PHYS-3301

Sep. 13, 2018

Lecture 6

n 4.1 The Atomic Models of Thomson and Rutherfordn 4.2 Rutherford Scatteringn 4.3 The Classic Atomic Modeln 4.4 The Bohr Model of the Hydrogen Atomn 4.5 Successes and Failures of the Bohr Modeln 4.6 Characteristic X-Ray Spectra and Atomic Numbern 4.7 Atomic Excitation by Electrons

Bohr’s different; he’s a football [U.S. soccer] player!

Ernest Rutherford, giving an uncharacteristic compliment to a theorist-Niels Bohr in this case.

CHAPTER 4Structure of the Atom

Experiments of Geiger and Marsden

n Rutherford, Geiger, and Marsden conceived a new technique for investigating the structure of matter by scattering a particles from atoms.

n Geiger showed that many a particles were scattered from thin gold-leaf targets at backward angles greater than 90�.

He2+ or 42He2+ indicating a helium

ion with a +2 charge (missing its two electrons). If the ion gains electrons from its environment, the alpha particle becomes a normal (electrically neutral) helium atom 4

2He.

Example 4.1

n The maximum scattering angle corresponding to the maximum momentum change

n Maximum momentum change of the α particle isor

n Determine θ by letting Δpmax be perpendicular to the direction of motion.

α particle is so much more massive than electron (~ 7000)

Elastic collision

It is impossible for an α particle to be deflected through a large angle by a single encounter with an electron.

Vector diagram illustrating the change in momentum Δpα of the αparticle after scattering from the electron

Elastic Collisions

1 1 2 2 1 1(Momentum Conservation:

) ( ) ( )fx fx ixm v m v m v+ =

1 1 12 2 21 1 2 2 1 12 2 2

Energy Conse

( ) ( ) (

rvation:

)fx fx ixm v m v m v+ =

21 1 2

1

( ) ( ) ( )fx ix fxm

v v vm

= −

1 1 12 2 221 1 2 2 2 1 12 2 2

1

[( ) ( ) ] ( ) ( )ix fx fx ixm

m v v m v m vm

− + =

1 21 1 2 1 22

22 222 2 2

11 2

1 12

[ ( ) 2 ( ) ( )

( ) ( ) ]

( )

ix ix fx

fx fx

ix

m v m v v

mv m v

m

m v

−

+ +

=

22 2 1

1

( ) [(1 )( ) 2( ) ] 0fx fx ixm

v v vm

+ − =

12 1 1

2 1 1 2

22( ) ( ) ( )1 /fx ix ixm

v v vm m m m

= =+ +

2 1 1 21 1 1 1

1 2 1 21

2( ) ( ) ( ) ( )fx ix ix ixm m m m

v v v vm m m mm

−= − =

+ +

Three Elastic Collisions

12 1

1 2

2( ) ( ) ;fx ixm

v vm m

=+

1 21 1

1 2( ) ( )fx ix

m mv v

m m−

=+

1 2 2 1 1: ( ) ( ) and ( ) (knock-o0 n)fx ix fxm m v v v= = =

1 2 2 1 1 1: ( ) 2( ) and (boost-ahe ( ) ( ) ad)fx ix fx ixm m v v v v>> = =

1 2 2 1 1: ( ) 0 and ( ) ( ) (bounce-of f) fx fx ixm m v v v<< = = −

n If an α particle were scattered by many electrons and N electrons results in average scattering angle

n The number of atoms across the thin gold layer of 6 � 10−7 m:

n Assume the distance between atoms is

and there are

That gives

Multiple Scattering from Electrons

Each atom occupies (5.9x1028)-1 m3

of space.

Even if the α particle scattered from all 79 electrons in each atom of gold, <q>total = 6.80

n even if the α particle scattered from all 79 electrons in each atom of gold

The experimental results were not consistent with Thomson’s atomic model.

n Rutherford proposed that an atom has a positively charged core (nucleus) surrounded by the negative electrons.

Rutherford’s Atomic Model

4.2 Rutherford Scattering The Assumptions

1. The scatterer is so massive that it does not recoil significantly; therefore the initial and final kinetic energies 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.

n Scattering experiments help us study matter too small to be observed directly.

n 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.

Rutherford Scattering

Rutherford Scatteringn Scattering experiments help us study matter too small to be

observed directly.n 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.

Rutherford Scattering

The Relationship Between the Impact Parameter b and the Scattering Angle

Figure 4.7 The relationship between the impact parameter b and scattering angle q. Particles with small impact parameters approach the nucleus most closely (rmin) and scatter to the largest angles. Particles within the range of impact parameters b will be scattered within Dq

n Any particle inside the circle of area πb02 will be similarly scattered.

n The cross section σ = πb2 is related to the probability for a particle being scattered by a nucleus.

n The fraction of incident particles scattered is

n The number of scattering nuclei per unit area .

Rutherford Scattering

= ntAs / A = nts = nt(pb2)

If we have a target foil of thickness t with n atoms/volume, the # of target nuclei per unit area = nt

n In actual experiment a detector is positioned from θ to θ + dθ that corresponds to incident particles between b and b + db.

n The number of particles scattered per unit area is

Rutherford Scattering Equation

Ni = total number of incidental particles

N(q) = Ni |df| / dA

The Important Points 1. The scattering is proportional to the square of

the atomic number 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.