# Conservation of momentum of a photon

1. May 5, 2015

### SU403RUNFAST

1. The problem statement, all variables and given/known data
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.

2. Relevant 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
3. 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)

2. May 6, 2015

### ehild

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

3. May 6, 2015

### SU403RUNFAST

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?

4. May 6, 2015

### ehild

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?

5. May 6, 2015

### SU403RUNFAST

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...

6. May 6, 2015

### ehild

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.

7. May 7, 2015

### SU403RUNFAST

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 in to 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

8. May 7, 2015

### SU403RUNFAST

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?

9. May 7, 2015

### ehild

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?

10. May 7, 2015

### SU403RUNFAST

Thanks for helping, I can do it now.