21: Lecture Lasers, Atoms, Molecules, Solids, etc. Review and Examples
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Lecture 21, p 1
Lasers
Photons are emitted when the electrons in atoms go from a higher state to a lower state Conversely, photons are absorbed when the electrons in atoms go from a lower state to a higher state
Emission
photon
photon
Absorption
Fermions and bosons:
Electrons, protons, and neutrons are fermions. Two identical fermions cannot occupy the same quantum state. (exclusion principle) Photons (and many atoms) are bosons. Unlike fermions, bosons actually prefer (to be explained soon) to be in the same quantum state. This is the physical principle on which lasers are based.
Lecture 20, p 2
Lasers
Suppose we have an atom in an excited state. Eventually (at some random time) it will emit a photon and fall to the lower state. The emitted photon will go in a random direction. This is called spontaneous emission.
Emission
photon
Suppose, however, that before the spontaneous emission occurs, another photon of the same energy (emitted by another atom) comes by. Its presence will stimulate the atom to emit its photon.
Photon emitted by some other atom
Two identical photons!
Stimulated emission
We now have two photons in the same quantum state: the same frequency, the same direction, and the same polarization. As they travel, they will stimulate even more emission.
Lecture 20, p 3
Lasers
Light Amplification by Stimulated Emission of Radiation
Mirrors at both ends
Laser operation (one kind):
Tube of gas with mirrored ends. Excite the atoms to the upper state (mechanism not shown).
100%
Tube of gas
99.99%
Spontaneous emission generates some photons. Photons that travel along the axis are reflected. Other directions leak out the sides. Because the amplification process is exponential, the axial beam quickly dominates the intensity. One mirror allows a small fraction of the beam out.
Lecture 20, p 4
Lasers
LASER: Light Amplification by Stimulated Emission of Radiation
What if you dont have an atom that emits the color you want? Make one (e.g., by stressing the crystal lattice). Did you know: Semiconductors dont naturally emit in the red. Charles Henry (UIUC 65) discovered how to combine layers of different semiconductors to make a quantum well (essentially a 1-D box for electrons). By adjusting the materials, one could shift the emission wavelengths into the visible. Nick Holonyak (UIUC ECE Prof) used this to create the first visible LEDs & visible laser diodes.
Act 1
The Pauli exclusion principle applies to all fermions in all situations (not just to electrons in atoms). Consider electrons in a 2-dimensional infinite square well potential. 1. How many electrons can be in the first excited energy level? a. 1 b. 2 c. 3 d. 4 e. 5 Hint: Remember the (nx,ny) quantum numbers.
2. If there are 4 electrons in the well, what is the energy of the most energetic one (ignoring e-e interactions, and assuming the total energy is as low as possible)? a. (h2/8mL2) x 2 b. (h2/8mL2) x 5 c. (h2/8mL2) x 10
Lecture 21, p 6
Solution
The Pauli exclusion principle applies to all fermions in all situations (not just to electrons in atoms). Consider electrons in a 2-dimensional infinite square well potential. 1. How many electrons can be in the first excited energy level? a. 1 b. 2 c. 3 d. 4 e. 5 The first excited energy level has (nx,ny) = (1,2) or (2,1). That is, it is degenerate. Each of these can hold two electrons (spin up and down).
(Note: On an exam, Id word this question a bit more carefully.)
2. If there are 4 electrons in the well, what is the energy of the most energetic one (ignoring e-e interactions, and assuming the total energy is as low as possible)? a. (h2/8mL2) x 2 b. (h2/8mL2) x 5 c. (h2/8mL2) x 10
Lecture 21, p 7
Solution
The Pauli exclusion principle applies to all fermions in all situations (not just to electrons in atoms). Consider electrons in a 2-dimensional infinite square well potential. 1. How many electrons can be in the first excited energy level? a. 1 b. 2 c. 3 d. 4 e. 5 The first excited energy level has (nx,ny) = (1,2) or (2,1). That is, it is degenerate. Each of these can hold two electrons (spin up and down).
2. If there are 4 electrons in the well, what is the energy of the most energetic one (ignoring e-e interactions, and assuming the total energy is as low as possible)? a. (h2/8mL2) x 2 b. (h2/8mL2) x 5 c. (h2/8mL2) x 10 Two electrons are in the (1,1) state, and two are in the (2,1) or (1,2) state. So, Emax = (12+22)(h2/8mL2).
Lecture 21, p 8
Magnetic Moments of Atoms
Electrons have a magnetic moment. Consequently, many atoms (e.g., iron) do as well. However, atoms that have completely filled orbitals never have a magnetic moment (in isolation). Why is this?
Lecture 21, p 9
Solution
Electrons have a magnetic moment. Consequently, many atoms (e.g., iron) do as well. However, atoms that have completely filled orbitals never have a magnetic moment (in isolation). Why is this? An orbital is a set of states, all with the same (n,). There are 2+1 m values and 2 ms values. When the orbital is completely full, the angular momentum of the electrons sums to zero, because for every positive value of m and ms, there is a corresponding negative one. Therefore, there cannot any be magnetic moment (which results from orbiting charge).
Lecture 21, p 10
Noble (Inert) Gases
Except for helium, the electronic configuration of all the noble gases has a just-filled p orbital. What is the atomic number of the third noble gas (neon is number 2)?
Lecture 21, p 11
Solution
Except for helium, the electronic configuration of all the noble gases has a just-filled p orbital. What is the atomic number of the third noble gas (neon is number 2)? This is just a counting problem: 2 +2 +6 +2 +6 = 18 Thats argon. 1s 2s 2p 3s 3p
neon
Lecture 21, p 12
Another Inert Gas Question
Why is it that (except for helium) the electronic configuration of all the noble gases has a just-filled p orbital? Can we understand helium also?
Lecture 21, p 13
Solution
Why is it that (except for helium) the electronic configuration of all the noble gases has a just-filled p orbital? Can we understand helium also?
Look at the order in which the states are filled. The atom just after each noble gas (including helium!) is the first to have an electron in the next n-shell, For example, sodium follows neon and is the first atom to have an electron in an n=3 state. The big energy gap makes it difficult for the noble gas electrons to form chemical bonds.
Lecture 21, p 14
Rare Earth (Lanthanide) Elements
The rare earth elements: Lanthanum (Z = 57) to Ytterbium (Z = 70), plus Lutetium (Z = 71) all have similar chemical properties. How can we explain this in terms of what we have learned?
Lecture 21, p 15
Solution
The rare earth elements: Lanthanum (Z = 57) to Ytterbium (Z = 70), plus Lutetium (Z = 71) all have similar chemical properties. How can we explain this in terms of what we have learned?
Element 56 (Barium) fills all the orbitals up to 6s. The next 14 elements (57-70) will put their electrons in the 4f orbital, which has a significantly smaller radius than 6s, and thus has little chemical effect. There is a similar effect (the Actinides) after element 88 (Radium, which fills 7s).
Caveat: Lanthanum (in violation of our mnemonic) and Lutetium put one electron in 5d, but have similar chemical properties nevertheless.
http://en.wikipedia.org/wiki/Lanthanide
Lecture 21, p 16
d
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Act 2
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Bonding state
Antibonding state
As d decreases, the energy of the bonding state decreases (until a minimum energy is reached), while that of the antibonding state does not. The reason is: a. The nuclei have a stronger interaction in the bonding state. b. The bonding state spreads out more, i.e., has less curvature. c. The electron is a fermion, and prefers the symmetric .
Lecture 21, p 17
d
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Solution
r
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Bonding state
Antibonding state
As d decreases, the energy of the bonding state decreases (until a minimum energy is reached), while that of the antibonding state does not. The reason is: a. The nuclei have a stronger interaction in the bonding state. b. The bonding state spreads out more, i.e., has less curvature. c. The electron is a fermion, and prefers the symmetric .
a) False. First, the interaction between the nuclei is repulsive. And in fact, this is somewhat shielded by the electron sitting in between, which actually reduces the repulsion (and lowers the energy). b) is right. The node in the antisymmetric state prevents the wavelength from increasing much. Also, in the bonding state the electron probability between the nuclei increases, pulling the nuclei toward it. c) The fact that the electron is a fermion is irrelevant. That only affects multielectron states.
Lecture 21, p 18
Act 3
3p band
Band gap
3s band
Consider a crystalline solid in which each atom contributes some electrons to the 3s band. Which situation can produce a conductor. a. Each atom contributes one 3s electron. b. Each atom contributes two 3s electrons. c. Each atom contributes three 3s electrons.
Lecture 21, p 19
Solution
3p band
Band gap
3s band
Consider a crystalline solid in which each atom contributes some electrons to the 3s band. Which situation can produce a conductor. a. Each atom contributes one 3s electron. b. Each atom contributes two 3s electrons. c. Each atom contributes three 3s electrons. For N atoms, the band can hold 2N electrons (spin up and spin down). So: a) gives us a half-filled band (conductor). b) gives a full band (insulator). c) is not possible. An atom can only have two 3s electrons.
Lecture 21, p 20
Band Widths
Why is the 3s band wider (bigger energy spread) than 1s, 2s, or 2p?
3s 2s, 2p 1s
a. Because n is larger. b. Because it overlaps with the 3p band (not shown). c. Because E3s is larger.
Lecture 21, p 21
Solution
Why is the 3s band wider (bigger energy spread) than 1s, 2s, or 2p?
3s 2s, 2p 1s
a. Because n is larger. b. Because it overlaps with the 3p band (not shown). c. Because E3s is larger. a) You might think this is equivalent to c), but beware. For larger n values, the order can be reversed. 6s has lower energy than 5p. b) may be true, but the presence of the 3p band does not affect 3s energy levels. c) Higher energy means that there is more tunneling. This causes the symmetric wave functions to have a larger energy shift.
Lecture 21, p 22
THE END
Best wishes in Physics 213 and all your other courses!
Lecture 21, p 23