PHYS 3313 – Section 001
Lecture #10
Monday, Sept. 30, 2013
Dr. Amir Farbin
•
•
•
•
•
Atomic Model of Thomson
Rutherford Scattering Experiment and
Rutherford Atomic Model
The Classic Atomic Model
The Bohr Model of the Hydrogen Atom
Bohr Radius
Monday, Sept. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
1
Announcements
• Reminder of Homework #2
– CH3 end of the chapter problems: 2, 19, 27, 36, 41, 47 and 57
– Due Wednesday, Oct. 2
• Mid-term exam
– In class on Wednesday, Oct. 16
– Covers from CH1.1 through what we finish on Oct. 9 + appendices
– Mid-term exam constitutes 20% of the total
– Please do NOT miss the exam! You will get an F if you miss it.
– Bring your own HANDWRITTEN formula sheet – one letter size
sheet, front and back
• No solutions for any problems
• No derivations of any kind
• Can have values of constants
Monday, Sept. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
2
Remind: Special Project #3
•
•
•
•
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 Wednesday, Oct. 9.
Monday, Sept. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
3
Thomson’s Atomic Model
 Thomson’s “plum-pudding” model
 Atoms are electrically neutral and have electrons in them
 Atoms must have an equal amount of positive charges in it to
balance electron negative charges
 So how about positive charges spread uniformly throughout a
sphere the size of the atom with, the newly discovered
“negative” electrons embedded in a uniform background.
 Thomson thought when the atom was heated the electrons
could vibrate about their equilibrium positions and thus
produce electromagnetic radiation.
Monday, Sept. 30, 2013
PHYS 3313-001, Fall 2013
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 particles from
atoms.
Geiger showed that many a
particles were scattered from
thin gold-leaf targets at
backward angles greater than
90°.
Monday, Sept. 30, 2013
PHYS 3313-001, Fall 2013
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?
•
•
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
'
•
'
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, Sept. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
6
Multiple Scattering from Electrons
• If an α particle were scattered by many electrons and 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
That gives
Monday, Sept. 30, 2013
q
total
= 2300 ( 0.016 ) = 0.8
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
7
Rutherford’s Atomic Model



< >total~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, Sept. 30, 2013
PHYS 3313-001, Fall 2013
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
and charges.
4. Only the Coulomb force is effective.
Monday, Sept. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
9
Rutherford Scattering

Scattering experiments help us study matter too
small to be observed directly by measuring the
angular distributions of the scattered particles


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, Sept. 30, 2013
PHYS 3313-001, Fall 2013
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 Δθ.
Monday, Sept. 30, 2013
PHYS 3313-001, Fall 2013
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, Sept. 30, 2013
Z1Z 2 e2
q
Z1Z 2 e2
q
b=
cot
=
cot
4pe 0 mva2i
2 8pe 0 KEi
2
PHYS 3313-001, Fall 2013
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
•
•
The number of scattering nuclei per unit area
Monday, Sept. 30, 2013
nt =
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
r N A N M t atoms
Mg
.
cm 2
13
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
2
2 angular
• The number of particles scattered into
the
the
N i nt æ e ö
Z12 Z 22
coverage per unit area is N (q ) = 16 çè 4pe ÷ø r 2 K 2 sin 4 (q 2 )
0
Monday, Sept. 30, 2013
PHYS 3313-001, Fall 2013
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, Sept. 30, 2013
PHYS 3313-001, Fall 2013
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, Sept. 30, 2013
e2
8pe 0 r
PHYS 3313-001, Fall 2013
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, Sept. 30, 2013
PHYS 3313-001, Fall 2013
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, Sept. 30, 2013
PHYS 3313-001, Fall 2013
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, Sept. 30, 2013
PHYS 3313-001, Fall 2013
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
e2 me ve2
Þ ve =
Fe =
=
2
r
4pe 0 r
1
e
4pe 0 mer
• So from above two equations, we can get
n
e
4pe 0 n 2
ve =
=
Þr=
mer
4pe 0 me r
me e2
Monday, Sept. 30, 2013
PHYS 3313-001, Fall 2013
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, Sept. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
21
The Hydrogen Atom
• The energies of the stationary states
E0
e2
En = ==- 2
2
n
8pe 0 rn
8pe 0 a0 n
e2
E0 =
e2
8pe 0 a0
=
(1.6 ´10
-19
C)
2
8p ( 8.99 ´10 9 N × m 2 C 2 ) × ( 0.53 ´10 -10 m )
= 13.6eV
where E0 is the ground state energy
 Emission of light occurs when the atom is
in an excited state and decays to a lower
energy state (nu → nℓ).
hf = Eu - El
where f is the frequency of a photon.
æ 1 1ö
f Eu - El E0 æ 1 1 ö
= =
=
- 2 ÷ = R¥ ç 2 - 2 ÷
2
ç
l c
hc
hc è nl nu ø
è nl nu ø
1
R∞ is the Rydberg constant. R¥ = E0 hc
Monday, Sept. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
22
Transitions in the Hydrogen Atom
• Lyman series: The atom will
remain in the excited state for a
short time before emitting a
photon and returning to a lower
stationary state. All hydrogen
atoms exist in n = 1 (invisible).
• Balmer series: When sunlight
passes through the atmosphere,
hydrogen atoms in water vapor
absorb the wavelengths (visible).
Monday, Sept. 30, 2013
PHYS 3313-001, Fall 2013
Dr. Jaehoon Yu
23