Photon frequency change taking recoil into account

[Uniquebum]

Homework Statement

Show that if atom's recoil in transition is [itex]\frac{p^2}{2m}[/itex], the emitted photon's frequency changes by a factor of [itex]1-\frac{\Delta E}{2mc^2}.[/itex]

Homework Equations

Recoil momentum [itex]p = \frac{hf}{c}[/itex]
m = mass of the atom
p = momentum of the electron

The Attempt at a Solution

When photon is absorbed
[itex]E_{atom} = mc^2 + E_{\gamma} = mc^2 + hf_0[/itex]

When the photon is emitted
[itex]\Delta E_{atom} = E_{atom} - E_{\gamma(emitted)} - E_r[/itex] where E_r = recoil energy.
Solving that only leads to
[itex]h(f_0 - f_1) = \Delta E_{atom} - mc^2 + \frac{p^2}{2m}[/itex]
Which doesn't look right... So, what am i doing wrong?

 

[mark726]

it is important to carefully analyze and understand the problem before attempting to solve it. In this case, we are given the information that the recoil momentum of an atom in transition is \frac{p^2}{2m}. This means that when a photon is emitted, the atom experiences a recoil momentum equal to \frac{p^2}{2m}. We also know that the recoil momentum of the atom is related to the frequency of the emitted photon by the equation p = \frac{hf}{c}.

To solve for the change in frequency of the emitted photon, we need to consider the energy conservation of the system. When a photon is absorbed by the atom, the energy of the system is given by E_{atom} = mc^2 + E_{\gamma} = mc^2 + hf_0, where f_0 is the frequency of the absorbed photon. When the photon is emitted, the energy of the system is given by E_{atom} - E_{\gamma(emitted)} - E_r, where E_r is the recoil energy of the atom.

Using the equation p = \frac{hf}{c} and the fact that the recoil momentum is \frac{p^2}{2m}, we can write E_r = \frac{p^2}{2m} = \frac{h^2f^2}{2mc^2}. Substituting this into the energy conservation equation, we get:

E_{atom} = mc^2 + hf_0 = mc^2 + hf - \frac{h^2f^2}{2mc^2}

Solving for f, we get:

f = \frac{hf_0}{mc^2 + hf_0 + \frac{h^2f^2}{2mc^2}}

Now, we can use the fact that \frac{p^2}{2m} = \frac{hf}{c} to get an expression for the recoil energy in terms of the frequency:

E_r = \frac{p^2}{2m} = \frac{h^2f^2}{2mc^2} = \frac{h^2}{2mc^2} \left(\frac{hf}{c}\right)^2 = \frac{h^2f^2}{2mc^2}

Substituting this into the energy conservation equation again, we get:

E