# Conservation of momentum of a photon

• SU403RUNFAST
In summary, the conversation discusses the release of a photon by a stationary atom when transitioning from an excited state to a lower state. The energy of the photon is determined by the energy difference between the two states and the conservation of energy and momentum. The kinetic energy of the recoiling atom must also be taken into account in the calculations. The equations used include E=pc for a photon, E^2=(pc)^2+(mc^2)^2 for the atom, and the equations for conservation of energy and momentum. The final goal is to determine the energy of the photon in terms of the energy difference between the two states.
SU403RUNFAST

## Homework Statement

A stationary atom releases a photon when it goes from excited state n to lower state n. Delta(E) is the energy difference between initial state En and final state En', consider kinetic energy of recoiling atom. Apply conservation of energy and momentum and solve for the energy of the photon hf.

## Homework Equations

Delta(E)=En-En'=hf=hc/lamda
E=pc for a photon due to no mass
E^2=(pc)^2+(mc^2)^2 pythagorean relation for momentum and energy

## The Attempt at a Solution

I know that the the conservation requires that the photon and the final atom have opposite momenta of the same magnitude p. P=p1+p2 is the conservation for momentum. P is the atom before, p1 is the atom after and p2 is the photon after. p1 is root of E^2-(mc^2)^2 all divided by c while p2 is E/c. so I put the momentum before as 0 for P and plugged in the other values, substituted E=hf in p2 and solved for hf alone to get
hf=-root of E^2-(mc^2)^2. So i was able to get the answer without using an energy conservation equation i am not sure if i did this right (didnt do E=E1+E2)

Is the energy of the atom the same as the energy of the photon? You used the same notation for both.

No it is not, the energy of the photon is pc or hf and the energy of the atom is root of pc^2 + mc^2 squared, this is the same notation?

teetercl said:
No it is not, the energy of the photon is pc or hf and the energy of the atom is root of pc^2 + mc^2 squared, this is the same notation?
You wrote that
E=p1c for a photon due to no mass
E^2=(p2c)^2+(mc^2)^2
and later, you said that E = hf and substituted for E in both equations. Did you not do that?

i only substituted for E in for the photon, you even changed my formula in your reply, the E for p1 was not substituted, that is why it is in the final answer still. But i believe i did not take into account the kinetic energy of the atom, 1/2mv^2...

I do not understand what you did. But you have to take the kinetic energy of the atom into account. And you do not need the relativistic equation for the atom. The energy difference ΔE between the exited state and the lover state is equal to the sum of the energy of the photon and the kinetic energy of the atom. Give the energy of the photon in terms of the energy difference ΔE.

Conservation of momentum i did 0=mv+delta(E)/c, conservation of energy i did mc^2=.5mv^2+delta(E), deltaE is the energy of the photon. Which equation do i solve first for v and plug into the other one? What is the energy of the atom at rest initially?(i put it as mc^2), I Need to solve for hf, which is deltaE

I have delta(E)=hf+(1/2)mv^2 for the equation, do i just solve for the velocity in terms of the momentums using p=mv? Is this the correct energy equation and is my previous momentum equation correct?

teetercl said:
I have delta(E)=hf+(1/2)mv^2 for the equation, do i just solve for the velocity in terms of the momentums using p=mv? Is this the correct energy equation and is my previous momentum equation correct?
You wrote two momentum equations previously, using the same notations p and E in both, but one equation refers to the atom, and on refers to the photon. In the equation for the atom, E is the total energy including the energy corresponding to the mass. You do not know it, you know the difference between energy levels. So determine hf in terms of ΔE .
The momentum of the atom will not be so large that the speed is close to the speed of light; you can use the classical approach, delta(E)=hf+(1/2)mv^2, in addition to conservation of momentum. What is the momentum of the photon in terms of its frequency?

Thanks for helping, I can do it now.

## 1. What is conservation of momentum and why is it important in regards to photons?

Conservation of momentum is a fundamental principle in physics that states the total momentum of a closed system remains constant. This means that the initial momentum of all objects involved in a collision or interaction must equal the final momentum. In the case of photons, conservation of momentum is important because it helps us understand how energy and momentum are transferred between particles during interactions.

## 2. How does conservation of momentum apply to photons?

Photons, as particles of light, also have momentum even though they have no mass. According to Einstein's theory of relativity, the momentum of a photon is equal to its energy divided by the speed of light. This means that when photons interact with other particles, their momentum must be conserved.

## 3. Can the momentum of a photon change?

Yes, the momentum of a photon can change during interactions with other particles. This can happen through processes such as absorption, where the photon's energy and momentum are transferred to the absorbing particle, or reflection, where the photon's momentum changes direction after bouncing off a surface. However, in all interactions, the total momentum of the system must remain constant.

## 4. How does conservation of momentum affect the behavior of light?

Conservation of momentum plays a crucial role in determining the behavior of light. For example, in the process of refraction, where light bends as it passes through a medium, conservation of momentum explains how the light changes direction while still maintaining its overall momentum. Additionally, the principle of conservation of momentum is essential in understanding the behavior of light in situations such as scattering and diffraction.

## 5. What are some real-world applications of conservation of momentum for photons?

Conservation of momentum has many practical applications, including in the field of optics and telecommunications. For instance, it is used in the design of lenses and mirrors to manipulate the direction of light. It is also crucial in the development of technologies such as fiber optics, which use the principle of total internal reflection to transmit information through light signals. Additionally, understanding conservation of momentum is essential in the study of various astronomical phenomena, such as the redshift of light from distant galaxies.

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