,
and O must weigh 16 times as much as H.
The suspicion that our continuous looking world, at a micro level, is discrete, is old. (Greeks philosophized about atoms, and invented the name) But ideas didn't really start "crystallizing" till the 1800's. Atoms and their structure are responsible for most of the properties of matter!
We all know about nuclei and electrons flying around them. But Maxwell had shown that rotating electrons radiate energy, so their orbits will decay, and we should all vanish within a tiny fraction of a section in a puff of light as electrons crash into the protons!
Armed with our "new" idea of photons, and quanta of energy, we will see how Bohr tried to put together a quantum theory of atoms. He didn't quite get it right, but it will make a convenient mental picture for us, and help us make a transition to more correct quantum 'intuition. (Ultimately, we will have to look deeper, and get more sophisticated than Bohr, to get a truly accurate atomic picture.) But first, we need to discuss some pre-Bohr history and background.
The simplest idea for atoms would be some sort of round, electrically neutral balls, which have no interaction or forces until they touch one another, and then they repel infinitely strongly. A sort of "billiard ball" picture of atoms. This could, in fact, explain gas -> liquid -> solid phase changes. Gas: these atoms flying around, bouncing off each other when they happen to hit. Liquid: Squeeze until the atoms are all touching, so you get an incompressible fluid. Solid: Squeeze and cool until the atoms form a closest packed lattice, so you get an incompressible and rigid solid. This picture is crude and effective, but has lots of problems. How big are these atoms? More important, what are they made of? The atom picture really took shape when we discovered the electrical nature of atoms.
Chemists had been studying elements and molecules and relative combinations. So
they knew, e.g., that 1 g of Hydrogen and 8 grams of Oxygen give 9 grams of
water. And, the hydrogen gas had twice the volume of the oxygen, so they
concluded that 2 hydrogens combine with every O to may a water molecule.
Conclusion: water is
,
and O must weigh 16 times as much as H.
Faraday, in 1831-34, experimented with electrolysis. Put electrical current
into solutions, and measure how much stuff you get out. So, e.g. electrolyzing
9 grams of water gives back 1 g of hydrogen and 8 grams of oxygen. Also, H
comes out at the negative pole, and O at the positive pole. Faraday also
measured how much charge the electrolysis took. He discovered that a fixed
amount of charge liberated elements in the same proportion as the chemist's
combinations. His interpretation: current is transported by ions,
molecular fragments, with particular amounts of electricity associated with
them, and electricity is quantized, it comes in units of some basic
charge. So, e.g.
.
He measured the amount of charge which releases 1 g of H (or 8 g of O) from
water, which is 96,495 C (= 1 Faraday) So, 2 Faradays liberates 2 g of H and 16
g of O.
Now, one can ask, how many H atoms dies it take to weigh exactly 1 g? Let's call this a mole. So, the chemists and Faraday, put together, tell us that
a mole of H atoms weights 1 g (definition)
a mole of H2 molecules weighs 2 g.
a mole of O atoms weighs 16 g
a mole of H2O molecules weights 18g.
1 Faraday = 96,485 Coulombs liberates one mole of H atoms, and 1/2 mole of O atoms. (1/2, because you need two charges for each O).
So, if only we know how much charge the basic unit was (i.e., the amount of
charge on one H+ ion), then we'd know how many particles make a mole.
The discovery of the electron. (J.J. Thomson, 1897)
This was a time when people were first studying X-rays, radioactivity, etc. They knew that one could make a stream of some kind of ray ("cathode ray") from negative electrodes in high voltage, low pressure discharge tubes. And Faraday had showed with electrolysis that molecules could be split into charged ions. In fact, X-rays passing through elemental helium seemed to ionize the He gas, and make it conductive. So, even neutral atoms apparently must possess minute internal electrical charges. The atom as "indivisible" was on its way out... And, the Curie's had observed that a gram of radium gives off 80 calories/hour. Was energy conservation being violated? Alpha particles were discovered coming out of pure metallic Uranium. Was mass conservation being violated? It was a time of great excitement and confusion... These cathode rays also generated a lot of debate -- The British side (Crookes, e.g.) advocated particles, while the Continental side (Lenard @ Bonn, e.g.) preferred waves/radiation. And of course, back then it had to be one or the other!!
Lenard observed cathode rays passing through foils, and continuing on thru gas. It was hard to believe that particles could penetrate a metal foil, and not be deflected. And, with a capacitor plate across his beam, he failed to observe a deflection. You'd certainly expect a beam of negatively charged particles to deflect in an electric field.
At this point, Thomson performed his famous experiments. He had a prejudice that the cathode rays were indeed charged particles, and set out to test the idea.
First: aim the beam at an electrometer cup/electroscope. He clearly observed the accumulation of negative charge. But, is that from the beam, or some secondary effect of the "wave"?
2nd: Put the cup to one side: no accumulation of charge. Now bend the beam with a strong magnetic field => charge enters the cup. (You can't bend photons, or "waves", with magnetic fields!)
3rd: Make a vacuum in the tube that is very good. As the vacuum improved, Thomson started to see a deflection of the beam, caused by the capacitor.
This had been part of Lenards problem - he didn't have "state of the art" vacuum technology. Why was the vacuum important? If there's gas in the tube, the cathode rays will ionize it, so it becomes conductive, and charges can migrate to cancel out the charge on the capacitor plates, so the beam in the middle doesn't notice any net sideways electric field. (At moderate vacuum, you do see a deflection, but after awhile it goes back to the center. Can you understand this effect now?)
4th: Apply an E field by charging up the capacitors, and observe the deflection of the beam:
Note:
(Too many unknowns - we know E and l, but not the other three yet)
5th: Observe that deflecting the spot does not smear it out. Implication? Cathode rays all have the same value of e/(mv^2).
Probably all exactly identical particles, with same initial energy(?)
6th: Now add a crossed B field (into the plane of the paper), and restore the reflection back down to zero:
Every quantity on the rhs is known or measurable, so we learn e/m for our "cathode rays". (Once again, deflecting the spot back to the center does not smear it out. So, vo is the same for all particles. Stronger evidence still that they're all identical.)
Result: e/m = 1.76*10^11 C/kg = 5.3 * 10^17 esu/g
Note: q/m found by Faraday's electrolysis experiments
varied with different ions, but the largest value of all was for Hydrogen itself, where q/m = 1 F/g = 96,000 C/g was about 2000 times smaller than Thomson's value. So, e/m of cathodes rays is big: is this because of a large e, or a small m?? Recall, electrolysis had showed that all ions have the same charge unit, so prejudice says e is constant, and m varies. Also, Lenard's arguments for the penetrating nature of cathode rays says they are very small, so again, probably m is small.
Final point: When you put other gases in the tube, and use other materials, e/m is still constant. Later, Thomson showed that photoelectrons (from the photoelectric experiment) have the same e/m, and again, true no matter what the filament is made of.
Later (1911), Millikan's oil-drop experiments found the charge, e, by itself. He directly observed that charges of his oil drops were quantized. So only at that point could we put everything together and learn that
e = 4.8 *10^-10 esu = 1.6*10^-19 C, and m_e = 9*10^-28 g,
(about 1/2000 the mass of a Hydrogen atom)
So now we can figure out the number of particles in a mole, as I advertised earlier: 1 Faraday = 1 mole of charge = 96,500 C.
So N_A = Avogadro's number = F/e = 6*10^23 atoms/mole.
This was the last link needed to figure out atomic masses, and sizes.
E.g.: If a material has density
= mass/volume, and one mole weighs A, then
,
so volume/atom =
,
i.e.
For example: Carbon weighs 3.5 g/cc, and A=12 g/mole, so
R = (1/2)(A/rho N_A)^(1/3) = 1 Angstrom
Another example: Platinum weighs 21.4 g/cc, and A=195 g/mole, so again R = 1 Angstrom (!)
This is one way, albeit very crude, of finding the rough size of atoms. Interestingly, heavy and light atoms seem to be about the same size.
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