# 4 Quantum Numbers for the Hydrogen Atom by kqm58610

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```									                             The End                                                       4 Quantum Numbers for the
• All homework solutions are now available                                                      Hydrogen Atom
• Check CULearn for all scores (HW, Clickers, Exams)                       For hydrogenic atoms (one electron), energy levels only depend
on n and we find the same formula as Bohr: En = −Z 2 ER / n2
– Any requested changes are due today
For multielectron atoms the energy also depends on ℓ.
• Final exam is Tuesday, December 15, 2009                                 There is a shorthand for giving the n and ℓ values.
4:30 pm – 7:00 pm in G125 (this room)
2p                   Different letters correspond
to different values of ℓ
• Bring Calculator! Arrive on time, check seating rows.                                                          s     p     d      f     g       h…
n=2                 ℓ=1
• Additional review help (in the Physics Help Room),                                                       l=   0      1     2      3     4       5
today    Friday 1:30 – 4:00 pm
The electrons also have another quantum number for
• I have to travel to Brookhaven National Lab next week
their spin (ms = +/- ½).
and so the exam will be proctored.

Clicker question                                                                                 Hydrogen energy levels
n = 1, 2, 3, … = Principal Quantum Number           En = −Z ER / n
2         2
ℓ=0        ℓ=1     ℓ=2
ℓ = 0, 1, 2, … n-1 = angular momentum quantum number                                 (s)        (p)    (d)
= s, p, d, f, …           L = l(l + 1)h                                n=3                                       E3 = − ER / 32 = −1.5 eV
3s     3p      3d
m = 0, ±1, ±2, … ±ℓ     is the z-component of
Lz = mh
angular momentum                                  n=2                          E2 = − ER / 22 = −3.4 eV
2s     2p
ms = ±1/2               from the spin contribution
A hydrogen atom electron is excited to an energy of −13.6/9 eV.
How many different quantum states could the electron be in?
E = −13.6/9 eV means n2 = 9 so n = 3
A.   1
B.   4              For n = 3, ℓ = 0,1,2 and thus
C.   9              l = 0, m = 0                      x 2 for spin!
D.   18             l = 1, m = -1, 0, +1                                  n=1            1s   E1 = − ER = −13.6 eV
E.   more than 18   l = 2, m = -2, -1, 0, +1, +2

Probability versus radius: P(r) = |Rnl(r)|2r 2                                 Higher n        average r bigger
more spherical shells stacked within each other
In spherical coordinates,                                                                       more nodes as function of r
the volume element has
an r2 term so probability                                                 n=1
increases with r2.                                                        l=0

Probability finding
Most probable radius for                                                                                             electron as function of r
the n = 1 state is at the                                                 n=2

ℓ=n-1 states (those with only                                             n=3
one peak) is at the radius                                                l=0
predicted by Bohr (n2 aB).
0.05nm

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Shapes of hydrogen wave functions:                                   Shapes of hydrogen wave functions:

ψ nlm (r ,θ , φ ) = Rnl (r ) f lm (θ ) g m (φ )                      ψ nlm (r ,θ , φ ) = Rnl (r ) f lm (θ ) g m (φ )
l=1, called p-orbitals: angular dependence (n=2)
Look at s-orbitals (l=0): no angular dependence
l=1, m=0: pz = dumbbell shaped.
n=1                           n=2
l=1, m=-1: bagel shaped around z-axis (traveling wave)
l=1, m=+1

1     r − r / 2 a0 ⎛   3       ⎞
n = 2, l = 1, m = 0 ⎯ ψ 211 =
⎯→                     e         ⎜−    cos θ ⎟
3               ⎜  4π       ⎟
2 6 a0 a 0           ⎝           ⎠
1 r − r / 2 a0 ⎛   3           ⎞   w/time dependence

n = 2, l = 1, m = 1 ⎯ ψ 211 =
⎯→                   e         ⎜−    sin θe iφ ⎟       eimφ+iΕt/h
3          ⎜  8π           ⎟
2 6a a 00          ⎝               ⎠

Superposition applies:
px=superposition (addition of m=-1 and m=+1)    Dumbbells
py=superposition (subtraction of m=-1 and m=+1) (chemistry)

Schrodinger finds quantization of energy and angular momentum:
Physics vs Chemistry view of orbits:                         n=1, 2, 3 …       l=0, 1, 2, 3 (restricted to 0, 1, 2 … n-1)
2p wave functions                   Dumbbell Orbits              En = − E1 / n 2                       | L |= l (l + 1) h
(Physics view)                      (chemistry)
(n=2, l=1)                                                     How does Schrodinger compare to what Bohr thought?
same
I. The energy of the ground state solution is ________
II. The angular momentum of the ground state solution is different
_______
px                        III. The location of the electron is different
_______
pz   py
m=1                  m=-1
m=0                     px=superposition                 a. same, same, same
(addition of m=-1 and m=+1)      b. same, same, different                    Bohr got energy right,
py=superposition                 c. same, different, different               but he said angular
(subtraction of m=-1 and m=+1)   d. different, same, different               momentum L=nh, and
e. different, different, different          thought the electron was
a point particle orbiting
around nucleus.

Can Schrodinger make sense of the periodic table?
Schrödinger’s Cat
A radioactive sample has a 50% chance of emitting an alpha particle.
If it decays, a Geiger detector triggers the release of poison killing a
cat in the box. Before opening the box, the cat is in a superposition
of wave functions: ψ = 1ψ           + 1ψ

When does the wave function collapse to either dead or alive?
No clear agreement. Interesting physics/philosophical question.

Schrodinger’s solution for multi-electron atoms
Need to account for all the interactions among the electrons
Must solve for all electrons at once! (use matrices)
V (for q1) = kqnucleus*q1/rn-1 + kq2q1/r2-1 + kq3q1/r3-1 + ….

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Good luck on
the final exam!

Thank you for all your
attention and hard work this
semester.

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