PHYS 3313 – Section 001
Lecture #12
Monday, Feb. 24, 2014
Dr. Jaehoon Yu
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Rutherford Scattering Experiment and
Rutherford Atomic Model
The Classic Atomic Model
The Bohr Model of the Hydrogen Atom
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
1
Announcements
• Quiz 2 results
– Class average: 26.7/50
• Equivalent to 53.4/100
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Previous quiz: 30/100
– Top score: 50/50
• Mid-term exam
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In class on Wednesday, Mar. 5
Covers CH1.1 – what we finish on Monday, Mar. 3 + appendices
Mid-term exam constitutes 20% of the total
Please do NOT miss the exam! You will get an F if you miss it.
BYOF: You may bring a one 8.5x11.5 sheet (front and back) of handwritten
formulae and values of constants for the exam
– No derivations or solutions of any problems allowed!
– No additional formulae or values of constants will be provided!
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
2
Reminder: Special Project #3
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A total of Ni incident projectile particle of atomic
number Z1 kinetic energy KE scatter on a target of
thickness t and atomic number Z2 and has n atoms
per volume. What is the total number of scattered
projectile particles at an angle  ? (20 points)
Please be sure to clearly define all the variables
used in your derivation! Points will be deducted for
missing variable definitions.
This derivation must be done on your own. Please
do not copy the book, internet or your friends’.
Due is this Wednesday, Feb.26 .
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
3
Prefixes, expressions and their meanings
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deca (da): 101
hecto (h): 102
kilo (k): 103
mega (M): 106
giga (G): 109
tera (T): 1012
peta (P): 1015
exa (E): 1018
Monday, Feb. 24, 2014
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deci (d): 10-1
centi (c): 10-2
milli (m): 10-3
micro ( ): 10-6
nano (n): 10-9
pico (p): 10-12
femto (f): 10-15
atto (a): 10-18
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
4
Experiments of Geiger and Marsden
• Rutherford, Geiger, and
Marsden conceived a new
technique for investigating the
structure of matter by scattering
a particle off atoms.
• Geiger showed that many
particles were scattered from
thin gold-leaf targets at
backward angles greater than
90°.
• Time to do some calculations!
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
5
Ex 4.1: Maximum Scattering Angle
Geiger and Marsden (1909) observed backward-scattered ( >=90o)  particles when a beam
of energetic  particles was directed at a piece of gold foil as thin as 6.0x10-7m. Assuming an
 particle scatters from an electron in the foil, what is the maximum scattering angle?
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The maximum scattering angle corresponds to the maximum momentum change
Using the momentum conservation and the KE conservation for an elastic
collision, the maximum momentum change of the α particle is
'
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'
M a va = M a va + me ve
1
1
1
M a va2 = M a va'2 + m2 ve2
2
2
2
D pa = M a va - M a va = me ve Þ Dpa -max = 2meva
'
'
Determine θ by letting Δpmax be perpendicular to the direction of motion.
q max
2me
Dpa -max 2me va
= 2.7 ´10-4 rad = 0.016
=
=
=
ma
pa
ma va
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
6
Multiple Scattering from Electrons
• If an α particle were scattered by many electrons, then N electrons
results in < >total ~ √N
• The number of atoms across the thin gold layer of 6 × 10−7 m:
é
æ mol ö ù é æ g ö ù
N Molecules
1
= N Avogadro ( molecules mol ) ´ ê
çè g ÷ø ú × ê r çè cm 3 ÷ø ú
cm 3
g
molecular
weight
û
ë
û ë
g ö
æ molecules ö æ 1mol ö æ
= 6.02 ´10 23 ç
×
×
19.3
÷
è mol ÷ø çè 197g ÷ø çè
cm 3 ø
= 5.9 ´10 22
molecules
28 atoms
=
5.9
´10
cm 3
m3
• Assume the distance between atoms is d = ( 5.9 ´10
-7
6
´10
m
and there are N =
= 2300 ( atoms )
)
28 -1 3
= 2.6 ´10 -10 ( m )
2.6 ´10 -10 m
This gives
Monday, Feb. 24, 2014
q
total
= 2300 ( 0.016 ) = 0.8
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
7
Rutherford’s Atomic Model



< >total~0.8*79=6.8o even if the α particle scattered
from all 79 electrons in each atom of gold
The experimental results were inconsistent with
Thomson’s atomic model.
Rutherford proposed that an atom has a positively
charged core (nucleus) surrounded by the negative
electrons.
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
8
Assumptions of Rutherford Scattering
1. The scatterer is so massive that it does not recoil
significantly; therefore the initial and final KE of
the  particle are practically equal.
2. The target is so thin that only a single scattering
occurs.
3. The bombarding particle and target scatterer are
so small that they may be treated as point masses
with electrical charges.
4. Only the Coulomb force is effective.
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
9
Rutherford Scattering
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Scattering experiments help us study matter too
small to be observed directly by measuring the
angular distributions of the scattered particles
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What is the force acting in this scattering?
There is a relationship between the impact
parameter b and the scattering angle θ.
When b is small,
r gets small.
Coulomb force gets large.
θ can be large and the particle can be repelled backward.
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
10
The Relationship Between the Impact Parameter b and the
Scattering Angle
a
The relationship between the impact parameter b and scattering angle  :
Particles with small impact parameters approach the nucleus most closely
(rmin) and scatter to the largest angles. Particles within a certain range of
impact parameters b will be scattered within the window Δθ.
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
11
Rutherford Scattering
• What are the quantities that can affect the scattering?
– What was the force again?
2
Z
Z
e
1 2
• The Coulomb force
F=
r̂e
2
4pe 0 r
– The charge of the incoming particle (Z1e)
1
– The charge of the target particle (Z2e)
– The minimum distance the projectile approaches the target (r)
• Using the fact that this is a totally elastic scattering
under a central force, we know that
a
a
N
– Linear momentum is conserved pi = p f + p
1 2 1 2 1 2
– KE is conserved
mva i = mva f + mvn
2
2
2
– Angular momentum is conserved mr 2v = mvaib
• From this, impact parameter
Monday, Feb. 24, 2014
Z1Z 2 e2
q
Z1Z 2 e2
q
b=
cot
=
cot
4pe 0 mva2i
2 8pe 0 KEi
2
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
12
Rutherford Scattering - probability
•
Any particle inside the circle of area πb02 will be similarly scattered.
•
The cross section  = b2 is related to the probability for a particle being scattered by
2
2
a nucleus.
æ
ö
ZZe
q
t: target thickness
ntp b 2 = p nt ç 1 2 cot ÷
2ø
n: atomic number density
è 8pe 0 KEi
The fraction of the incident particles scattered is
target area exposed by scatterers
f=
total target area
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The number of scattering nuclei per unit area
Monday, Feb. 24, 2014
nt =
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
r N A N M t atoms
Mg
.
cm 2
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Rutherford Scattering Equation
• In an actual experiment, a detector is positioned from  to  +
d that corresponds to incident particles between b and b + db.
K=
1
mi v02
2
• The number of particles scattered into the the angular coverage
2
2
per unit area is
N i nt æ e ö
Z12 Z 22
N (q ) =
16 çè 4pe 0 ÷ø r 2 K 2 sin 4 (q 2 )
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
14
The Important Points
1. The scattering is proportional to the square of the
atomic numbers of both the incident particle (Z1)
and the target scatterer (Z2).
2. The number of scattered particles is inversely
proportional to the square of the kinetic energy
of the incident particle.
3. For the scattering angle  , the scattering is
proportional to 4th power of sin( /2).
4. The Scattering is proportional to the target
thickness for thin targets.
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
15
The Classical Atomic Model
As suggested by the Rutherford Model, an atom consisted of
a small, massive, positively charged nucleus surrounded by
moving electrons. This then suggested consideration of a
planetary model of the atom.
Let’s consider atoms as a planetary model.
• The force of attraction on the electron by the nucleus and
Newton’s 2nd law give
mv 2
1 e2
Fe = -
4pe 0 r
2
er =
r
er
where v is the tangential speed of an electron.
• The total energy is E = K +V =
Monday, Feb. 24, 2014
e2
8pe 0 r
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
-
e2
4pe 0 r
=-
e2
8pe 0 r
16
The Planetary Model is Doomed
• From classical E&M theory, an accelerated electric charge radiates
energy (electromagnetic radiation) which means total energy must
decrease. Radius r must decrease!!
Electron crashes into the nucleus!?
• Physics had reached a turning point in 1900 with Planck’s hypothesis
of the quantum behavior of radiation.
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
17
The Bohr Model of the Hydrogen Atom – The assumptions
• “Stationary” states or orbits must exist in atoms, i.e., orbiting electrons do
not radiate energy in these orbits. These orbits or stationary states are of
a fixed definite energy E.
• The emission or absorption of electromagnetic radiation can occur only in
conjunction with a transition between two stationary states. The frequency,
f, of this radiation is proportional to the difference in energy of the two
stationary states:
•
E = E1 − E2 = hf
• where h is Planck’s Constant
– Bohr thought this has to do with fundamental length of order ~10-10m
• Classical laws of physics do not apply to transitions between stationary
states.
• The mean kinetic energy of the electron-nucleus system is quantized as
K = nhforb/2, where forb is the frequency of rotation. This is equivalent to the
angular momentum of a stationary state to be an integral multiple of h/2
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
18
How did Bohr Arrived at the angular momentum quantization?
• The mean kinetic energy of the electron-nucleus system is quantized as
K = nhforb/2, where forb is the frequency of rotation. This is equivalent to
the angular momentum of a stationary state to be an integral multiple of
h/2.
nhf
1 2
• Kinetic energy can be written K =
= mv
2
2
• Angular momentum is defined as
L = r ´ p = mvr
• The relationship between linear and angular quantifies
•
v = rw ; w = 2p f
1
1
1
nhf
Thus, we can rewrite K = mvrw = Lw = 2p Lf =
2
2
2
2
h
h
,where
=
2p L = nh Þ L = n
=n
2p
2p
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
19
Bohr’s Quantized Radius of Hydrogen
• The angular momentum is
L = r ´ p = mvr = n
n
• So the speed of an orbiting e can be written ve =
me r
• From the Newton’s law for a circular motion
2
e
1 e2 meve
Þ ve =
Fe =
=
2
r
4pe 0 r
4pe 0 mer
• So from above two equations, we can get
n
e
4pe 0 n 2
ve =
=
Þ r=
2
mer
4pe 0 mer
mee
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
2
20
Bohr Radius
• The radius of the hydrogen atom for stationary states is
4pe 0 n 2
rn =
2
mee
2
= a0 n 2
Where the Bohr radius for a given stationary state is:
4pe 0
=
a0 =
2
me e
2
( 8.99 ´10 N × m C ) × (1.055 ´10 J × s )
( 9.11 ´10 kg ) × (1.6 ´10 C )
9
2
-34
2
-31
-19
2
2
= 0.53 ´10-10 m
• The smallest diameter of the hydrogen atom is
d = 2r1 = 2a0 »10-10 m » 1A
– OMG!! The fundamental length!!
• n = 1 gives its lowest energy state (called the “ground” state)
Monday, Feb. 24, 2014
PHYS 3313-001, Spring 2014
Dr. Jaehoon Yu
21