Energy of a Photon: Frequency, Intensity, Momentum

In summary, the conversation discusses the relationship between the energy and wavelength of a photon and how changing the frequency of a light source affects the number of photons emitted and the energy flux. It also explores the concept of a photon being both a wave and a particle, and the implications of its momentum and energy. The conversation also delves into the concept of inertial mass and how it relates to photons. The question of how higher frequency translates into higher energy is also discussed, with the explanation that a more energetic photon vibrates at a higher rate.
  • #1
Super Kirei
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I was reading an elementary treatment of the photoelectric effect and it says that the energy of a photon is related to it's wavelength by λ=h/p where p is the momentum. So this would imply that changing the frequency of a light source while keeping the intensity constant means that at higer frequencies fewer more energetic photons are emitted than at lower frequencies so that the energy flux remains constant. But what does it mean for a photon to be more energetic? The speed of light is fixed at c, so how does its momentum vary?
 
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  • #2
Hi Super Kirei! :smile:
Super Kirei said:
… But what does it mean for a photon to be more energetic?

a photon is a wave as well as a particle

you're probably familiar with the doppler shift … if you move towards a wave, its frequency increases, and it appears more energetic (because more wave-crests are "bumping" into you each second)

if you move towards a photon, its frequency increases, and it appears more energetic

a photon is (usually) produced when an electron changes orbit round an atom or molecule …

the loss of energy of the electron equals the energy of the photon produced (in the frame of reference of the atom or molecule)

all photons are produced at the same speed, but with different energies :smile:
 
  • #3
On a small additional note.. and just to make things a bit more fun, light doesn't have a constat speed. The one you refer to is the speed of light in vacuum. When light passes through matter or has other things affecting it, it changes speed and loses or gains energy like everything else. There have even been experiments which have brought light to a virtual stand still :)
 
  • #4
I'm still trying to reconcile the billiard ball conception of the photon with the wave conception of a photon. So, if a photon has momentum, as the de Broglie equation seems to imply, then I imagine that would imply it has mass. But if it has mass then it's momentu varies with velocity and it's momentum is related to it's energy so that higher energy means higher momentum which means higher velocity, so that can't happen. But for a wave there's no problem. If I start moving towards a wave it seems to be more energetic because more wave crests are hitting me per second, the wave's fequency relative to me has shifted up because my velocity relative to the wave has increased. But that's not possible for a light wave. So how does the higher frequency translate into higher energy? Or is this one of those brute fact things you just have to accept by virute of experimental evidence, intuition be damned?
 
  • #5
Super Kirei said:
If I start moving towards a wave it seems to be more energetic because more wave crests are hitting me per second, the wave's fequency relative to me has shifted up because my velocity relative to the wave has increased. But that's not possible for a light wave. So how does the higher frequency translate into higher energy?

that's right, the wave's fequency relative to you has shifted up … so why do you not accept that that means greater energy? :confused:
 
  • #6
Super Kirei said:
I was reading an elementary treatment of the photoelectric effect and it says that the energy of a photon is related to it's wavelength by λ=h/p where p is the momentum.
The relationship between the photon's energy and its momentum is E = pc or p = E/c. Substitute in λ=h/p to obtain λ= h(1/p) = h(c/E) or λ = hc/E. This gives the expression of the wavenght in terms of its energy.

Super Kirei said:
So this would imply that changing the frequency of a light source while keeping the intensity constant means that at higer frequencies fewer more energetic photons are emitted than at lower frequencies so that the energy flux remains constant.
That depends on what one means by intensity. It has always been my understanding that the intensity of a beam of light refers to the number of photons per second. So an increase in photon frequency does not change the beam intensity.

Super Kirei said:
But what does it mean for a photon to be more energetic? The speed of light is fixed at c, so how does its momentum vary?
For a photon to be more energetic means that the photon carries more energy with it. The more energy the photon carries the smaller its wavelength and the larger its frequency.

Super Kirei said:
I'm still trying to reconcile the billiard ball conception of the photon with the wave conception of a photon. So, if a photon has momentum, as the de Broglie equation seems to imply, then I imagine that would imply it has mass.
Photons, which have zero rest mass, have momentum. However there is the concept of inertial mass which refers to the quantity m in p = mv (some people refer to this as relativistic momentum). I know this seems weird but its straight textbook physics, strange as it may be.

Super Kirei said:
But if it has mass then it's momentu varies with velocity and it's momentum is related to it's energy so that higher energy means higher momentum which means higher velocity, so that can't happen.
The inertial mass of a photon depends, not on velovity, but on energy or momentum. Vary the energy or momentum of a photon and you've changed it's inertial mass.

Super Kirei said:
But for a wave there's no problem. If I start moving towards a wave it seems to be more energetic because more wave crests are hitting me per second, the wave's fequency relative to me has shifted up because my velocity relative to the wave has increased. But that's not possible for a light wave. So how does the higher frequency translate into higher energy? Or is this one of those brute fact things you just have to accept by virute of experimental evidence, intuition be damned?
You can think of a more energetic photon as one which vibrates more. The higher the rate of vibration the greater its energy.
 
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  • #7
Or is this one of those brute fact things you just have to accept by virute of experimental evidence, intuition be damned?
If your intuition tells you that only massive particles can have momentum than yes you have to change your intuition, because (funny thing) only massive particles can have zero momentum. "Massless particle with zero momentum" is nonsense.
 
  • #8
Super Kirei said:
I'm still trying to reconcile the billiard ball conception of the photon with the wave conception of a photon.

Don't try. Whatever a photon "really is", it is not a tiny billiard ball!
 
  • #9
I think I get it now. For a billiard ball flying in space, to have greater momentum is to go faster. For a photon, to have greater momentum is to be more massive which is a function of the momentum which is a function of frequency so that a blueshift in frequency will result in a more energetic and hence more massive photon. Is the equation E=pc a fundamental experimental result, or is it derived from a more fundamental relationship?
 
  • #10
jtbell said:
Don't try. Whatever a photon "really is", it is not a tiny billiard ball!

Quite right! I couldn't agree more. Feynman really hits it on the head when he wrote in his lectures, V-III page 1-1
"Quntum meechanics" is the description of the behavior of matter and light in all its details and, in particular, of the happenings on the atomic scale. Things on a very small scale behave like nothing that you have any direct experience about. They do not behave like waves, they do not behave like particles, they do not behave like cloudsm or billiard balls, or weighs on springs, or anything you have ever seen.
Newton thought that light was made up of particles, but then it was discovered that it behaves like a wave. Later, however (in the beginning of the twentieth century), it was found that light indeed sometimes behave like a particle. Historically, the electron, for example, was thought to behave like a particle, and then it was found that in many respects it behaved like a wave. So it really behaves like neither. Now we have given up. We say: "It is like neither."
I think tha pretty much sums up how i feel about it. Some people still refer to it as a particle though. I supoose that because its so difficult to hard to explain it every time. Aurhor Eddington used the term wavicle to descibe things. Not something I'd do but I can see the point.
 
  • #11
Boston_Guy said:
I think tha pretty much sums up how i feel about it. Some people still refer to it as a particle though. I supoose that because its so difficult to hard to explain it every time. Aurhor Eddington used the term wavicle to descibe things. Not something I'd do but I can see the point.
The term "particle" in my vocabulary always means quantum mechanical particle. Classical Newtonian particles should be called corpuscles. I think clinging to the obsolete definition of particle just creates confusion.
 
  • #12
Dead Boss said:
The term "particle" in my vocabulary always means quantum mechanical particle. Classical Newtonian particles should be called corpuscles. I think clinging to the obsolete definition of particle just creates confusion.

I agree.
 
  • #13
Boston_Guy said:
You can think of a more energetic photon as one which vibrates more. The higher the rate of vibration the greater its energy.

Does this energy remain the same when the possible paths, between emission and final detection, are increased/decreased (i.e. added or blocked)?
 
  • #14
San K said:
Does this energy remain the same when the possible paths, between emission and final detection, are increased/decreased (i.e. added or blocked)?
Yes.
 
  • #15
Boston_Guy said:
That depends on what one means by intensity. It has always been my understanding that the intensity of a beam of light refers to the number of photons per second. So an increase in photon frequency does not change the beam intensity.

This is not correct. Intensity is a well-defined concept in physics; we are not free to re-define it at our whim. Intensity is defined as energy per unit area per unit time, or power per unit area. The SI units of intensity are Watts/meter^2, or Joules/(second*meter^2). So the intensity of a beam of light depends on both the number of photons per unit area per second and on the energy per photon (which is of course proportional to the frequency).
 
  • #16
phyzguy said:
This is not correct. Intensity is a well-defined concept in physics; we are not free to re-define it at our whim. Intensity is defined as energy per unit area per unit time, or power per unit area. The SI units of intensity are Watts/meter^2, or Joules/(second*meter^2). So the intensity of a beam of light depends on both the number of photons per unit area per second and on the energy per photon (which is of course proportional to the frequency).
You're correct of course. I now realize the error I made. I was thinking about the photo electric effect and something that must be floating around in the back of my mind. What a goof on my part. Sorry! :(
 
  • #18
Super Kirei said:
I was reading an elementary treatment of the photoelectric effect and it says that the energy of a photon is related to it's wavelength by λ=h/p where p is the momentum. So this would imply that changing the frequency of a light source while keeping the intensity constant means that at higer frequencies fewer more energetic photons are emitted than at lower frequencies so that the energy flux remains constant. But what does it mean for a photon to be more energetic? The speed of light is fixed at c, so how does its momentum vary?

The speed of light is fixed. However, the relativistic mass of the photon is not fixed.
The photon has zero rest mass. However, it has a relativistic mass that is given by the equation for equivalence of energy and mass. The relativistic mass varies with the inertial frame of the observer.
The rest mass has a meaning only in the inertial frame of reference where the speed of light is zero. However, there is no physical inertial frame where the speed of light is zero.
The kinetic energy, E, of a photon is the relativistic mass of the photon, M, times the speed of light, c. Therefore,
E=Mc
The rest mass of the photon is zero. Therefore, an observer traveling in the rest frame of the photon would see a photon of zero relativistic mass. However, nothing can move at the speed of light relative to an observer in any inertial frame. Therefore, the rest mass is zero. Momentum is not proportional to the rest mass. It is proportional to the relativistic mass.
It is like one is on the highway with a 65 MPH speed limit, where every vehicle travels at the speed limit. There are three vehicles: a motorcycle, a sedan and a truck. Which has the greatest momentum and why?
 
  • #19
Darwin123 said:
The rest mass has a meaning only in the inertial frame of reference where the speed of light is zero. However, there is no physical inertial frame where the speed of light is zero.
The kinetic energy, E, of a photon is the relativistic mass of the photon, M, times the speed of light, c. Therefore,
E=Mc
The relativistic mass of any particle is defined by its momentum over the velocity. i.e. relativistic mass is defined as the m in p = mv or m = p/v. For a photon E = pc and m = p/c or m = (E/c)/c → m = E/c2. You missed a factor of c.
 
  • #20
Boston_Guy said:
The relativistic mass of any particle is defined by its momentum over the velocity. i.e. relativistic mass is defined as the m in p = mv or m = p/v. For a photon E = pc and m = p/c or m = (E/c)/c → m = E/c2. You missed a factor of c.

Sorry. I meant:
p=Mc
where "p" is the momentum of the photon, "M" is the relativistic mass of the photon and "c" is the speed of light in the vacuum.
The units always have to match on both sides of the equation. Hence, I should have caught my mistake just from the units.
Still, the variation in momentum comes from the relativistic mass, "M". Not the speed of light, "c".
 

FAQ: Energy of a Photon: Frequency, Intensity, Momentum

What is the relationship between the frequency of a photon and its energy?

The energy of a photon is directly proportional to its frequency. This means that as the frequency of a photon increases, its energy also increases. This relationship is described by the equation E = hf, where E is the energy of the photon, h is Planck's constant, and f is the frequency.

How does the intensity of light affect the energy of a photon?

The intensity of light does not have a direct effect on the energy of a single photon. The intensity of light refers to the amount of energy carried by a beam of light, while the energy of a photon is determined by its frequency. However, increasing the intensity of light can increase the number of photons present, which can ultimately increase the overall energy of the light beam.

What is the momentum of a photon?

The momentum of a photon is equal to its energy divided by the speed of light. This relationship is described by the equation p = E/c, where p is the momentum, E is the energy, and c is the speed of light. Photons have a unique property where their momentum is directly proportional to their frequency, rather than their mass.

Can the energy of a photon be changed?

No, the energy of a photon is a fundamental property that cannot be changed. The energy of a photon is determined by its frequency, which is a characteristic of the electromagnetic wave it belongs to. However, the energy of a photon can be absorbed or emitted by an atom through interactions with electrons, leading to changes in the atom's energy level.

How do we measure the energy of a single photon?

The energy of a single photon can be measured using a device called a photon detector. This device works by converting the energy of the photon into an electrical signal, which can then be measured. The most common type of photon detector is the photomultiplier tube, which uses a photocathode to convert photons into electrons, which are then amplified and measured.

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