# Interplay of energy between photon, electronic energy levels and kinetic energy

I can understand that atoms have quantized energy levels for its electrons, but an atom's translational kinetic energy is continuous. As such, why is the absoprtion spectrum not continuous? That is to say, why can't the excess energy from the photon beyond what is needed to promote electrons be simply converted into kinetic energy?

On a related topic, can an excited atom simply move its electron to a lower energy level and convert that into translational kinetic energy of the atom (the same atom itself)? If so, why can't it convert only a fraction of that energy and emit the remaining fraction as a photon (thereby also producing a continuous emission spectrum)? After all, an excited atom can collide another atom and pass that energy as kinetic energy to that other atom.

Because both momentum and energy have to be conserved when the atom absorbs the photon.

Suppose we have an atom with a mass $$m$$ and the energy difference between the initial and final levels is $$E$$. The photon energy is $$hf$$, so conservation of energy gives us:

$$hf = E + \tfrac12 m v^2 \tag{1}$$

where $$v$$ is the speed of the atom after absorbing the photon. However a photon also has a momentum $$hf/c$$ so conservation of momentum gives us:

$$\frac{hf}{c} = mv \tag{2}$$

and combining equations (1) and (2) we find:

$$E + \tfrac12 m v^2 = mvc$$

which gives us a quadratic for the final velocity of the atom:

$$v^2 - 2vc + \frac{2E}{m} = 0$$

or:

$$v = c\left(1 \pm \sqrt{1 - 2E/mc^2} \right)$$

If we assume $$E \ll m$$ we can expand the square root using a binomial approximation to get:

$$v = c\left(1 \pm \left(1 - \frac{E}{mc^2} \right) \right)$$

and we can ignore the solution greater than the speed of light since it's unphysical, so we end up with:

$$v = c \frac{E}{mc^2} \tag{3}$$

(I've written it this way because $$mc^2$$ is the rest energy of the atom so the equation makes clear that the key factor is the ratio of the excitation energy to the rest energy.)

So there is one, and only one, possible velocity the atom can have after absorbing the photon. That's why the absorption line is sharp and not a continuum.

The velocity after absorbing the photon is generally negligibly small. For example consider a hydrogen atom absorbing the $$10.2$$ eV photon required for the $$1s \to 2p$$ transition. Using equation (3) gives us $$v \approx 3$$ m/s, and the kinetic energy associated with this velocity is only about $$10^{-8}$$ eV.

• For $mc^2=1GeV$, Eq. (3) gives v=3 m/s or $v/c=10^{-8}$. This leads to $E_{recoil}=10^{-16} \cdot 1e^9/2 = 0.00000005 eV$, strengthening your point. – my2cts Mar 29 at 9:15
• Oops, I put the mass of the atom in as 511keV and that's the mass of an electron not a hydrogen atom! Some quick editing is required!!! – John Rennie Mar 29 at 9:17
• @John Rennie A quick follow-up question: why is it not possible for a photon of energy $E >10.2eV$ to hit the atom, and after the collision we have a photon of energy $E-10.2eV$ and an excited atom (with the atom traveling at some negligible velocity such that the momentum is conserved)? – suncup224 Mar 29 at 14:37
• The interaction of the photon with the atom is analogous to a resonance. A mixture of the hydrogen 1s and 2p orbitals has an oscillation frequency, and energy from the photon is transferred to this resonance when the photon frequency matches the atom frequency. It is still possible to transfer energy to the atom when the photon frequency doesn't match the transition frequency, but it is very improbable and in practice we never see it happening. – John Rennie Mar 29 at 15:39