Chapt2 : Configuration of the Atom: Rutherford’s Model

2.1 Background

2.2 Emergence of the Rutherford’sModel 2.3 Rutherford scattering Formula

2.4 Experimental test of the RF Formula.

2.5 Summary of the Nuclear planet model

2-1 Background

1.  The discovery of the electron 2.  The charge and mass of the electron 3.  The size of Atom

H⊙

A ,B +

+ -

C D E

P1 P2

•  Thomson’s discharge tube

– Cathode rays(C)⇒Slits(A,B)⇒paralle plates(D,E)⇒screen

– D,E with E ⇒ rays P1 → P2 ⇒ negative charged

– With H ⇒ rays P2 → P1 ⇒ Hev=Ee ⇒ v=E/H – Without E ⇒ rays radius r ⇒ mv2/r= Hev ⇒ e/m=v/Hr

2.1.1 Discovery of the electron

•  In the late 1800s several experiments were performed on electrical discharges in gases at low pressure.

•  It was established that a stream of some kind of “rays” – called cathode-rays – was emitted by any electrode at high negative potential in a vacuum tube. –  What were these rays?

Thomson’s Gas discharge tube

2.1.2 Charge & Mass of Electron

•  Oil-drop experiment by Millikan : – Measured charge ：e = 1.6×10-19 C

⇒electron’s mass me = 9.1×10-31 kg – Millikan found charge is quantized

• Any charge can only be an integral multiple of e (smalled charge )

given ⇒

p

em

11836

e

p

mm

=

•  Oil drop experiment (2) – Atom is neutral，with negative charged electrons， ⇒ there must be postive charged matter( proton)

Atom = Postive + Negative + Neutral

2.1.3 Avogadro’s Constant

•  NA gives the number of atoms in one Mole

•  Atomic mass unit [u] – 1[u] ≡ The mass of one 12C /12

–  Atomic mass MA [u] = atomic quantity [u] = Mass number A [u]

–  1g= u NA ; F=e NA R= k NA – Macro -- NA --- Micro

 

] 克[1066.1] 克[112

克12 24.

−×==×

=AA NN

2.1.4 The size of atoms

•  Atomic radius – Atomic volum =

= Mass of an atom/atomic density =

⇒ radius r =

order of magnitude：r ~ 10-10 m = 1 Å

343rπ

/ AA Nρ

31)

N4A3(

Aπρ

2.2 Emergence of Rutherford’s Model

1.  Atomic charges’ distribution? 2.  α scattering 3. Scattering experiments

2.2.1 How are charges distributed in atoms

•  Thomas atomic model postive charges are distributed uniformly over the whole atom,

electrons are embedded

•  Rutherford model Atom is composed of by

nuclei and circling around

electrons

α粒子散射实验：

探测器

α粒子 金原子

2.2.2 α scatting experiment

• results：

– Most α particles have scattering angle ： ~ 2º - 3º

– few of 1/8000 α with SA：> 90º

•  puzzled : like a gun bullet was bounced back by a paper

2.2.3 Quantitative explaination

•  Thomson’s model – Force of Positive charge Ze of atom on the α particle

2

20

2

30

1 24

1 24

Ze r Rr

FZe r r RR

πε

πε

⎧≥⎪

⎪= ⎨⎪ <⎪⎩

R r

Radius of atom

The Max. Force of Positive charge Ze of atom on the α particle

– Scat. Angle

–  Momentum change~ F t ( 2R/v)

2

20

1 24

ZeFRπε

=p’

p

pΔθ

pp

θΔ

=

•  So we have

20

2

51.44f

2 /(4 )2 /12

2 /0.1nm 3 10 rad(mMeVMeV)

Ze Rp FR vp m v m v

Z ZE E

αα

α α

πε

−

Δ= =

×≈ ≈ ×

Eα

R 2

04eπε

– Electron and α particle’s (head on )

P and E conservation à

2 2 20 1 0 1

1 1 1,2 2 2e e e em v m v m v m v m v m vα α α α= + = +

In coming α Outgoing α

Outgoing electron

0 00

2 2 2ee

m v m vv vm m m

α α

α α

→ = ≈ =+

0 1 02e e ep m v m v m v m vα α→Δ = − = ≈

2.2.3 Electron’s effect

therefore

– Momentum change

0 1 0

0

2e e ep m v m v m v m vp m v

α α

α

Δ = − = ≈

=

0 4

0

2 2 1100

040

e em v mm v m

pp α α

−→ ≈ = ≈ ≈Δ

Deflection angle of αis very tiny Almost no contribution

– α-particle on Gold atom

– One scattering angle – Multiple scatterings do not result large angle • Random in direction

– Thomson model contradicts with experiment result !

5MeVEα = Z=79

5 4 33 10 rad<10 rad<10 radp Z Zp E Eα α

θ − − −Δ= ≈ ×

310 rad−

Thomson model predicts:

•  SA 3°probabilty less than 1%

•  SA greater than 90°, about 10-3500。

模拟实验

10-3500 ---------- 1/8000

•  most with small SA，about 1/8000 with SA>90°；

•  Very few with 180°。

Positive charges locate on the central of atom

R

C

2.3 Rutherford Scattering Formula

1.  Derivation of Coulomb formula

2.  Derivation of Rutherford formula

2-3-1 Coulomb scattering formula

•  Scattering of particle with E, Z1e by a target nucleus of charge Z2e

瞄准距离 碰撞参数

散射角

•  Coulomb scattering formula

2 2ab ctg θ=

21 2

04Z Z ea

Eπε=

Impact parameter

Scattering factor

•  Assumptions：

1.  Single scattering 2.  Point charge and only Coulomb force 3.  Extranuclear elctrons may be negnected 4.  Target nucleus is rest

•  According to 2

01 22

0

d4 dZ Z e vF ma r m

r tπε= → =

vv v v

2 dd

mr Ltφ= Central forceà L conservation

201 2

20

d4 d

dd

Z Z e vr mr t

φφπε

=vv

2 20 01 2 1 2

0 02

1d d d4 4d

d

Z Z e Z Z ev rL

t

rmr

φ φπε πεφ

= =v v v

201 2

0

d ;4Z Z edv r

Lφ

πε=∫ ∫

v v d f i uf iv v v v v e= − = −∫v v v v vv

22 ;1 12 2i f i fE m v m v v v v= = → = =

v v v v 2 sin2f iv v v θ

− =v v

0

0( cos sin )d 2cos

2sin cos2 2

r d i jj iπ θ θ

φ φ φθ

φθ− ⎛ ⎞+⎜ ⎟

⎝ ⎠= + =∫ ∫

vv vv

uev

2 21 2 1 2

0 0

1 1sin cos cos2 4 2 4 2

Z Z e Z Z evL mvb

θ θ θπε πε

= =

cot2 2ab θ

=2

1 2

04Z Z ea

Eπε=

2

2E mv=

à

b ~ θ ; b↑à θ↓; b ↓ à θ ↑

2-3-2 Rutherford’s formula (1)

α particle : b ~ b + db à θ ~ θ - dθ α:b~b+db ring area à θ ~ θ-dθ hollow cone

prob. on the ring of

•  Target foil with thickness t, A，number density n，mass density ρ

•  Area of the ring •  α particle on the ring

2 | |b dbπ

2

2

4

2 | | 2 1cot csc2 2 2 2 2

2 sin

16 sin2

b db a a dA Aa d

A

π π θ θθ

π θ θθ

= −

=

•  Hellow cone cubic angle ~ dθ

2

2 sin ;

2 sin

ds r rddsd dr

π θ θ

π θ θ

=

Ω = =

2

4

2 | |

16 sin2

b db a dA A

πθ

Ω=

à

•  Many rings in a foil : nucleus ~ring; •  volum: At; number of rings: Atn •  α on Atn ,with the same θ •  The prob. Of one α hiting on the foil with θ ~ θ -dθ

2

4( )16 sin

2

a d n td ApA

θθ

Ω=

•  N α scattering on a foil θ ~ θ -dθ

•  Differential Cross-section

The Ns of scattered particles per number of atoms,per unit area of the target per incident particle per solid angle

2221 2

4 40

14 416 sin sin

2 2

Z Z eaN d ddN nAt ntNEA θ θπε

⎛ ⎞Ω Ωʹ = = ⎜ ⎟

⎝ ⎠

221 2

40

1 14 4 s

(i2

)n

( )C

Z Z edNNntd E

dd θπσ θ

σε

θ⎛ ⎞ʹ

≡ ≡ = ⎜ ⎟Ω Ω ⎝ ⎠

•  Understanding the D-cross section

–  For per unite area of the target nucleus,each incident particle,each unite cubic angle , the number of scatterings

– Effective scattering area of every atom into the solid angle element in θ direction

– Dimension : area (m2/sr) – Cross-section: b):

221 2

40

( ) 1 1( )4 4 sin

2

CZ Z ed dN

d Nntd Eσ θ

σ θθπε

⎛ ⎞ʹ≡ ≡ = ⎜ ⎟

Ω Ω ⎝ ⎠

28 2 31 21 10 ,1 10 mb m mb− −= =

2-4 Experimental verification

1.  Geiger-Marsden Experiment 2.  Nuclear size

Geiger-Marsden Experiment

à

•  fixing(α source, target)

•  Fixing (α source,material of target,angle )

•  same(target,angle)

•  same(α source,angle, nt)

221 2

40

( ) 1 1( )4 4 sin

2

CZ Z ed dN

d Nntd Eσ θ

σ θθπε

⎛ ⎞ʹ≡ ≡ = ⎜ ⎟

Ω Ω ⎝ ⎠

4

1 ,sin

2

dNd θʹ∝

Ω

dN tdʹ∝

Ω

2

1dNd E

ʹ∝

Ω

22~dN Z

dʹ

Ω

((a) Side view (b) 俯视图

•  R:放射源

•  F:散射箔

•  S:闪烁屏

•  B:圆形金属匣

•  A:代刻度圆盘

•  C光滑套轴

•  T抽空B的管

•  M显微镜

实验装置和模拟实验

模拟实验

模拟实验

•  The experiment verified above relations!

2-4-2 Nuclear size estimation (1)

22

2

00

0

1 212 4

12 m

mm

m

mv

mv

Z Z eE mv

vr

r Lb mπε

⎧=⎪

⎨⎪ =

=

=⎩

+

222 2 21 2

0

; 02 4m m m m

Z Z eLr r r ar bmE Eπε

= + − − =

2 2 22201 2

200

, 14 22

m v bZ Z ea bE m mvπε

= =⋅

2.4.2 Estimation of the Size(2)

•  solve 2 2 0m mr ar b− − =2 2

22

2

4 , 02

41 (1 1 cot )2 2 2 2

1(1 )2 sin

2

m m

m

a a br r

a a b ara

a

θ

θ

± += > →

= + = + +

= +

+

180 1.20fmmr aθ = → =o :

cot2 2ab θ

=

2.5 On Nuclear model

•  Significances：

– Nuclear struncture of atom

– New method ——— black body method •  difficulties:

– Stable (acceleratingàradiation) – Identical (atom ß> solar system, IC) – Regeneration