# Using recoil of atom to find correct in wavelength of emitted photon

1. Mar 9, 2013

### Fluxthroughme

So firstly, the book supplies the answer of $\Delta\lambda = \frac{h}{2mc}$

I use energy and momentum conservation. I say the total energy is that of the incoming photon, which is then changed to energy of the photon released, and kinetic energy of the atom, from the recoil. I say the same thing, but for momentum, with momentum conservation. Thus I get the equations I'm using. However, the three ideas I can come up with all give incorrect answers. 2 and 3 are obviously closer, as we know the correction is very small, but it's not what I'm looking for.

Thus, I cannot figure out how to get the correct answer. I am unsure where I'm going wrong. Thanks for any help.

2. Mar 9, 2013

### Staff: Mentor

Which incoming photon do you mean? The atom emits a photon, it does not receive one.
You need both energy and momentum conservation here. The net momentum of the system (photon+atom) is zero.

3. Mar 9, 2013

### Fluxthroughme

To emit a photon, it needs to be in an excited state. I'm saying the photon it absorbs is the one that puts it in the excited state, giving it the energy needed to emit in the first place.

4. Mar 9, 2013

### Staff: Mentor

It can get excited in other ways, too. Anyway, the excited atom is at rest initially here.

5. Mar 10, 2013

### Fluxthroughme

The below is what I get when I try it with the atom at rest:

Even now, I still run into a problem. I appreciate that the quantity in brackets is extremely close to 1, but is it safe to just ignore it?

6. Mar 10, 2013

### Staff: Mentor

Right, that is exactly the approximation you need here.
$$\frac{\lambda_f}{\lambda_r} = \frac{\lambda_r-\Delta \lambda}{\lambda_r} = 1-\frac{\Delta \lambda}{\lambda_r} \approx 1$$

7. Mar 10, 2013

### Fluxthroughme

Ahh, I see. Thanks for clearing that up.