Atomic Spectra
Rutherford's model of a heavy, small, +'ly charged nucleus surrounded by Z light electrons is appealing, but is flawed in that it doesn't explain why the electrons don't radiate away their energy and fall in. Also, any real theory of atoms should explain the details of atomic spectra. If you take a low pressure gas, and excite it (e.g. with current), it will glow. This glow differs from blackbody glow - it depends on the specific material. The emitted radiation typically has very specific characteristic wavelengths.
The simplest of all is Hydrogen. By the late 1800's, a series of visible lines were known:
6563 A (Red)
4861 (Turquoise)
4341 (Blue)
4102 (Violet)
3970 (Deep Violet)
In 1885, J.J. Balmer (a Swiss high school teacher) discovered that these numbers fit the formula
.
(n=3,4,5,...) (Try it yourself!)
What an amazing and lucky guess! (Would you have deduced that?!) Or, rewriting:
.
This number, 109,700 cm^-1 is called the Rydberg constant, R_h.
Remember, that our quantum theory of light says that
.
Perhaps this gives a clue as to what's going on. If the Hydrogen atom
had certain preferred energies, and the emitted light is what comes out when
the atom shifts from one such energy state to another, then the Balmer series
tells us that Hydrogen has energies of
.
In this way, shifting from e.g. E_4 to E_2 would release an energy of
and so on. In fact, this predicts other possible energies/wavelengths,
e.g. n=6 -> n=5, with
,
which is not in the Balmer series, but is indeed a Hydrogen
line.
These are all Balmer lines, with increasing frequency, except the one at the
bottom, which is not a Balmer line, but does exist. It is part of the
Lyman series:
Knowing a (small?) number of energy levels predicts lots of possible transition energies. And, one predicts certain combination rules. E.g., (Rydberg-Ritz, 1900)
..
Neils Bohr in 1913 tried to piece together Rutherford's results, and the quantization of light, and the clues from spectra, to make the first Quantum Model of the atom. He invoked 2 main postulates:
(1) Atoms have "stationary states". These are Newtonian electron orbits, but they do not radiate.
(2) Radiation is emitted only when an electron changes orbits. The photon energy is just the classical energy difference of the initial and final states. So,
.
These postulates may seem like cheats - you are concerned about why the electron doesn't radiate, and Bohr says "Because it doesn't". Bear in mind that this model is just a *stepping stone* to the true quantum theory, which will do a much better job at satisfying us! (The Bohr model is most decidedly not quantum mechanics) But, let us at least first see how these simple postulates yield the Balmer formula. They will have even more consequences too! Bohr used a funny mix of classical and quantum ideas, but for now we stick to these simple hypotheses, and use experiment to guide our way... First, let's find the "stationary states". What are the classical, Newtonian orbit equations for hydrogen?
For an electron (charge -e) circling a proton (charge +e), using cgs units, we have
.
The total energy of the electron is given by kinetic + potential energy,
(minus sign on the potential, because they attract each other, or in other
words you must do work to get the electron pulled out to infinity). So,
combining with the equations above,
.
This negative sign is good -- it means the electron is bound to the
proton.
Bohr's postulate #1 says that only certain of these orbits are allowed, or stationary. Bohr had a theoretical (quantum) argument for which orbits are allowed, which we will discuss soon. Let's first figure it out empirically, using Balmer's formula.
Certain radii will be allowed (by assumption), and let us number them, r_1,
r_2, r_3, ... This means that the energy corresponding to that radius is
.
If you move from one orbit to another, you emit a photon with
.
But, our empirical spectral formula gave us (recall)
,
(Note! I made a sign error in the last part of that formula. The n_f and n_i terms need to be reversed, because of the minus sign I pulled outside. Sorry about that)
which implies that
a_0 is known as the "Bohr radius". We also have, combining earlier equations, that
.
Now, since R_H is known from Mr. Balmer, the lowest energy level (n=1) and smallest orbit radius can be computed.
.
(which is a reasonable number for the size of a hydrogen atom) Also,
the lowest energy level can be computed:
.
(Note! That last formula needs a minus sign in front of the 13.6 eV, clearly. Sorry 'bout that)
Also note: E_infinity = 0, so 13.6 eV is also the number we obtain for the ionization energy of Hydrogen.
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