# Is the photon energy required to cause an atomic transition $\Delta E+\Delta KE$, where $\Delta E$ is the “transition energy”?

An atom "at rest" can absorb a photon, and while some of this energy goes into increasing the energy level of the electron, momentum must be conserved, and so some energy must also increase the kinetic energy of the atom. The amount of $KE$ delivered to the atom is easy to calculate.

Therefore, is the energy to cause a transition $\Delta E+\Delta KE$, where $\Delta E$ is the "transition energy"?

So that no commenters cover old ground, please see this question.

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Possibly a duplicate of physics.stackexchange.com/q/52379/2451 and links therein. –  Qmechanic Nov 25 '13 at 1:03

Yes, the photon energy has to be equal to the total energy of the atom before the transition (including the kinetic energy) minus the total energy of the atom after the transition (including the kinetic energy).

In practice, the kinetic energy of the atom is negligible. The mass of a nucleus is at least 1 GeV/$c^2$ or so (the mass of the proton) while the photon's energy is a billion times smaller, 1 eV/$c^2$ or so. This implies that the atom (mostly nucleus) gets the momentum kick of order $p=$1 eV/$c$ which means that its kinetic energy is of order $p^2/2m\sim (1eV)^2/GeV\sim 10^{-9}eV$. This recoil effect therefore modifies the calculated transition energy (photon energy) by something like one billionth of the original value which is lower than many other subleading corrections. But yes, it's measurable.

These recoil energies become more important in the case of gamma rays emitted by nuclei in nuclear reactions and their proper incorporation is an important part of the Pound-Rebka experiment verifying general relativity.

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learned something today: I was not aware of such precision tests of GR. thanks. –  anna v Feb 11 '13 at 11:09

Yes. Lubos answered quite well, so I will simply expand on his answer with the calculation of the energy transfer. The answer, once you know that the energy is conserved and you are in the non-relativistic limit, is obvious. The advantage of this derivation is that you can see the steps and see where the recoil energy will be non-negligible (when the approximations I use break down).

The key ratio governing the recoil energy is $E_\gamma / M c^2$, where $E_\gamma$ is the photon energy and $M$ is the effective mass of the recoiling atom. I say effective mass because this may differ from the actual mass of an atom. For instance the Mössbauer effect is an instance where the recoil of atoms in a solid is greatly suppressed because all of the atoms are bonded together and essentially move together, multiplying the effective mass to something macroscopic. This is what allows certain gamma ray energies to be measured to great precisions, and is key to the experiment Lubos briefly mention.

Here follows the derivation, which will assume $E_\gamma \ll M c^2$. The usual energy and momentum conservation laws hold in quantum mechanics. So: $$\begin{array}{lcl} \left(p_{\gamma}+p_{\mathrm{atom}}\right)_{\mathrm{before}}&=\left(p_{\gamma}+p_{\mathrm{atom}}\right)_{\mathrm{after}}\\\left(E_{\gamma}+E_{\mathrm{atom}}\right)_{\mathrm{before}}&=\left(E_{\gamma}+E_{\mathrm{atom}}\right)_{\mathrm{after}} \end{array}$$

Working in the rest frame of the original atom $p_{\mathrm{atom before}}=p_{\mathrm{\gamma after}}=E_{\mathrm{\gamma after}}=0$ and $E_{\mathrm{atom before}}=m_a c^2$ where $m_a$ is the mass of the unexcited atom. So

$$p_a \equiv p_{\mathrm{atom after}}=p_{\mathrm{\gamma before}}=\frac{E_{\mathrm{\gamma}}}{c}$$

and

$$E_{\mathrm{atom after}} = E_\gamma + m_a c^2 = \sqrt{p_a^2 c^2 + (m_{a'} c^2)^2}$$

where I have used the correct relativistic formula relating energy and momentum $E^2 = p^2 c^2 + m^2 c^4$, which reduces to the usual formula $E = m c^2 + \frac{p^2}{2m}$ in the limit of small velocities. Plugging in $p_a$ gives

$$E_\gamma + m_a c^2 = \sqrt{E_{\mathrm{\gamma}}^2 + (m_{a'} c^2)^2}$$

Note also that $m_a'\neq m_a$: the mass of the atom is different. This is because it has extra energy due to excitation, and energy is mass. The change in mass is $\delta m = m_a' - m$, which is going to very much small than the original mass: $\delta m \ll m_a$ as long as $E_\gamma \ll m_a c^2$. In fact it is the same order of magnitude as $E_\gamma$. So I can pull out the $m_a c^2$ out of the root and make some simplifying approximations:

$$\begin{array}{lcl} E_\gamma + m_a c^2 &=& \sqrt{E_{\mathrm{\gamma}}^2 + (m_{a'} c^2)^2} \\ &=& m_a c^2 \sqrt{\left(\frac{E_{\mathrm{\gamma}}}{m_a c^2}\right)^2 + \left(1 + \frac{\delta m}{m_a}\right)^2} \\ &\approx& m_a c^2 \left[ 1 + \frac{1}{2} \left( \left(\frac{E_{\mathrm{\gamma}}}{m_a c^2}\right)^2 + 2 \frac{\delta m}{m_a} \right) \right] \\ &=& m_a c^2 + \frac{1}{2} \frac{E_{\mathrm{\gamma}}^2}{m_a c^2} + \delta m c^2 \end{array}$$

So

$$E_\gamma = \frac{1}{2} \frac{E_{\mathrm{\gamma}}^2}{m_a c^2} + \delta m c^2$$

Now the naive estimate, ignoring recoil energy, would be that $\delta m c^2 = E_\gamma$. In reality it is a little less than this because not all of the photon energy goes into raising the internal energy of the atom. The difference is

$$E_\gamma - \delta m c^2 = \frac{1}{2} \frac{E_{\mathrm{\gamma}}^2}{m_a c^2} = \frac{p_a^2}{2m_a}$$

which is not at all surprising, and barely worth the effort, but it's nice to see it come from first principles.

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