Does light just reflect off the nucleus of atoms?

Perhaps you could be a little more specific in what you are thinking. It's not a simple answer.

Generally though, light does not interact with nucleii. If it did, it might convert a proton into a neutron. More commonly, it will
- get absorbed by the electrons in their orbitals, kicking them up into a higher energy state.
- get absorbed by the bonds between atoms in a molecule and converted into vibrations
- a bunch of other things.
 
DaveC426913 said:
Perhaps you could be a little more specific in what you are thinking. It's not a simple answer.

Generally though, light does not interact with nucleii. If it did, it might convert a proton into a neutron. More commonly, it will
- get absorbed by the electrons in their orbitals, kicking them up into a higher energy state.
- get absorbed by the bonds between atoms in a molecule and converted into vibrations
- a bunch of other things.
Click to expand...
So the reflection of light is more due to electron interactions than passive nuclaic interactions.
 
CeasarSalad1979 said:
So the reflection of light is more due to electron interactions than passive nuclaic interactions.
Click to expand...
Pretty much. Most things that happen to atoms tend to happen to the electrons.
Reflection is an aggregate phenomemon; it happens above the atomic level where the light acts as a wavefront.
 
CeasarSalad1979 said:
So the reflection of light is more due to electron interactions than passive nuclaic interactions.
Click to expand...
Yes.

What we normally call "light" is electromagnetic radiation in the visible region of the spectrum. There is however a form of highly energetic electromagnetic radiation, of very much shorter wavelength than visible light, which does interact with atomic nuclei. We call these γ (gamma) rays. γ-rays can be absorbed and emitted by atomic nuclei.
 
I have a question, which I think may be relevent here: Light, when it interacts with "certain metals" will eject electrons completely. This of course gives rise to the photoelectric effect, known about since the second half of the 19th centuary, but only fully explained by considering light energy as being quantized.
My questions are: what are the properties of the so-called "certain metals"? I know they were originally the alkali metals, but my chemistry is not strong enough to attach any real meaning to that fact

Second: presumably these so-called certain metals will have their properties changed by the loss of electrons?? If so, do they still qualify as "certain metals" in the original sense?
 
QuarkHead said:
I have a question, which I think may be relevent here: Light, when it interacts with "certain metals" will eject electrons completely. This of course gives rise to the photoelectric effect, known about since the second half of the 19th centuary, but only fully explained by considering light energy as being quantized.
My questions are: what are the properties of the so-called "certain metals"? I know they were originally the alkali metals, but my chemistry is not strong enough to attach any real meaning to that fact

Second: presumably these so-called certain metals will have their properties changed by the loss of electrons?? If so, do they still qualify as "certain metals" in the original sense?
Click to expand...
This is ejection from the topmost electron, in energy terms, in the partly filled conduction band in the bulk metal. The "work function" is the name given to the energy needed to do it for each metal. I found a table of them here: http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/photoelec.html
and you are right that it is lowest for alkali metals. This will be for the same reason the first ionisation energy is lowest for the alkali metals, viz. the alkali metals are starting to populate with electrons the s orbital of the next principal quantum number up, so they are outside the last closed shell and only see a low effective nuclear charge. That results in the alkali metals having one electron that is weakly bound, which is what makes them prone to form cations.

The work function is however not the same as the first ionisation energy, which is the energy needed to strip the first electron off a neutral atom of the metal. Interestingly, I found on comparing them that the work function is typically only about half the first ionisation energy. I found this at first glance counterintuitive, seeing as atoms in the bulk metal, being in a bound state, are of lower energy than isolated atoms would be. However, I suppose that's because we are taking an uppermost electron from the conduction band: there are many more electrons lower down in the band, so there is overall a lower energy state in the metal than in the isolated atom. But I've never made the comparison before, so I'm speculating a bit here.

Since the photoelectric effect is an effect in bulk metals, the metal left behind is a still a metal, just one fractionally deficient in electrons.
 
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exchemist said:
This is ejection from the topmost electron, in energy terms, in the partly filled conduction band in the bulk metal. The "work function" is the name given to the energy needed to do it for each metal. I found a table of them here: http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/photoelec.html
and you are right that it is lowest for alkali metals. This will be for the same reason the first ionisation energy is lowest for the alkali metals, viz. the alkali metals are starting to populate with electrons the s orbital of the next principal quantum number up, so they are outside the last closed shell and only see a low effective nuclear charge. That results in the alkali metals having one electron that is weakly bound, which is what makes them prone to form cations.

The work function is however not the same as the first ionisation energy, which is the energy needed to strip the first electron off a neutral atom of the metal. Interestingly, I found on comparing them that the work function is typically only about half the first ionisation energy. I found this at first glance counterintuitive, seeing as atoms in the bulk metal, being in a bound state, are of lower energy than isolated atoms would be. However, I suppose that's because we are taking an uppermost electron from the conduction band: there are many more electrons lower down in the band, so there is overall a lower energy state in the metal than in the isolated atom. But I've never made the comparison before, so I'm speculating a bit here.

Since the photoelectric effect is an effect in bulk metals, the metal left behind is a still a metal, just one fractionally deficient in electrons.
Click to expand...
Having thought more about this, and having read a bit more, I think I can add something to explain better the difference between ionisation energy and work function. The latter is the energy to remove an electron from the surface of the metal. At the surface, there is incomplete bonding of the electrons, since there are only +ve charged "cores" (atoms minus the outer electrons that are delocalised into the metallic bonding system) on one side of them. So these electrons are only "half-bound", as it were. That's one reason why it is easier to liberate one than to take one out of the outermost orbital of an isolated atom. There are also, apparently - I did not know this, but it makes obvious sense - a lot of detailed surface effects, arising from oxide films and so, on that can radically change the ease with which a surface electron can be liberated.

So the relationship between work function and ionisation energy is not entirely straight forward.
 
exchemist said:
Having thought more about this, and having read a bit more, I think I can add something to explain better the difference between ionisation energy and work function. The latter is the energy to remove an electron from the surface of the metal. At the surface, there is incomplete bonding of the electrons, since there are only +ve charged "cores" (atoms minus the outer electrons that are delocalised into the metallic bonding system) on one side of them. So these electrons are only "half-bound", as it were. That's one reason why it is easier to liberate one than to take one out of the outermost orbital of an isolated atom. There are also, apparently - I did not know this, but it makes obvious sense - a lot of detailed surface effects, arising from oxide films and so, on that can radically change the ease with which a surface electron can be liberated.

So the relationship between work function and ionisation energy is not entirely straight forward.
Click to expand...
Yes, this makes sense. Thanks. I know that ionzing radiation is far more energentic than the visible light one typically sees in the photoelectic effect, though I do know that this effect is more efficient at higher energies. Which raises the question: where is the transition between the two, or does that question makes no sense?
My yellowing Phys. Chem. text points out that ionization energy is in general equal to the energy of the incicdent radion less the kinetic energy of the expelled electron. Thus, ionizations that require large energies will leave the expelled electrons with lower kinetic energy and vice versa
 
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QuarkHead said:
Yes, this makes sense. Thanks. I know that ionzing radiation is far more energentic than the visible light one typically sees in the photoelectic effect, though I do know that this effect is more efficient at higher energies. Which raises the question: where is the transition between the two, or does that question makes no sense?
My yellowing Phys. Chem. text points out that ionization energy is in general equal to the energy of the incicdent radion less the kinetic energy of the expelled electron. Thus, ionizations that require large energies will leave the expelled electrons with lower kinetic energy and vice versa
Click to expand...
The series limit of an atomic spectrum gives one the wavelength needed to just ionise an electron. For example with hydrogen the Lyman series, which is concerned with transitions to and from the quantum number n=1 (ground state), has a series limit at a wavelength of around 90nm (900Å),which is well into the UV but not into the X-ray region. The term "ionising radiation"normally conjures up radiation from atomic. nuclei (α, β, γ) rather than mere UV, but sufficiently energetic UV can in fact ionise atoms.

There's an explanation here: https://chem.libretexts.org/Bookshe...Molecules/Hydrogen's_Atomic_Emission_Spectrum

The ionisation energy of H is 13.6eV which is higher than for most elements, so in general UV of longer wavelength should be enough to ionise the first electron from most elements.
 
Moderator note: Ceasarsalad1979 has been permanently banned (sock puppet of a previously-banned user).
 
The opening question is not completely inane, even though the opening poster might have intended it to be so.

The nucleus of an atom is very small. Therefore, only light of a small enough wavelength will interact with it, most of the time. Atomic nuclei have their own quantised energy levels, analogous to the atomic energy levels associated with the electrons in the atom. However, the nuclear energy levels are much more widely spaced than atomic energy levels. Therefore, for a nucleus to absorb a photon typically requires that the photon have around 1000 times the energy of a photons of visible light, which puts the photons into the x-ray or gamma ray region of the spectrum for nuclear absorption. In comparison, many atoms absorb and emit visible light by means of their electronic energy levels - a fact that should be obvious when you look at the world around you.

As for the photoelectric effect, in that case we're dealing with electrons in the bulk of a metal that consists of many billions upon billions of individual nuclei. In a bulk metal, the electrons exist in energy bands. The "work function" is the minimum energy required to remove an electron from the highest-energy band (the conduction band) of a particular metal.
 
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