Phys 233
Thurs 10/4
Q9 Atoms
Tues. 10/9
Wed. 10/10
Thurs 10/11
Study Day
Day 9, Q9: Understanding Atoms
1
RE-Q9; Lab Intro 2nd Draft
HW5: Q8: S2, S6, S8; Q9: S2, S4, S5; Lab Notebook & Analysis
RE-Q10; Lab Theory 2nd Draft
Q10 Schrod Eqn
Equipment
o Ppt.
Note: I’ll be out Monday, but here Tuesday.
Q9 Understanding Atoms
Q9.1 Radial and Angular Waves
Q9.2 The Periodic Table
Q9.3 Selection Rules
Q9.4 Stimulated Emission and Lasers
Questions?
So Far
Q9 Understanding Atoms
My perspective on the chapter. As Moore bills, this chapter isn’t getting at fundamental
explanations, but tries to make plausible some results that every good physicists and chemist is
familiar with.
Q9.1 Radial and Angular Waves
“it turns out that” (I hate seeing that phrase in texts – it means “I’m not going to explain this
but”)
The atom
2
Radial wavefunctions for l = 0 states
n = nr + l
Phys 233
Day 9, Q9: Understanding Atoms
These variables relate to the shape of the square of the wavefunction.
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Phys 233
Day 9, Q9: Understanding Atoms
nr is the number of bumps in the radial direction
2l is the number of bumps around the atom at a given radius
Q9B.1 – The square of a hydrogen energy eigenfunction with n = 4 and l = 2 has how many
radial bumps?
How many complete sinusoidal waves around the atom does the wavefunction make?
Number of radial bumps is nr = n – l = 4-2 = 2.
Number of compete wavelengths around is l=2.
n – specifies the energy level (n = 1, 2, 3, …)
E = E0 /n2
l – specifies the magnitude of the orbital angular momentum (l = 0, 1, …, n – 1)
L2 = l (l + 1) ħ2
m – specifies the z component of the orbital angular momentum (m = – l, …, + l)
Lz = mħ
ms – specifies the z component of the electron’s spin
Sz = msħ
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Phys 233
Day 9, Q9: Understanding Atoms
Q9T.3 - One of the spectroscopic symbols listed below specifies an energy eigenstate of the
hydrogen atom that does not exist. Which is the impossible state?
A. 6f
B. 8p
C. 3f
D. 7s
E. 5d
F. 12f
Translating from the spectroscopic notation to angular momenta, s = 0, p = 1, d = 2, f = 3
so the options are
n
l
6
3
8
1
3
3
if n is the number of radial bumps, nr, plus angular waves, l, l < n
7
0
5
2
12
3
Q9T.1 - If all of the n = 3 energy eigenstates in a multielectron atom are filled with electrons,
how many electrons are in those states?
A. 2
B. 4
C. 8
D. 16
E. 18
F. Other (specify)
For a given n, there are n possible angular momenta / number of angular wavelengths.
For each of these, there are 2l+1 possible orbital orientations.
For each of these, there are 2 possible spin alignments
n =3
l=2
l=1
l=0
lz =m=
lz =m=
lz =m=
2
1
0
1
0
0
-1
-1
-2
so, we have 9 *2(spin up, spin down) = 18 possible state
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Phys 233
Day 9, Q9: Understanding Atoms
5
Q9.2 The Periodic Table
Q9B.3 – Use figure Q9.3 to predict the ground-state electron configuration of a manganese atom
(Z = number of protons = 25).
1s22s22p63s23p64s23d5 either that or 1s22s22p63s23p63d7 (depends on which is
energetically lower in the atom, 3d or 4s.)
Q9S.3 – Imagine that we excite the lone outer 4s electron of potassium to a level above the 4s
state. Using figure Q9.3, identify an excited state that is metastable.
The 3d would be metatable; it could decay to the 3p except that that’s full; it can’t decay to the
4s by just one photon without also flipping spin since that would constitute a l = -2. (with a
change of spin that is aligned, could make back up +1 so the photon would still carry away just
1.)
Q9.3 Selection Rules
The photon has a spin quantum number of s = 1 and
Sz = +ħ or –ħ (0 isn’t possible for quantons moving at v = c) – Beth’s experiment
Transitions (emission or absorption) with Δl = ±1 are “allowed” (more probable)
Other transitions are “forbidden” (much less probable) because they require the electron spin to
flip
Phys 233
Day 9, Q9: Understanding Atoms
6
Q9T.6 - An electron in a hydrogen atom is in a 5d energy level. Which of the states below could
it not decay to?
A. 2p
B. 4f
C. 3p
D. 2s
E. 4p
F. All are possible
Let’s translate the letters into angular momenta: s = 0, p = 1, d = 2, f = 3. When the electron
drops to a lower state, it emits a photon (and generally just one photon). A photon has angular
momentum sph = 1. Depending on its spin orientation, emitting that photon could be taking
away or adding angular momentum to the electron. So starting with an l = 2, the electron can
only transition to an l = 1 or l=3 state (p or f) not an l=0 state (s.)
Q9.4 Stimulated Emission and Lasers
Stimulated emission is a lot like adsorption, but in reverse. The book rather qualitatively
suggests how jiggling an apple tree at resonance can make an apple fall. Giving it a little more
detail, if the electron is initially in the higher of two states and it experiences an oscillating
electric field of just the frequency associated with the difference between that and the lower
state, then in that oscillating potential, the electron’s wavefunction is best described by a linear
combination of the two states, so it’s probability density / charge density will oscillate at that
frequency and thus radiate a photon of its own – dropping the electron into the lower state.
Of course, this description presupposes that the electron is initially in the higher state and that the
lower state is initially unoccupied. That’s where the “population inversion” comes in, and
metastable states help with that. The book describes the situation for He Ne lasers: electron
current excites the Helium atoms, in the process of de-exciting, they excite the Ne atoms from
the 2p all the way to the 5s which is metastable in that it can’t drop to the nearest open state – 4d,
but instead must progress to the 3p, then 3s, and finally 2p.
Phys 233
Day 9, Q9: Understanding Atoms
7
Q8S.9 – The potential energy function for a proton or neutron in an atomic nucleus can be
crudely modeled as being a “box” that is about 4 fm = 4 x 10-15 m wide. Imagine that we put 12
neutrons into such a box. Find the minimum total energy of the 12 neutrons in this case. (Hint:
Only two neutrons can occupy each energy level. Explain why.)
Approximately, how much lower would this minimum total energy become if we changed one
half of the neutrons to protons?
For the first question, we’ve got 2 neutrons in each energy level thanks to their being fermions
(having half-integer spin) and the Pauli Exclusion Principle. So, with 12 neutrons, we’re filling
up through the 6th level:
Etotal 2 E1 2 E2 2 E3 2 E4 2 E5 2 E6
Etotal
2 E1 2 E1 2 2
Etotal
2 E1 1 4 9 16 25 36
As for E1, E1
2 E1 32
h2
8m p L2
2 E1 4 2
2 E1 5 2
2 E1 6 2
182E1
2
hc
8 m p c 2 L2
1240eVnm
2
12.8MeV
2
8 939.57 106 eV 4 10 6 nm
So, the total amount of energy (measured relative to the bottom of the well) is 2,330MeV.
If half of these were protons, since they have roughly the same mass, and neglecting the
electric repulsion (which isn’t so bad since we’d probably pack these so protons weren’t right by
each other), we’d essentially have two boxes, one for protons and one for neutrons, and so we’d
only have 6 quantons in each – 2 boxes of six would have
Etotal 2 2 E1 2 E2 2 E3
2 E1 2 2
Etotal
2 2 E1
Etotal
4 E1 1 4 9
so only 4/13ths as much energy.
2 E1 32
56E1
716.8MeV