely small, the change in momentum of the atom is negligible.

Zz.
The change in atomic momentum is equal to the photon momentum.
The associated change in kinetic energy is much smaller than the photon's energy.
 
  • #7
my2cts said:
The change in atomic momentum is equal to the photon momentum.
The associated change in kinetic energy is much smaller than the photon's energy.

I'm not sure what you mean by "atomic momentum". The change in the orbital angular momentum quantum number is due to the photon's orbital momentum. But it also has a linear momentum k. This is what is absorbed by the atom and caused it to negligibly recoil. In photoabsoroption in solids, the lattice ions absorbed this linear momentum.

Zz.
 
  • #8
ZapperZ said:
I'm not sure what you mean by "atomic momentum".
Mv, where M is the atomic mass.
The change in the orbital angular momentum quantum number is due to the photon's orbital momentum.
I believe it is due to the photon's intrinsic momentum aka spin.
But it also has a linear momentum k. This is what is absorbed by the atom and caused it to negligibly recoil.
Mv=hk
In photoabsoroption in solids, the lattice ions absorbed this linear momentum.
The entire crystal moves with momentum hk after the absorption.
Indeed, this involves negligible kinetic energy.
 
  • #9
my2cts said:
Mv, where M is the atomic mass.

I believe it is due to the photon's intrinsic momentum aka spin.

Mv=hk

The entire crystal moves with momentum hk after the absorption.
Indeed, this involves negligible kinetic energy.

Then what are you disputing or arguing against with my post?

This is very puzzling.

Zz.
 
  • #10
Your question about the meaning of "atomic momentum" surprised me but your account is adequate.
One thing, I would call the angular momentum of a photon "spin" and not "orbital" momentum.
 

1. What is photon momentum/energy?

Photon momentum/energy is a concept in physics that refers to the amount of momentum and energy carried by a single photon, which is a fundamental particle of light. It is a measurement of the particle's mass and velocity, and is often expressed in units of kilograms per meter per second (kg*m/s) for momentum and joules (J) for energy.

2. Is photon momentum equal to its energy?

Yes, according to the theory of relativity, the momentum of a photon is equal to its energy divided by the speed of light. This is expressed in the equation p = E/c, where p is momentum, E is energy, and c is the speed of light (3.00 x 10^8 m/s). Therefore, photon momentum is directly proportional to its energy.

3. How is photon momentum/energy measured?

Photon momentum/energy can be measured through various experimental techniques, such as the Compton effect, which involves measuring the change in momentum and energy of a photon after it collides with a stationary particle. Other methods include using the photoelectric effect or the photoelectric equation, which relates the energy of a photon to the work function of a material.

4. What is the significance of photon momentum/energy?

Photon momentum/energy plays a crucial role in understanding the behavior of light and other electromagnetic radiation. It helps explain phenomena such as the photoelectric effect, where photons transfer their energy to electrons, and the Doppler effect, where the momentum of a photon can be affected by the relative motion of its source and observer. Additionally, the concept of photon momentum/energy is vital in the development of technologies such as solar cells and lasers.

5. Can photon momentum/energy change?

Yes, the momentum and energy of a photon can change through interactions with other particles or through changes in its own properties, such as its frequency or wavelength. For example, when a photon is absorbed by an atom, its energy and momentum are transferred to the atom's electrons, causing them to move and potentially emitting a new photon with different properties. Similarly, when a photon passes through a medium, such as a lens or a prism, its momentum and energy can be altered due to the medium's refractive index.

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