Post on 10-May-2018
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
• 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−
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
• 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
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− −= =
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
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
=