# Recoil hydrogen emitting photon

1. Jan 29, 2012

### Al X

1. The problem statement, all variables and given/known data
A hydrogen atom at rest emits a photon with wavelength 410.1 nm.
First, calculate the recoil of the hydrogen atom when emitting the photon. (recoil = momentum)
Second, calculate the recoil of the hydrogen if we replace the electron mass with the reduced mass μ.
Third, find the relationship between the two above.

2. Relevant equations
$E=\frac{hc}{\lambda}=pc$

$p=mv$

$\mu = \frac{m_{e}m_{p}}{m_{e}+m_{p}}$
where me is the electron mass and mp is the proton mass (the core is a proton of course).

3. The attempt at a solution
Alright, the photon has a momentum equal to $\frac{hc}{\lambda c}$ in some random direction. As the momentum is conserved, the hydrogen atom must have gotten the same momentum in the opposite direction, that's the recoil in the first part of the exercise.

HERE COMES THE PROBLEM:
When replacing the electron mass with the reduced mass, the momentum of the hydrogen should be exactly the same as the same type of photon is emitted(?). Thus, if you divide the two recoils with eachother, you'll end up with 1. And that's not the answer to this exercise.

Really hope you could help me out with this one, I've been stuck on this for some time now and it's the last exercise on a paper due tomorrow.

2. Jan 29, 2012