Transcript n - Purdue Physics

CHAPTER 6
Structure of the Atom
The Atomic Models of Thomson and
Rutherford
Rutherford Scattering
The Classic Atomic Model
The Bohr Model of the Hydrogen Atom
Successes & Failures of the Bohr Model
Niels Bohr (1885-1962)
Homework due Friday Oct. 16th
Chapter 6: 2,15, 20, 24, 28
Rutherford’s Atomic Model
Experimental observation
of many large angle
scattering events!
Experimental results were
not consistent with
Thomson’s atomic model.
Rutherford proposed that an
atom has a small positively
charged core (nucleus)
surrounded by the negative
electrons.
Geiger and Marsden
confirmed the idea in 1913.
Ernest Rutherford
(1871-1937)
The Classical Atomic Model
Consider an atom as a planetary system. Like gravity, the force on the
electron an inverse-square-law force. This is good.
Now, by Newton’s 2nd Law:
1 e2 mv 2
Fe 

2
4 0 r
r
where v is the tangential velocity of the electron:
v
e
4 0 mr
So:
K  12 mv 2 
Kinetic energy
1
2
e2
4 0 r
This is negative, so
the system is bound,
which is good.
e 2
The potential energy is: V 
4 0 r
The total
e2
e2
e 2
E  K V 


energy is then:
8 0 r 4 0 r 8 0 r
The Planetary Model is Doomed!
Because an accelerated electric charge continually radiates energy
(electromagnetic radiation), the total energy must continually decrease.
So the electron radius must continually decrease!
The
electron
crashes
into the
nucleus!
Physics had reached a turning point in 1900 with Planck’s hypothesis
of the quantum behavior of radiation, so a radical solution would be
considered possible.
The Bohr Model of the Hydrogen Atom
Bohr’s general assumptions:
n=2
n=1
1. Electrons reside in stationary states,
and do not radiate energy. They have
well-defined energies, En. Transitions can
occur between them, yielding light of
energy:
E = En − En’ = hn
En > En’ : Emission
En < En’ : Absorption
n=3
2. Classical laws of physics do not apply to
transitions between stationary states, but they do apply elsewhere.
3. The angular momentum of the nth state is: nħ
where n is called the Principal Quantum Number.
Angular
momentum is
quantized!
Consequences of the Bohr Model
The angular momentum is:
L  mvr  n
So the velocity is:
But:
v
Solving for rn:
v  n / mr
e
4 0 mr
So:
rn  n 2 a0
n2 2
e2

2 2
mr
4 0 mr
where:
a0
4 0
a0 
me 2
2
a0 is called the Bohr radius. It’s ½ the diameter of the Hydrogen
atom (in its lowest-energy, or ground, state).
Bohr Radius
The Bohr radius,
4 0
a0 
me 2
2
is the radius of the ground state of the
Hydrogen atom:
4 0 2
(1.055 10-34 J  s) 2
-10
a0 


0.53

10
m
2
2
9
2
2
-31
-19
me
8.99 10 N  m /C 9.1110 kg 1.6 10 C 
The ground state of the Hydrogen atom has a diameter:
2r1  2a0  10 10 m
The Hydrogen
Atom Energies
Use the classical
e2
result for the
E
energy:
8 0 r
and:
4 0 n2
rn 
me2
2
So the energies of the stationary
states are:
e2
e2
En  

80 rn
80 a0 n 2
or:
En =  E0/n2
where E0 = 13.6 eV.
4 0
a0 
me 2
2
The Hydrogen Atom
Emission of light occurs when the atom is in an excited state
and decays to a lower energy state (nu → nℓ).
hn  Eu  E
where n is the frequency of a photon.
1


n
c

e2
En  
80 a0 n 2
hn En  E
 1 1

 R  2  2 
hc
hc
 n nu 
R∞ is the Rydberg constant.
me4
R 
(4 )3 c 0 2
Transitions
in the
Hydrogen
Atom
The atom will remain in
the excited state for a
short time before
emitting a photon of
energy hn and returning
to a lower stationary
state. In equilibrium, all
hydrogen atoms exist in
the ground state, n = 1.
Also, hydrogen in
the ground state
(n = 1) can absorb
a photon of
energy hn and
make a transition
to an excited
state.
The Correspondence
Principle
Bohr’s correspondence
principle is rather obvious:
In the limit where classical and
quantum theories should agree,
the quantum theory must reduce
the classical result.
The Correspondence Principle
The frequency of the radiation emitted nclassical is equal to
the orbital frequency norb of the electron around the nucleus.
n classical  n orb
v
1
nclassical 
2 r 2
1/2
 e


3 
4

mr
0


2
me 4 1

4 0 2 h3 n3
This should agree with the frequency of the
transition from n + 1 to n (when n is very large):
n Bohr 
E0
h
1
1 

 n 2 (n  1) 2 


v
e
4 0 mr
4 0 n2
rn 
me2
En = hnn =  E0 /n2
E0  n 2  2n  1  n 2  E0  2n  1 



 n 2 (n  1) 2 
2
2
h  n (n  1)
h



For large n: n Bohr
2nE0 2 E0

 3
4
hn
hn
Substituting for E0:
n Bohr
E0 
e2
8 0 a0
me4 1

 n classical
2 3
3
4 0 h n
4 0
where: a0 
me 2
2
2
Fine Structure Constant
The electron’s velocity in the Bohr model:
n 1 e 2
vn 

mrn n 40 
In the ground state,
v1 = 2.2 × 106 m/s ~ 1% of the speed of light.
The ratio of v1 to c is called the fine structure constant.
v1

e2
1
 


c ma0c 40c 137
Successes and Failures of the Bohr Model
Success:
The electron and
hydrogen nucleus
actually revolve
about their mutual
center of mass.
The electron mass is replaced
by its reduced mass:
me M
me
e 

me  M 1  me
M
The Rydberg constant for infinite nuclear mass, R∞, is replaced by R.
e
e e 4
1
R
R 
R 
3
2
m
me
4

c

(
4

)
e
0
1
M
This modification improved
the theory’s accuracy!
Limitations of the
Bohr Model
The Bohr model was a great
step in the new quantum
theory, but it had its limitations.
Failures:
Works only for single-electron (“hydrogenic”) atoms.
Could not account for the intensities or the fine structure of
the spectral lines (for example, in magnetic fields).
Could not explain the binding of atoms into molecules.