Lets say I have an atom of known mass who goes from excited state to the ground state whose energy is $14.4 keV$ lower.

I know that the emitted energy $14.4keV$ got converted into an energy of a photon and kinetic energy of the Atom - this means energy of the photon $E_\gamma < 14.4keV$. I tried to calculate it like this and got a nonsense:

\begin{align} E_{1} &= E_{2}\\ \sqrt{ {E_{0~Fe}}^2 + {p_1}^2c^2 } &=\sqrt{ \left(E_{0~Fe} - E_\gamma\right)^2 + {p_2}^2 c^2 } \longleftarrow \substack{\scriptsize \boxed{p_1 = 0}~\boxed{p_2 = E_\gamma /c}}\\ \sqrt{ {E_{0~Fe}}^2 + 0 } &= \sqrt{ \left(E_{0~Fe} - E_\gamma\right)^2 + \frac{{E_\gamma}^2}{c^2} c^2 } \\ E_{0~Fe} &= \sqrt{ \left(E_{0~Fe} - E_\gamma \right)^2 + {E_\gamma}^2 }\\ {E_{0~Fe}}^2 &= {E_{0~Fe}}^2 - 2E_{0~Fe}E_\gamma +{E_\gamma}^2 + {E_\gamma}^2 \\ {E_\gamma}^2 + (-E_{0~Fe})E_\gamma + 0 &= 0\\ &\Downarrow\\ E_\gamma &= 0\\ E_\gamma &= E_{0~Fe} \end{align}

None of the solutions to the quadratic equation make sense. Can anyone give me a hint where did i go wrong?


After i read some of your comen I did rethink the situation and wrote equation like this. But still energy of a photon $E_\gamma$ is too large. Shouldnt it be smaller than $14.4keV$?

\begin{align} E_1&=E_2\\ E_{0~Fe} + \Delta E &= \sqrt{{E_{0~Fe}}^2 + {p_2}^2c^2}\\ E_{0~Fe} + \Delta E &= \sqrt{{E_{0~Fe}}^2 + \frac{{E_\gamma}^2}{c^2}c^2}\\ E_\gamma &= \sqrt{\left(E_{0~Fe} + \Delta E\right)^2-{E_{0~Fe}}^2}\longleftarrow{\substack{\text{Lets assume that we have a "Fe" atom}\\\text{whose rest energy is $E_{0~Fe}=53GeV$}}}\\ E_\gamma &= \sqrt{\left(53\times10^9eV + 14.4\times10^3 eV\right)^2 - \left(53\times10^9 eV\right)^2}\\ E_\gamma &\approx 3.91\times10^7 eV \end{align}


By even improving the equation I got better result which I think must be the right one.

\begin{align} E_1&=E_2\\ E_{1~atom}&=E_{2~atom} + E_\gamma\\ E_{0~Fe} + \Delta E &= \sqrt{{E_{0~Fe}}^2 + {p_2}^2c^2} + E_\gamma\\ E_{0~Fe} + \Delta E &= \sqrt{{E_{0~Fe}}^2 + \frac{{E_\gamma}^2}{c^2}c^2} + E_\gamma\\ E_{0~Fe} + \Delta E - E_\gamma &= \sqrt{{E_{0~Fe}}^2 + {E_\gamma}^2}\\ \substack{\text{square of}\\\text{a trinomial}}\longrightarrow(E_{0~Fe} + \Delta E - E_\gamma)^2 &= {E_{0~Fe}}^2 + {E_\gamma}^2\\ \big((E_{0~Fe} + \Delta E) - E_\gamma\big)^2 &= {E_{0~Fe}}^2 + {E_\gamma}^2\\ (E_{0~Fe} + \Delta E)^2 - 2(E_{0~Fe} + \Delta E)E_\gamma + {E_\gamma}^2 &= {E_{0~Fe}}^2 + {E_\gamma}^2\\ (E_{0~Fe} + \Delta E)^2 - 2(E_{0~Fe} + \Delta E)E_\gamma &= {E_{0~Fe}}^2\\ E_\gamma &= \frac{(E_{0~Fe} + \Delta E)^2 - {E_{0~Fe}}^2}{2(E_{0~Fe} + \Delta E)}\\ E_\gamma &= \frac{\left(53\times10^9eV + 14.4\times10^3eV\right)^2 - \left(53\times10^9eV\right)^2}{2\left(53\times10^9eV + 14.4\times10^3eV\right)}\\ E_\gamma &= 14.399 keV \end{align}

Is it possible that atom takes so little energy?

  • $\begingroup$ Why is there a "+" sign between $E^2$ and $p^2$? $\endgroup$
    – Ali
    Jul 21, 2013 at 20:14
  • $\begingroup$ Well isn't the Lorentz invariant ${E_0}^2 = E^2 - p^2c^2$ so I solved this for full energy $E$ and got: $E^2 = {E_0}^2 + p^2c^2$ then $E = \sqrt{{E_0}^2 + p^2c^2}$ This is what i used on the energy conservation eq. at the beginning (left and right side). $\endgroup$
    – 71GA
    Jul 21, 2013 at 20:27
  • $\begingroup$ I am going to delete any further comments with block formatted equations in them. $\endgroup$ Jul 21, 2013 at 20:28
  • 1
    $\begingroup$ But the initial energy that you have written is the one for an atom at rest. Maybe you should add the 14.4 keV due to the excited state to the initial energy. $\endgroup$
    – neutrino
    Jul 21, 2013 at 21:07
  • 1
    $\begingroup$ At the right hand side you have to add the energy of the emitted photon. You are considering the energy of the atom only, I think. $\endgroup$
    – neutrino
    Jul 21, 2013 at 22:30

1 Answer 1


The atom takes so little kinetic energy becuse it is far more massive $ ( E_{0~Fe}=53 \frac{GeV}{c^{2}})$ than the electron $(E_{0}^{e-}=0.511 \frac{MeV}{c^{2}} )$.

  • 1
    $\begingroup$ This effect is even more pronounced with optical photon absorption and re-emission - so much so that conservation of momentum for the atom photon system often has little practical meaning - because it is so hard to find atoms that are free enough from other interactions. The atom's interactions with other things around it utterly swamps the photon-atom interaction. $\endgroup$ Jul 22, 2013 at 7:17
  • $\begingroup$ One more question. If the atom is very hardly bound the whole crystal takes the momentum $p$ right? In this case am I alowed to swap the rest energy of the iron atom $E_{0~Fe}$ with rest energy of a whole crystal $E_{0~crystal}$? $\endgroup$
    – 71GA
    Jul 22, 2013 at 10:31

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