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Fri. 11.10 Quantization, Quiz 11 Lecture Evals RE 11.e
Mon. Review for Final (1-11) HW11: Pr’s 39, 57, 64, 74, 78
Sat. 9 a.m. Final Exam (Ch. 1-11)
Using Angular Momentum The measure of motion about a point
Magnitude
sinrp
r
p
p
r
Direction
Orient Right hand so fingers curl from the axis and with motion,
then thump points in direction of angular momentum.
rpL rp
=
Magnitude and Direction
=
pole
polebllr
p
ropeballF
handballF
Fr AfromAabout
)()(where,
net
AaboutAaboutLdt
d)()(
Angular Momentum Principle
Torque
Quantifies motion
about a point
Interaction that
changes motion
about a point
Magnitude
(yet another cross product)
FrAA
sinFrAA
FrA
FrA sin sinFrA
Making sense of the
factors and cross-product
Direction
(yet another cross product)
F
Ar
A
Zero-Torque Systems
Initial Final
tLL aveirotfrot
..
0.. irotfrot LL
frotirot LL ..
ffii II
mass farther
from axis: Ii larger
i smaller
mass closer to axis:
If smaller
i larger
Spinning Skater
Three Fundamental Principles Angular Momentum:
net
AaboutAaboutLdt
d)()(
(linear) Momentum:
net
Fpdt
d Energy:
net
WE
Example all three together! Say we have a uniform 0.4 kg puck with an 8 cm radius. A
24 cm string is initially wrapped around its circumference. If it’s on a frictionless surface
and a 10 N force is applied to the end of the string until it’s unwound…
R = 0.08m
l = 0.24 m
R
d = ? 2
21 mRI puck
f =?
vcmf = ?
a. What will be its rate of rotation when the string is fully unwound?
Energy Principle
WEtotal
Ftrans rFEK
int
ldFKrF rotcm
FlFdIIFd if
2
212
21
FlI f
2
21
I
Flf
2
m
Fl
RmR
Flf
222
21
?t
Three Fundamental Principles Angular Momentum:
net
AaboutAaboutLdt
d)()(
(linear) Momentum:
net
Fpdt
d Energy:
net
WE
Example all three together! Say we have a uniform 0.4 kg puck with an 8 cm radius. A
24 cm string is initially wrapped around its circumference. If it’s on a frictionless surface
and a 10 N force is applied to the end of the string until it’s unwound…
R = 0.08m
l = 0.24 m
R
d = ? 2
21 mRI puck
f
vcmf = ?
a. What will be its rate of rotation when the string is fully unwound? m
Fl
Rf
2
b. How long was the force applied?
?t
Angular Momentum Principle
(axis through final location of cm) dtL
tLL if
tLLLL rotitransirotftransf
)()( ....
tFrI aFf
tRFI f
Torque and final angular velocity in –z direction
F
ml
F
m
Fl
RRm
F
Rm
RF
mR
RF
It
fff
2
2
2
2
21
RF
It
f
RF
mR f
2
21
F
Rm f
2 F
m
Fl
RRm
2
2
F
ml
Three Fundamental Principles Angular Momentum:
net
AaboutAaboutLdt
d)()(
(linear) Momentum:
net
Fpdt
d Energy:
net
WE
Example all three together! Say we have a uniform 0.4 kg puck with an 8 cm radius. A
24 cm string is initially wrapped around its circumference. If it’s on a frictionless surface
and a 10 N force is applied to the end of the string until it’s unwound…
R = 0.08m
l = 0.24 m
R
d = ? 2
21 mRI puck
f
vcmf = ?
a. What will be its rate of rotation when the string is fully unwound? m
Fl
Rf
2
b. How long was the force applied?
?t
F
mlt
You try:
c. How quickly is the puck finally sliding, vcm f?
d. How far has the puck moved, d?
Understanding the Hydrogen Spectrum:
Bohr’s step toward Quantum Mechanics
3.0
2 eV
2.8
5 e
V
2.5
5 e
V
21.6
9 e
V
K+U
Ground State
Excited
States
n = 1
n = 3 n = 4 n = 5 n =
-13.6 eV
-1.51 eV -0.85 eV -0.54 eV
r
eU
o
2
41
2
6.13
n
eVUK n
211
6.13 eVUK
etc.
0
Hydrogen Energy Levels
n = 2 -3.40 eV 222
6.13 eVUK
K+U
Hydrogen Excitation: 1st in ground state
Ground State
Excited
States
n = 1
n = 2
n = 3 n = 4 n = 5 n =
-13.6 eV
-3.40 eV
-1.51 eV -0.85 eV -0.54 eV
2
6.13
n
eVUK n
0
Marks the “occupied state”
K+U
Hydrogen Excitation: 2nd Adsorbs energy from Collision
excites to 2nd state
Ground State
Excited
States
n = 1
n = 2
n = 3 n = 4 n = 5 n =
-13.6 eV
-3.40 eV
-1.51 eV -0.85 eV -0.54 eV
2
6.13
n
eVUK n
0
K+U
Hydrogen Excitation: 3rd Looses Energy by photon emission,
de-excites to ground state
Ground State
Excited
States
n = 1
n = 2
n = 3 n = 4 n = 5 n =
-13.6 eV
-3.40 eV
-1.51 eV -0.85 eV -0.54 eV
2
6.13
n
eVUK n
0
Photon Emission
eVE
eVeVE
EE
ph
ph
Hph
2.10
1
6.13
2
6.1322
hfEph
fHzeVs
eV55.2
10125.4
2.1015
Why?
Why?
Some combo of relevant constants?
hfE ph
r
eU
o
2
412
21 vmK e
2
21
2
41
2 hn
emoe
Why?
2
6.13
n
eVUK n
Rephrasing Classical Energy Expression for Orbiting Electron 2
21
2
41
2 hn
emUK oe
n
r
evmUK
oe
2
412
21 Specifically for Circular Motion:
2
2
41
2
r
e
r
vm
oe
netFdt
pd
so
r
evm
oe
2
412
22
21 vmvmUK ee
2
2vem
Rephrase orbital kinetic
in terms of L:
vrmL e so
How to eliminate r:
vm
Lr
e
Naturally,
rm
ev
eo
2
412
vm
Lm
e
e
e
o
2
41
vL
eo
2
41
Target expression:
22
41
2L
eUK
o
emso
22
41
2L
eUK oem Works if
2
hnL Why?
L
ev
o
2
41
Why : De Broglie’s Contribution 222
mcpcEFrom Einstein:
For photon (being massless) pcE From experiments hf
frequency – wavelength – wave-speed relation
cf
so c
hfp
so h
p De Broglie’s big idea: what if this is true for particles too
– some kind of wave associated with momentum
Then (for circular motion) means prL rh
L
22
41
2L
eUK oem
22
41
2 r
m
h
eoe
2
hnL
Works if 2
nr
Why?
Return to our K+U expression:
nr2
circular waves demo
Whatever these waves are, they must ‘fit’ the orbit
Today we understand the waves to relate to the probability
Bohr Radii
We’d found that vrmL e
Along with only specific L values and
K+U values, there are only specific radii
08_Bohr_levels.py
L
ev
o
2
41
where
2
hnL
so r
L
emL
oe
2
41
2
2
41
2
nem
r
oe
2
41
2
em
Lr
oe
2
h
n
08_Bohr_levels.py
2
2
41
2
nem
r
oe
1) r = (8.5×10-30 meter)n2
2) r = (5.0×10+23 meter)n2
3) r = (4.8×10-1 meter)n2
4) r = (5.3×10-11 meter)n2
5) r = (1.2×10-38 meter)n2
sJ 1005.1 -34 kg 109 -31
em
C 106.1 -19e
229
0
C/mN 1094
1
Fri. 11.10 Quantization, Quiz 11 Lect Evals RE 11.e
Mon. Review for Final (1-11) HW11: Pr’s 39, 57, 64, 74, 78
Sat. 9 a.m. Final Exam (Ch. 1-11)
