# Why is absorption spectrum a line spectrum?

The minimum energy required to excite a hydrogen electron is 10.2 eV. When photons of energy spread over a continuous range of wavelengths fall on a sample of hydrogen, why are only those photons absorbed which have energies corresponding to exact energy differences between the allowed states? For a transition requiring energy $E=h\nu_0$, a photon with frequency $\nu>\nu_0$ can be absorbed, with the remaining energy $\Delta E=h\nu-h\nu_0$ appearing as kinetic energy of the atom. This would result in a continuous spectrum, however - with no missing frequencies.

• From my deleted answer - The Formation of Spectral Lines – user94040 Oct 14 '15 at 17:12
• Photons are not only absorbed at exact energies corresponding to exact energy differences between allowed states. Photoionisation happens for energies above 13.6 eV in cold hydrogen. – Rob Jeffries Oct 14 '15 at 17:34
• If I understand your question correctly, the main point is why a transition at frequency $\omega_0$ cannot accept light at frequency $\omega=\omega_0+\Delta>\omega_0$, with the remainder energy $\hbar\Delta$ being spent to accelerate the atom. The question is a bit unclear and this point is relatively hard to get from it, so it would help if you modify the question to make this clearer. – Emilio Pisanty Oct 14 '15 at 17:54

What you're describing does happen to some extent. It's called "Doppler broadening": absorption and emission lines in hot materials are wider (in wavelength) than absorption and emission lines in cool materials, because atoms in the hot material occupy a broader range of velocities.

An atom at rest can't absorb any old photon and convert the extra energy into energy of motion, because such collisions must conserve both energy and momentum. Let's start with the decay of an excited atom by photon emission. An $n=1$ excited hydrogen atom at rest has energy $$E_\text{excited} = m_\text{ground}c^2 + \Delta E \approx \rm 1\,GeV + 10\,eV$$ and decays to a ground-state atom nearly at rest and a photon, with energies $$E_\text{ground} = mc^2 + \frac{p^2}{2m} \approx \rm 1\,GeV \qquad \it E_\gamma = h\nu \approx \rm 10\,eV$$ Conservation of energy tells us that $$\Delta E = \frac{p^2}{2m} + h\nu$$

By conservation of momentum we must have the same magnitude of momentum $p$ for the atom and for the photon. The photon's momentum is $p_\gamma = h\nu/c$, so we get $$\Delta E = \frac{1}{2m} \left(\frac{h\nu}{c}\right)^2 + h\nu = h\nu \left( \frac{h\nu}{2mc^2} + 1 \right) \approx h\nu$$ So you can see that the atom does get a kick from emitting the photon, but it's at the parts-per-billion level of the photon's energy.

You're asking about absorption. In the center-of-mass reference frame, every absorption must look exactly like the process we've just gone through, only with the initial and final states switched. If you want your ground-state hydrogen atom to absorb a 12 eV photon, it must be "running away" from the photon so the the Doppler-shifted energy in the center-of-mass frame corresponds to 10 eV.

This is different from elastic collisions between billiard balls, because there is no photon in the final state, and different from inelastic collisions like the "ballistic pendulum" because the atom can only absorb energy from the electromagnetic field, and thus momentum from the electromagnetic field, in specially-sized lumps.

We can be even more specific. Let's start with an incident photon of unspecified energy and a ground-state atom at rest, $$E_\text{ground} = mc^2 \qquad E_\gamma = h\nu$$ and try to end up with a "kicked" excited atom, $$E_\text{excited} = mc^2 + \Delta E + \frac{p^2}{2m}.$$ From conservation of energy we have \begin{align} E_\text{excited} &= E_\text{ground} + E_\gamma \\ \Delta E + \frac{p^2}{2m} &= h\nu. \tag{A} \end{align} By the same logic as above we have $|p| = h\nu$ (though now the directions are parallel, rather than antiparallel, since we are not in the center-of-momentum frame), which gives \begin{align} \Delta E &= h\nu \left( 1 - \frac{h\nu}{2mc^2} \right) \\ 0 &= \Delta E - h\nu + \frac{1}{2mc^2} (h\nu)^2 \\ h\nu &= mc^2 \left( 1 \pm \sqrt{ 1 - \frac{2\Delta E}{mc^2} } \right) \end{align} This quadratic equation has exactly two solutions: $h\nu \approx 2mc^2$ (a relativistic photon interaction that we're not interested in) and the one we expected, $h\nu \approx \Delta E$. If we keep two terms in the binomial expansion for the square root we get \begin{align} \frac{h\nu}{mc^2} &\approx 1 - 1 + \frac12 \frac{2\Delta E}{mc^2} + \frac18 \left(\frac{2\Delta E}{mc^2} \right)^2 \\ h\nu & \approx \Delta E \left( 1 + \frac12 \frac{\Delta E}{mc^2} \right), \tag{B} \end{align} which is more than the "transition energy" by an extra amount $\epsilon = (\Delta E)^2/2mc^2$. That's the only extra energy that you're allowed to impart to the atom at rest, and it's required by momentum conservation and small enough to usually be negligible.

• The extra factor of energy is negligible if one visualises this as being gained as kinetic energy by the atom. What if the $\epsilon$ appears partially as kinetic energy of another photon, and partially as kinetic energy of the atom? Writing energy of that photon as $h\nu'$, the momentum conservation gives $\frac{h\nu}{c} = \frac{h\nu'}{c} + p$ (I am retaining symbols apart from $\nu'$ from your answer) and energy conservation $h\nu = h\nu' + \Delta E + \frac{p^2}{2m}$. The solution is the almost the same as yours, only now $h(\nu - \nu') \approx \Delta E$. – Charles Oct 16 '15 at 10:18
• This means photon of any frequency can be absorbed, with release of another photon simultaneously? – Charles Oct 16 '15 at 10:19
• @Charles Sounds like you are asking now about atom-photon scattering, as in Rayleigh scattering (without exciting the atoms) or Raman scattering (involving low-energy excitations in molecules or bulk materials). – rob Oct 16 '15 at 23:21
• If you can articulate it a little more clearly, that might make a good follow-up question. – rob Oct 17 '15 at 16:09
• Does this phenomenon of scattering also occur when atoms are irradiated with light spread over a continuous range of wavelengths? And does this (in accordance with the equations in the above comments) contradict the actual laboratory absorption spectra observations - considering that the absorption spectra observed are line spectra (meaning that only photons of specefic frequencies are absorbed), whereas the equations predict that a photon of any frequency can be absorbed when accompanied by release of another photon? – Charles Oct 21 '15 at 12:26

The minimum energy to photoionise a hydrogen atom depends what energy level it is in. If it is in the ground state ($n=1$), then that energy is 13.6 eV. If it is in the first excited state ($n=2$) then it is 3.4 eV. Photons with energies higher than this are capable of interacting with the atoms through photoelectric absorption and this does lead to a continuous absorption spectrum with sharp edges at the energies corresponding to the photoionisation energies of the atom in its various states.

For example in "cold" hydrogen, with all the atoms in the $n=1$ level, then there is continuous photoelectric absorption of photons with energies higher than this. If the hydrogen is warmer, so that an appreciable fraction of the atoms are in the $n=2$ level then there is a continuous absorption spectrum up to the first ionisation edge at 13.6eV (91 nm), then a drop and then more continuous absorption that rises to the second ionisation edge at 3.4 eV (365 nm).

Your reference to 10.2 eV may be confusion with Lyman alpha absorption corresponding to photons that instigate a transition from the $n=1$ to $n=2$ levels. This resonant process is caused because photons at that energy have a high probability of interacting with the atom and causing this transition. Such photons would also be capable of causing photoionisation of atoms in the $n=2$ level. So one would have continuous photoelectric absorption plus a sharp line absorption.

However, if the hydrogen is "cold" such that all the hydrogen atoms are in the $n=1$ state then the absorption spectrum would be (i) a sharp spike at 10.2 eV (121 nm, Lyman alpha), then a gap until 91 nm and then continuous absorption at higher energies (smaller wavelengths). So probably your real question is why is there a gap in the absorption spectrum? The reason for this is quantum mechanics and conservation of energy. The energy levels in a hydrogen atom are quantised. The electron is not allowed to have any old energy. Therefore unless the atom is being excited almost exactly into one of the other allowed energy "eigenstates" or the electron is being photoionised into a continuum of free energy states outside the atom, then the absorption simply doesn't take place and the hydrogen atom is "transparent" to the radiation.

Or, I may have misinterpreted your question. If you are asking why 10.2eV of the photon energy cannot be absorbed with the rest converted into the KE of the atom, then as Chris White says, you cannot conserve energy and momentum simultaneously apart from a unique photon energy that is very close to 10.2 eV. If we let the atom be at rest initially, then conservation of energy and momentum give $$E + mc^2 = mc^2 + 10.2 + K$$ $$\frac{E^2}{c^2} = 2mK = 2m (E-10.2),$$ where $E$ is the photon energy in eV, $K$ is the atomic kinetic energy after the interaction and $m$ is the atomic rest mass.

This gives a quadratic $$E^2 - 2mc^2 E + 20.4mc^2 = 0$$ with solutions $$E = mc^2 \left[ 1 \pm \sqrt{1 - \frac{20.4}{mc^2}} \right]$$

For a H-atom with $mc^2 \simeq 939$ MeV, this says that only a photon with energy very close to 10.2 eV (10.2005 eV) can be absorbed.

• The OP says "with the remaining energy being left over as its kinetic energy" so I think they're envisioning Ly-alpha absorption + giving the whole atom a kick with whatever energy remains in order to conserve energy. This particular scenario isn't forbidden by quantum mechanics but by conservation of momentum. – user10851 Oct 14 '15 at 17:41
• To be honest I was hoping you'd work out the details, because I didn't ;) Intuitively I think this is just an overconstrained system (the kick velocity can come from energy conservation or momentum conservation, but they might disagree). If there is no problem, then the question becomes much deeper, because something more subtle must prevent this process. – user10851 Oct 14 '15 at 19:00
• @ChrisWhite Backtracking - think I have it. – Rob Jeffries Oct 14 '15 at 19:24
• There's no relativistic reason that forbids the absorption, cf. this question. Suppose an atom of mass $m$ is struck by a photon of energy $\hbar\omega$ and momentum $\hbar k=\hbar \omega/c$. Its final momentum will be $p=\hbar k$, its final rest energy will be $mc^2+\hbar\omega_0$ to account for the transition, and its final energy will be $\sqrt{(mc^2+\hbar\omega_0)^2+p^2c^2}$. Energy conservation then demands$$mc^2+\hbar\omega=\sqrt{(mc^2+\hbar\omega_0)^2+p^2c^2}.$$ Some algebra will then give a unique allowed frequency $\omega$... – Emilio Pisanty Oct 14 '15 at 20:35
• given by $\hbar\omega=\hbar\omega_0+\frac{\hbar^2\omega_0^2}{2mc^2}$. Here $\omega$ is the photon frequency measured in the lab frame, and $\hbar\omega_0$ is the energy of the transition in the new rest frame. – Emilio Pisanty Oct 14 '15 at 20:37

An photon with energy equal to or more than 10.2 eV can be absorbed - with the remaining energy being left over as its kinetic energy.

it seems you're asking: if there's an atom with a transition at frequency $\omega_0$, and you have a photon at frequency $\omega>\omega_0$, why can't the atom absorb the photon anyways to excite the transition, and take the extra energy $\hbar(\omega-\omega_0)$ as kinetic energy?

I will address this and not the reason for the existence of discrete lines in the first place. There are two things going on here:

• When the atom absorbs a photon of frequency $\hbar\omega$, that photon also has a momentum $\hbar k=\hbar\omega/c$ and the atom also needs to absorb it. This does always happen, but the absorbed momentum is typically very small and the atoms need to be very cold, and hence slow, for this to matter. Indeed, the way you get them that cold is via Doppler cooling, which exploits exactly this mechanism.

For more details on how to describe the momentum kick in photon absorption, in what situations it's relevant, and why it's not in other cases, see How does one account for the momentum of an absorbed photon?

• Having absorbed this amount of energy, the atom then cannot absorb any more. It would be conceivable to have a sort of two-step process, where the atom first absorbs the main chunk into its electronic state (plus a bit of kinetic energy), and then deals the rest of the energy into kinetic energy. However, this cannot happen, for two reasons:

• The centre-of-mass motion cannot engage with the electromagnetic field, because it is not charged. You need to split the atom up into its charged constituents to get it to interact with the field, and this means exciting internal, electronic transitions. More importantly, though,

• even if it were charged (like, say, if you had a $\mathrm{He}^+$ ion), it still cannot absorb any energy from an electromagnetic field. This is fundamentally because of relativistic reasons: a free particle simply cannot absorb the light-like four-momentum of the photon (which means that the photon has no mass, so its energy $E$ and its momentum $p$ are related via $E^2-p^2c^2=0$) without changing its internal state.

This last bit is really the crucial part. If you have a particle with rest mass $m$ and you hit it with a photon of energy $\hbar\omega$ and momentum $\hbar k=\hbar\omega/c$, you'll get the particle going away with a momentum $p=\hbar k$ and a new rest energy of $mc^2+\Delta E$, where $\Delta E$ is the change in internal energy of the particle. The total energy of the system is then $$mc^2+\hbar\omega=\sqrt{(mc^2+\Delta E)^2+p^2c^2}.$$ Doing some algebra, you get that the photon's energy must be $$\hbar\omega=\Delta E+\frac{(\Delta E)^2}{2mc^2},$$ which is bigger than $\Delta E$ by a term which directly describes the kinetic energy from the absorbed momentum. If there is no change in the internal state, on the other hand, then $\Delta E$ and the atom cannot absorb any energy from the photons, which prevents the second step of our clever one-two plan from taking place.

(Source.)

Find above the spectrum of hydrogen (without fine structure).

Hydrogen (and all other atoms) can absorb or emit electromagnetic radiation (photons) of energy $E=hf$ that corresponds precisely to the (total) energy difference $E_n-E_m$ between the Quantum Mechanically allowed hydrogen energy levels, which are given by:

$E_n=\frac{-13.6\:\mathrm{eV}}{n^2}$ where $n$ is the Principal Quantum Number.

So if you illuminate hot hydrogen gas with white, visible light, only photons that correspond to allowed total energy transitions are absorbed, in this case belonging to the Balmer Series. These absorptions show up as dark ('missing') parts of the originally white light.

As you can see, the $10.2\:\mathrm{eV}$ (transition from $n=1$ to $n=2$) you mentioned actually belongs to the Lyman Series, which lies in the Ultraviolet part of the electromagnetic radiation spectrum.

• If I understand the OP correctly, he's asking why a transition at frequency $\omega_0$ cannot accept light at frequency $\omega=\omega_0+\Delta>\omega_0$, with the remainder energy $\hbar\Delta$ being spent to accelerate the atom. – Emilio Pisanty Oct 14 '15 at 17:47
• @EmilioPisanty: it's not really how I understood but I could be wrong. I have to admit the photoelectric effect didn't really occur to me, in the context of OP's question. – Gert Oct 14 '15 at 18:59
• Emilio's suggested version is certainly more interesting. – Rob Jeffries Oct 14 '15 at 19:00
• @RobJeffries: no contest. – Gert Oct 14 '15 at 19:01

You can understand everything that the hydrogen atom does if you solve the Schroedinger equation in the presence of a weakly oscillating electric field. The solution shows that for a random oscillation frequency (below the ionization frequency), almost nothing happens. But for certain critical frequencies, the wave function ocscillates strongly in resonance with the field. You can interpret this as an oscillating charge cloud.

Now you have to use Maxwell's equations to understand what happens next. An oscillating charge density in the presence of an oscillating field will both absorb and emit energy (electromagnetic radiation) depending on the phase relationship between the oscillating field and the oscillating charge cloud.

Maxwell's equations will then give you the correct rates of energy exchange for both the emission and absorption effects. The hydrogen atom behaves exactly like a tiny classical antenna. It is not necessary to make any ad hoc assumptions about quantization of energy levels.

why are only those photons absorbed which have energies corresponding to exact energy differences between the allowed states?

The observation of spectral lines was a surprising observational fact.. The quantum mechanical theory of atomic physics developed slowly, and is a self consistent one, and describes and predicts a lot more data than spectral lines.

So the answer to "why" is "because" the model fitting innumerable data describes the transitions by photons of fixed energy equal to the difference of the atomic levels as the other answers have explained.

An photon with energy equal to or more than 10.2 eV can be absorbed - with the remaining energy being left over as its kinetic energy.

If you are thinking of kicking an electron completely out, ionizing the atom, you have the wrong energy. In this case there can be a continuum absorption , the extra energy of the photon transferred to the kinetic energy of the electron, as another question has answered.

Since its energy changes, so should its frequency.

The interacting photon completely disappears. If it raises the energy level of the electron nothing is left over, and the same if it kicks the electron out, no photon is left over to change frequency.

This would result in a continuous spectrum, however - with no missing frequencies.

There will be a continuum of absorption frequencies over the ionization energy, the electron can be kicked out with continuum photons once their energy is larger thant 13.6 ev

In simple terms, an electron is both a particle and a wave. Therefore, an electron in an atom must sit in one of a determined number of orbitals.

The same is observed with sound waves. Any frequency is possible for a travelling wave, but a sound wave in a cymbal reflects and "loops back on itself" so it must correspond to one of the cymbal's resonant frequencies. See here https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membrane It's a very rough analogy, yet it has very clear parallels with the various orbital shapes (s,p,d) found in atoms.

In a cymbal, each vibrational mode has continuously variable energy. In an atom, where each orbital is occupied by exactly one electron (or two of opposite spin) the energies are fixed.

A photon with energy equal to or more than 10.2 eV can be absorbed with the remaining energy being left over as its kinetic energy.

Kinetic energy of what? The atom cannot suddenly and arbitrarily change its velocity, as this would violate the law of conservation of momentum. On the other hand, a molecule is able to change the kinetic energy of its atoms by entering a different vibrational state. As vibrational states are also quantised, this results in vibrational fine structure, where the line is split into several lines (which are very close together.) https://en.wikipedia.org/wiki/Molecular_vibration

There is also some broadening of the lines for doppler reasons. An atom moving toward the light source will absorb slightly lower frequency light, and one moving away from it slightly higher frequency light. From the atoms' points of view / frames of refernece, though, all are absorbing at the same frequency.

• The atom cannot suddenly change its velocity, as this would violate the law of conservation of momentum, if one completely forgets that the photon also has momentum, that is. – Emilio Pisanty Oct 14 '15 at 19:51
• @Rob Jeffries I already changed to suddenly and arbitrarily. The momentum of a photon is entirely dependent on its frequency, so the atom does not have the freedom to dissipate arbitrary excess energy as kinetic energy. That was my point. – Level River St Oct 14 '15 at 20:37