# How does recoil happen when an atom emits a photon?

Suppose an excited atom at rest emits a photon. The atom will experience a recoil and will end up moving in the opposite direction with a certain speed. But how does the atom acquire this speed, exactly? Does it gain speed smoothly, is it an infinite acceleration exactly when it emits the photon, or are we just not allowed to ask this question?

Let us try a back of the envelope computation, so we get an idea of the numbers you would be speaking of. Let us assume as the OP says there is a photon, $$\gamma$$, emitted from a Hydrogen atom (simplest atom). Ignoring the details of how an electronic transition may happen, the initial momentum is 0, so the following equation holds: $$0 = p_\gamma + p_H = \hbar \vec{k} + m_H \vec{v_H},$$ where $$k$$ is the emitted photon's wave number, $$m_H$$ the mass of the H-atom and $$v_H$$ its acquired speed. We will see a posteriori that the non-relativistic version of momentum is fine. Solving for $$|v_H|$$ since the process happens along a single axis $$|v_H| = \frac{\hbar |k|}{m_H} = \frac{h}{\lambda m_H}$$

We can do an order of magnitude estimate for the above, with $$h\sim 10^{-34}$$ and $$m_H \sim 10^{-27}$$ this is $$|v_H| \sim \frac{1}{\lambda} \, \times 10^{-7} \,\text{m}^2\,\text{Hz}$$

This means in this units that for the situation to lead to sizeable speeds, $$\lambda < 10^{-7} \text{m}$$ which is already the lower end of the visible spectrum. This is close to the Lyman limit for Hydrogen. So shorter wavelengths are not occurring in the simplest of atoms which has the simplest energy levels and the biggest chances for more energetic transitions given its spacing. So we are speaking of extremely small speeds in general.

Consider then the uncertainty principle applied to energy and time $$\Delta t \Delta E \geq h$$

For the process above assume you know the energy of the photon and the speed of the atom and therefore the energy completely well, so that your uncertainty in the energy approaches zero, then $$\Delta t \geq \frac{h}{\Delta E} \rightarrow \infty$$ The uncertainty on how long the process took approaches infinity and you cannot determine any estimate on the acceleration. You can imagine in the practice a precision of meV for the process above, this in Joules is $$10^{-22}$$ and implies that $$\Delta t \geq 10^{-12}\,\text{sec}$$

One can interpret the frequency of the wavelength emitted as the formation time of the photon. For the case of a Lyman line we are looking at the order $$10^{-14}\,\text{sec}$$, which is below your precision so the experiment will appear to be instantaneous (approximately).

• What if you don't care about measuring the energy of the photon? No matter what that energy is, it will always cause a recoil. So forget about the photon and just measure the atom's speed as precisely as you can. Does it discontinuously jump? Commented Dec 14, 2021 at 15:48
• Yes as far as I know the theory says that the recoil when related to electron transitions for example, should be as discrete es the electron transition... but then bottleneck will always be the uncertainty principle. Perhaps somebody closer to experiment than me can shed some more light on practical obstacles. Commented Dec 14, 2021 at 16:14