Announcements
• Reading next class 5.7
• Homework 6 due Wednesday.
• Exam 2 next week (Thursday)
– Inclusive there will be some SR.
– Modern physics
– Photons and PE
– Atomic spectra
– Bohr atom etc.
Atoms/ Balmer/ Bohr atom
A neon lamp emits a red line. Sodium emits a yellow line.
What accounts for this difference?
a.The electrons in the discharge lamp hit the neon atoms
with higher kinetic energy than the electrons hit the sodium
atoms
b.The electrons in the discharge lamp hit the neon atoms
with less speed than the electrons hit the sodium atoms
c.The energy spacing between the electronic energy levels
that are responsible for these lines are smaller in the neon
atom than in the sodium atom.
d.The energy spacing between the electronic energy levels
that are responsible for these lines is larger in the neon
atom than in the sodium atom.
What we know so far about atoms:
Rutherford: Atoms have a tiny, but very dense
core surrounded by a cloud of electrons.
Discharge lamps: Atoms struck by fast
electrons emit light of distinct colors.
Energy levels: Electrons in atoms are found only
in discrete energy levels. When they jump
down to a lower level, a photon is emitted
carrying away the energy.
p
n nn
p p p
nn nn
p
120V
ΔE
ΔE=hf
e
Different Atoms: Different types of atoms have different level
structures (seen by the distinct set of colors they emit):
Neon: Strong red line; Sodium: strong yellow line …
increasing
energy
What do the vertical arrows represent?
allowed energies in an atom
“ground state”
a. arrows on right represent absorption of light,
arrows on left represent emission.
b. arrows on both left and right represent emission
c. arrows on both left and right represent absorption
d. arrows on left represent absorption of light,
arrows on right represent emission.
e. I don’t know. I have never seen such a diagram.
increasing
energy
allowed energies
ground state
d. arrows on left represent absorption of light,
arrows on right represent emission.
e
e
e
e
Energy
Energy level diagrams represent energy levels the electron can
go to. Different height = different energy
e
For Hydrogen,
transitions to
ground state in
deep ultraviolet!
No light emitted with
colors in this region
because no energy
levels spaced with
this energy.
e
An single electron bashes into an atom in a discharge lamp
Electron leaves hot filament with
nearly zero initial kinetic energy
-2 eV
- 10 V +
-3 eV
-6 eV
-10eV
If atom fixed at the center of the tube,
list all the possible photon energies (colors) that you might see?
A. 1eV, 2eV, 3eV, 4eV, 7eV, 8eV
B. 4eV, 7eV, 8eV
C. 1eV, 3eV, 4eV
D. 4eV
E. Impossible to tell.
An single electron bashes into an atom in a
discharge lamp
- 10 V +
d
-2 eV
-3 eV
D
-6 eV
If atom fixed at this point in tube,
-10eV
list all the possible photon energies (colors) that you might see?
A. 1eV, 2eV, 3eV, 4eV, 7eV, 8eV
B. 4eV, 7eV, 8eV
C. 1eV, 3eV, 4eV
D. 4eV Answer is D. Electron only gains about 5eV!
E. Impossible to tell.
Electron energy = qV = e(Ed),
where E is the electric field = (battery V)/(total distance D),
and d is the distance it goes before a collision.
When an emission line appears very bright that
means:
a.Multiple photons are emitted for each single
electron transition
b.More electron transitions are occurring each
second
c.The electronic energy levels are farther apart
and thus the line appears brighter
d.a and b
e.b and c
Why do I have a problem with this question?
When an emission line appears very bright (to
an intelligent observer) that means:
a.Multiple photons are emitted for each single
electron transition
b.More electron transitions are occurring each
second
c.The electronic energy levels are farther apart
and thus the line appears brighter
d.a and b Answer is b. Each electron transition produces
exactly one photon with energy equal to energy
e.b and c difference between electron energy levels. More
electron transitions, more photons, brighter line.
But this assumes we’ve corrected for what
“apparent” brightness is.
Applications of atomic spectroscopy
1. Detecting what kind of atoms are in a material.
(excite by putting in discharge lamp or heating in flame
to see spectral lines)
2. Detecting what the sun and the stars are made of:
Look at the light from a star through a diffraction grating.
See what lines there are; Match up to atoms on earth.
telescope
star
diffraction
grating
3. Making much more efficient lights!
Incandescent light bulbs waste >90% of the electrical
energy that goes into them! (<10% efficient)
Streetlight discharge lamps (Na or Hg) ~80%
efficient. Fluorescent lights ~ 40-60% efficent.
Atomic spectra in astronomy
L
Let’s investigate the internal
working of a galaxy
L
R
R
longer wavelength
lower frequency
Spectral lines from
atoms in stars
Hubble and the big bang
Spectral lines from Hydrogen
Edwin Hubble, PNAS March 15, 1929 vol. 15 no. 3 168-173
Application:
Designing a better light
What is important?
(Well, we have to be able to see the light!)
What color(s) do you want?
(red, green & blue is what our eyes can see)
How do you excite the atoms to desired level?
( Electron collisions)
How to get electrons with desired energy when
hit atoms? What determines energy of electrons?
Incandescent light (hot filament)
Temperature = 2500-3000K
Hot electrons jump
between many very
closely spaced levels
(solid metal). Produce all
colors. Mostly infrared at
temp of normal filament.
>90% is worthless
Infrared radiation (IR =
longer than ~700nm)
IR
P
λ
~10% of energy is
useful visible light
Discharge lamp
Energy levels in
isolated atom:
Only certain
wavelengths
emitted.
HV
Right atom, right pressure
and voltage, mostly visible light.
Streetlight discharge lamps
(Na or Hg) 80% efficient.
Florescent Lights.
Discharge lamp, but want to have it look white.
White = red + green +blue
40-60% efficient (electrical power ⇒ visible light)
How to do this?
Converting UV light into visible photons with “phosphor”.
Phosphor converts 180 nm UV to red+ green+blue.
180nm  6.9 eV energy per photon
633nm (red)  2 eV / photon
532nm (green)  2.3 eV / photon
475nm (blue)  2.6 eV / photon
phosphor wastes 20-30% energy ⇒ heat
}
6.9 eV
Florescent Lights. Discharge lamp, White= Red + green +blue
40-60% efficient (electrical power⇒visible light)
Converting 180nm UV light into visible photons with “phosphor”.
phosphor
coating
Hg
180 nm far UV
Hg
Hg
Hg
Hg
e
Hg
discharge lamp/flor. tube
120 V
real phosphors more than just 3
phosphor wastes 20-30% energy
⇒ heat
energy of electron
in phosphor molecule
Summary of important Ideas
1) Electrons in atoms are found at specific energy levels
2) Different set of energy levels for different atoms
3) one photon emitted per electron jump down between energy
levels. Photon color determined by energy difference.
4) If electron not bound to an atom: Can have any energy. (For
instance free electrons in the PE effect.)
Hydrogen
Lithium
Energy
Electron energy levels in 2
different atoms: Levels have
different spacing (explains
unique colors for each type of
atom.
(not to scale)
Atoms with more than one
electron … lower levels filled.
Now we know about the energy
levels in atoms. But how can we
calculate/predict them?
 Need a model
Step 1: Make precise, quantitative observations!
Step 2: Be creative & come up with a model.
Step 3: Put your model to the test.
Balmer series:
A closer look at the
spectrum of hydrogen
656.3 nm
410.3 486.1
434.0
Balmer (1885) noticed wavelengths followed a progression
91.19nm

1 1
 2
2
2 n
where n = 3,4,5, 6, ….
As n gets larger, what happens to wavelengths of
emitted light?
 λ gets smaller and smaller, but it approaches a limit.
Balmer series:
A closer look at the
spectrum of hydrogen
656.3 nm
410.3 486.1
434.0
Balmer (1885) noticed wavelengths followed a progression
91.19nm
So this gets smaller  
1 1
 2
2
Balmer correctly
2 n
predicted yet
undiscovered
spectral lines.
where n = 3,4,5,6, ….
gets smaller as n increase
gets larger as n increase,
but no larger than 1/4
limit  4 * 91.19nm  364.7nm
λ gets smaller and smaller, but it approaches a limit
Hydrogen atom – Rydberg formula
Does generalizing Balmer’s formula work?
Yes!  It correctly predicts additional lines in HYDROGEN.
Rydberg’s general formula
Hydrogen energy levels
91.19nm

1
1
 2
22
mm n
Predicts  of nm transition:
n
(n>m)
m
(m=1,2,3..)
m=1, n=2
Hydrogen atom – Lyman Series
Rydberg’s formula
91.19nm

1
1
 2
2
m n
Hydrogen
energy
levels
Predicts  of nm transition:
n
(n>m)
m
(m=1,2,3..)
0eV
Can Rydberg’s formula
tell us what ground
state energy is?
-?? eV
m=1
Balmer-Rydberg formula
Hydrogen energy levels
0eV
91.19nm

1
1
 2
2
m n
Look at energy for a transition
between n=infinity and m=1
0eV
0
hc  1
1
Einitial  E final 

 2 2
 91.19nm  m n 
hc
hc
1
 E final 
2
91.19nm m
-?? eV
1
Em  13.6eV 2
m
Balmer/Rydberg had a mathematical formula to describe
hydrogen spectrum, but no mechanism for why it worked!
Why does it work?
Hydrogen energy levels
Balmer’s formula
656.3 nm
410.3 486.1
434.0
91.19nm

1
1
 2
2
m n
where m=1,2,3
and where n = m+1, m+2
m=1, n=2
The Balmer/Rydberg formula is a
mathematical representation of an
empirical observation.
It doesn’t explain anything, really.
How can we calculate the energy levels
in the hydrogen atom?
 A semi-classical explanation of the
atomic spectra (Bohr model)