A bunch of basic questions on electrons


by Helicobacter
Tags: basic, bunch, electrons
Helicobacter
Helicobacter is offline
#1
Aug20-11, 04:58 PM
P: 159
Feel free to answer any subset. I'm not a science person, so apologies in advance for some of the absurdity.

1. When heat up some piece of matter (e.g., a rock), do the the electrons (of the atoms of which the rock is composed of) start moving faster? If so, then why do we observe temperature as a continuous phenomenon - or, at least, pseudo-continuous with millions of different possibilities (e.g., 25.743 C°, 245.24565435 C° etc.) - after all, the elements that are in the rock are bound by a very low energy levels (say, n=3). Therefore, the angular momentum can only be 0, 1, 2, which would mean that only three different temperatures are possible (assuming you would introduce the heat uniformly throughout the entire rock). What am I misunderstanding?
Another reason I am wrong is because I've read that temperature does not have an upper limit, however, the angular momentum does have an upper limit...

2. I've seen pictures of electron clouds that appear to diffuse outwards in a continuous fashion (like a picture of the bivariate normal distribution) to positive infinity and negative infinity. I've also seen picture where it's just a geometric shape (a sphere or a oval sphere or a spherical cone) that's bounded in space. Which one is more accurate? Asking in another way: For a particular atom in Melbourne, is there a nonzero probability (albeit extremely small) of its electron appearing in the US (while the nucleaus stays in Melbourne and nobody does anything to it)?

3. Is there a mathematical pattern that produces the standing waves of the subshells we know of? Or is it just arbitrary and we don't understand it yet?

4. What does it take to kick out an electron? How often does it happen in our daily lives? What kind of wavelength and what kind of intensity is required to kick out an electron of an hydrogen atom, say?

5. Why are noble gases and other non-bonded atoms colorless? Why is that only after bonding the molecules assume color?

6. Why do we not have color photographs of molecules? Shoulnd't the wavelength be visible at that scale?

7. Why do the planets of our solar system have such large variation in color (blue, red etc.)? The visible spectrum is so extremely small - why does the emission/absorption just happen to take place in that range? Or are the emission/absorption characteristics so extremely idiosynchartic for different matters at all wavelength ranges?

8. When I look at the 2p (x,y,z) subshells I notice that when you put them together, then there will be some overlapping of the clouds (does that mean that the overlapping regions have additive probabilities of the two subshells)?

9. I once asked why the electrons and protons don't collide due to the magnetic attraction. Someone said that the laws at the atomic scale are different. However, I then read on wikipedia that electrons obey electromagnetic rules. What gives?

Thanks!
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espen180
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#2
Aug20-11, 06:00 PM
P: 836
1. Yes and no. The temperature is due to the motion of the atoms, not their electrons. The temperature only appears continous in systems with a large number of atoms because the gaps between the enery levels become very small.

2. The diffuse cloud is more accurate. The "bubbles" you mention are constructed such that there is a certain probability of finding the electron inside them.

3. We have a theory which predicts the states of quantum systems and is in agreement with experiment. If this is what you mean, then yes, we understand it.

4. This happens all the time. You could say it happens in any chemical redox-reaction.

5. The colour of gases arise from their absorption/emission spectra. You can read about this in any basic physics textbook or on wikipedia.

6. See above. You can only define the colour of an object from its emission spectrum. A molecule absorbs and emits light as a whole, so you cannot obtain a colour photograph of one the same way you can for macroscopic objects.

8. Not neccesarily. It depends on the coefficients in their sum. This question may be difficult to answer precisely without knowledge of linear algebra.
Bill_K
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#3
Aug20-11, 06:22 PM
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9. The electrons and protons in an atom most certainly do "collide." Or at least they overlap. The s-wave orbital for an electron is nonzero at the origin, and consequently there is a nonzero probability of finding the electron inside the nucleus. In most cases there is no interaction simply because there's not enough energy available to do so.

For example in the hydrogen atom, the electron and proton cannot unite to form a neutron because the neutron is heavier and outweighs the proton and electron combined. For other atoms, such as Al-26, the electron may indeed be captured, and one of the protons is turned into a neutron causing a neutrino to be emitted.

jtbell
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Aug20-11, 10:18 PM
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A bunch of basic questions on electrons


Quote Quote by Helicobacter View Post
3. Is there a mathematical pattern that produces the standing waves of the subshells we know of? Or is it just arbitrary and we don't understand it yet?
These waves arise from solving the Schrödinger equation. The solutions for the hydrogen atom are well-understood and are described with varying levels of detail in many "modern physics" and quantum mechanics textbooks, from the second-year university level on upward. Also on many Web sites, for example this one:

http://hyperphysics.phy-astr.gsu.edu...um/hydsch.html
m.e.t.a.
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#5
Aug21-11, 11:42 AM
P: 112
Quote Quote by Helicobacter View Post
1. ...Why do we observe temperature as a continuous phenomenon?
To expand on espen180's answer: For a single atom or molecule in vacuum (far from any other atoms or electromagnetic fields) the energy levels available to its electrons are indeed few and highly discrete. However, if you bring more atoms in to create a crystalline lattice, new electron energy levels are created. This is because each electron now feels the potential of not only its parent nucleus, but of all other nuclei and electrons in the lattice. The number of new electron energy levels is proportional to the number of atoms in the lattice. For a crystal of reasonable mass, there will be in excess of [itex]10^{20}[/itex] atoms, and a correspondingly large number of electron energy levels. These energy levels are so numerous and closely spaced as to merge into a near continuum of energy levels, called a band. See Wikipedia. For materials other than crystalline solids, the relationship between the number of atoms/molecules and the number of available electron energy levels is probably more complex than a simple linear proportionality, but I would presume that the basic principle is the same.

4. What does it take to kick out an electron? How often does it happen in our daily lives? What kind of wavelength and what kind of intensity is required to kick out an electron of an hydrogen atom, say?
Take a look at, or derive, the electron energy levels for the Bohr hydrogen atom:

[tex]{E_n} = - \frac{{{m_e}{e^4}}}{{8{\pi ^2}\varepsilon _0^2{\hbar ^2}{n^2}}} = - \frac{{13.6{\rm{ eV}}}}{{{n^2}}}[/tex]
You can see that it takes a photon of at least 13.6 eV to liberate a ground-state electron from a hydrogen atom (photoelectric effect). This corresponds to a photon of 91 nm, which is in the ultraviolet.

7. Why do the planets of our solar system have such large variation in color (blue, red etc.)? The visible spectrum is so extremely small - why does the emission/absorption just happen to take place in that range? Or are the emission/absorption characteristics so extremely idiosynchartic for different matters at all wavelength ranges?
The visible spectrum (~ 400-700 nm) corresponds to the range of photon energies ~ 1.8-3.1 eV. This also happens to be the approximate range in which the absorption/emission spectra of most materials are at their most detailed (i.e. have the most spikes). This is not a coincidence—electron transition energies are in the ~ 0.1-10 eV range. Since every material has a unique electronic absorption spectrum, it is likely that (say) a trichromatic eye or camera sensitive to visible light will perceive two different materials as different colours. Hence it is not so unlikely that the planets and moons of the solar system, each having a different surface composition/atmosphere etc., should appear differently coloured when viewed in visible light by the human eye.

Just beyond the visible spectrum you have IR and UV. Near-IR and near-UV can absorbed and emitted electronically by most materials just like visible light, but (if I remember correctly) the absorptivity/emissivity in this range is typically lower. Deep-IR can only be absorbed and emitted thermally, so in this range you see very little visual contrast between materials of similar temperature. Far-UV and X-rays tend to interact via scattering. I am not sure how much variation in "colour" one would see when viewing the world in this band, but probably not very much. Microwaves and radio waves are absorbed weakly or not at all. Again, I am not sure how much variation in "colour" would be seen between objects when viewed in microwave, but I would guess very little.

As for whether the absorption/emission spectrum of every material is extremely idiosyncratic across all wavelengths, I am fairly sure the answer is no: the absorption/emission spectrum is at generally its most characteristic in the visible part of the spectrum. But, as I say, I am not positive about this. Good question!
Helicobacter
Helicobacter is offline
#6
Aug21-11, 11:03 PM
P: 159
Many thanks for your answers. Especially the "visible spectrum sensitivity" answer I found very revealing, meta. Some follow-up:

Quote Quote by espen180 View Post
1. Yes and no. The temperature is due to the motion of the atoms, not their electrons. The temperature only appears continous in systems with a large number of atoms because the gaps between the enery levels become very small.
Quote Quote by m.e.t.a. View Post
To expand on espen180's answer: For a single atom or molecule in vacuum (far from any other atoms or electromagnetic fields) the energy levels available to its electrons are indeed few and highly discrete. However, if you bring more atoms in to create a crystalline lattice, new electron energy levels are created. This is because each electron now feels the potential of not only its parent nucleus, but of all other nuclei and electrons in the lattice. The number of new electron energy levels is proportional to the number of atoms in the lattice. For a crystal of reasonable mass, there will be in excess of [itex]10^{20}[/itex] atoms, and a correspondingly large number of electron energy levels. These energy levels are so numerous and closely spaced as to merge into a near continuum of energy levels, called a band. See Wikipedia. For materials other than crystalline solids, the relationship between the number of atoms/molecules and the number of available electron energy levels is probably more complex than a simple linear proportionality, but I would presume that the basic principle is the same.
So, all else being equal, if you increase the temperature of an object, will the average principal quantum number of its atoms increase? If this question is absurd because you can only view the object as one collective of atoms, what will happen to the orbital as you increase temperature? What about the angular momentum? If it's not temperature, what determines what angular momentum an atom will adopt?

espen, I don't understand why you say "yes and no" when you say that the atomic vibrations are mainly responsible for the heat, and not the electron clouds. Or are they the same phenomenon?

Quote Quote by espen180 View Post
3. We have a theory which predicts the states of quantum systems and is in agreement with experiment. If this is what you mean, then yes, we understand it.
Quote Quote by jtbell View Post
These waves arise from solving the Schrödinger equation. The solutions for the hydrogen atom are well-understood and are described with varying levels of detail in many "modern physics" and quantum mechanics textbooks, from the second-year university level on upward. Also on many Web sites, for example this one:
http://hyperphysics.phy-astr.gsu.edu...um/hydsch.html
I phrased my question in such a generic way in which it was impossible to infer my intentions. I was looking at the different angular momentum shapes of the orbitals at N=3. So, when you vary the angular momentum from 0...2, is there an "elegant"/simple equation which will give you the new electron field by just varying over l=0 to 2?

Quote Quote by espen180 View Post
4. This happens all the time. You could say it happens in any chemical redox-reaction.
Quote Quote by m.e.t.a. View Post
Take a look at, or derive, the electron energy levels for the Bohr hydrogen atom:

[tex]{E_n} = - \frac{{{m_e}{e^4}}}{{8{\pi ^2}\varepsilon _0^2{\hbar ^2}{n^2}}} = - \frac{{13.6{\rm{ eV}}}}{{{n^2}}}[/tex]
You can see that it takes a photon of at least 13.6 eV to liberate a ground-state electron from a hydrogen atom (photoelectric effect). This corresponds to a photon of 91 nm, which is in the ultraviolet.
Does the photon (that kicks out the electron) get absorbed by the atom and replace the kicked-out electron? When I compare the properties of electrons vs. photons I see vast differences: How can the photon just "emulate" being an electron? After all, the forces at that range are extremely picky/sensitive to the orbiting particle's properties.
On the other hand, if it does not get kicked out, the object under light exposure would get ionized very fast (unless the kicked-out electrons will be absorbed by a neighboring ionized atom of the wood immediately thereafter).

What gives?

For a typical object (say, a piece of wood in direct sunlight), roughly at what frequency does this happen (i.e., an electron is kicked out of an orbital) and roughly what % of the woods surface atoms are affected? From your equation it looks like it would happen way over 99.9% of the objects surface at a extremely high frequency. But what is the approximate mean time for an individual atom from one photoelectric event to the next given typical sunlight exposure?


Quote Quote by espen180 View Post
6. See above. You can only define the colour of an object from its emission spectrum. A molecule absorbs and emits light as a whole, so you cannot obtain a colour photograph of one the same way you can for macroscopic objects.
I still don't understand how we cannot see the color at that range. What determines at which scale you start seeing colors?

Quote Quote by Bill_K View Post
For example in the hydrogen atom, the electron and proton cannot unite to form a neutron because the neutron is heavier and outweighs the proton and electron combined. For other atoms, such as Al-26, the electron may indeed be captured, and one of the protons is turned into a neutron causing a neutrino to be emitted.
This is something I never understood. What causes the transformation of elementary particles on collision? What triggers the "I take this amount of this particle and that amount of the other particle and give you this particle and so much of kinetic energy." It seems extremely unlikely that the reacting particles just happen to have the same amount of property x as the resulting particle...and why do they not just kick off each other with conversion of energy?

Quote Quote by m.e.t.a. View Post
Far-UV and X-rays tend to interact via scattering.
What does this mean?


Also, I saw that mass of elementary particles is given in eV. Since when is mass provided in volts? Is this because of e=mc^2?
espen180
espen180 is offline
#7
Aug22-11, 03:06 AM
P: 836
Quote Quote by Helicobacter View Post
espen, I don't understand why you say "yes and no" when you say that the atomic vibrations are mainly responsible for the heat, and not the electron clouds. Or are they the same phenomenon?
The temperature is dependent on the kinetic energy of the contituent atoms. However, there can be energy transfer to the atoms' electrons, to the average energy of the electrons is dependent on the temperature.

Quote Quote by Helicobacter View Post
I phrased my question in such a generic way in which it was impossible to infer my intentions. I was looking at the different angular momentum shapes of the orbitals at N=3. So, when you vary the angular momentum from 0...2, is there an "elegant"/simple equation which will give you the new electron field by just varying over l=0 to 2?
Yes, there is such an equation. You can view it here: http://en.wikipedia.org/wiki/Hydrogen_atom#Wavefunction

Quote Quote by Helicobacter View Post
Also, I saw that mass of elementary particles is given in eV. Since when is mass provided in volts? Is this because of e=mc^2?
An electron volt is a unit of energy, equivalent to 1.602*10-19 J, which is equivalent to roughly 1.8*10-36 kg
Galron
Galron is offline
#8
Aug22-11, 05:11 AM
P: 45
Positive and negative infinity are limits, they don't physically exist. Infinities of temperature are a consequence of the integral calculus and entropy, nothing there is really physically infinite either, although some people will tell you they are; I wouldn't believe physics allows for infinities in any energy system though if I was you, only that a number sometimes has peculiarities that reality does not exhibit infinitely.

This is something I never understood. What causes the transformation of elementary particles on collision? What triggers the "I take this amount of this particle and that amount of the other particle and give you this particle and so much of kinetic energy." It seems extremely unlikely that the reacting particles just happen to have the same amount of property x as the resulting particle...and why do they not just kick off each other with conversion of energy?
Why quantisation is pretty much an obvious logical deduction, when energy is transferred it can be lost in several ways, all of which are usually just conceptual issues that boill down to things like momentum etc.

The energy of lifting a book onto a shelf is really no different that other forms of energy it's just a way to explain context. Energy types are more of a philosophy of kinds than an absolute condition of reality. A way of delineating reactions and interactions.
Bhar Quar
Bhar Quar is offline
#9
Aug22-11, 09:03 AM
P: 1
Quote Quote by Helicobacter View Post
Quote Quote by m.e.t.a.
Far-UV and X-rays tend to interact via scattering.
What does this mean?
I think m.e.t.a. is hinting towards Compton scattering(wiki it).


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