第一章 原子的位形：卢瑟福模型

2 Ze 2 1 D= 2 4 πε 0 mv 0 2
b θ α
v v v v m r0 × v0 = m r × v − bv 0 dϕ 1 2 ⇒ mbv 0 = mr ( − ) ⇒ = 2 ( 1) dt dt r d ϕ v v dv dv z F =m ⇒ Fz = F sin ϕ = m dt dt 2 2 Ze F = 2 4πε 0 r
∆Ω (N
“
” α
α )
dw ∆N = ⋅ N ⋅ ∆Ω dΩ 2 1 2 Ze 2 ∆Ω =( ) ⋅( 2 ) ⋅n⋅t ⋅ N ⋅ 4 πε 0 mv 0 4 θ sin 2 D 2 ∆Ω = ( ) ⋅n⋅t ⋅ N ⋅ 4 4 θ sin 2
§4
D 2 −4 θ ∆N = ntN ( ) sin dΩ 4 2
2
rm
R
r
mM µ= , m+M
µ
EK → EK (
(c )
) )
2
θ → θc (
Dc θc Z1 Z 2 e b= cot , Dc = (c) 2 2 4πε 0 Ek
E K( L) E K( c )
EK
( L)
m (c) = 1 + E K M
214P o
N A
dN +d dN/N
§ § § §
θ t 2 sin 2 EK
Z
2 −4
EK
Z,t
4
∆N ' sin θ / 2 = C ;
∆N ' ∝ t ;
∆N ' ∝ E ∝ v ;
∆N' ∝ Z ;
2
t
−2 K
EK
−4
Z
nt
∆s ∆Ω = 2 r dw ∆N = ⋅ N ⋅ ∆Ω dΩ
θ > 45 θ < 45
o o
α Z 1920 (77.4) (46.3) (29.3)
(r=rm
v’ .
v0 v0 D θ mv0 b = mv' rm ⇒ v' = b = cot rm rm 2 2
1 2 Ze 1 2 2 mv' + = mv0 2 4πε 0 rm 2
:
2
z1e rm z2e
D rm = (1 + 1 ) sin θ 2 2
=
rm=D
D
2
1 2 Ze D= 2 4 πε 0 mv 0
2 Ze dv z = 4πε 0 mv 0 b
2
∫π
θ
sin ϕ d ϕ
2 Ze v0 sin θ = ( 1 + cos θ ) 4 πε 0 mv 0 b
⇒
θ D θ 2 Ze b= cot = cot 2 2 2 2 4 πε 0 mv 0
2
1 2 Ze D= 2 4 πε 0 mv 0
2
θ
θ+dθ
dθ • • • • • • θ r • • • • • • • • • •• •
rsin θ
dA 2π r sin θ ⋅ rd θ dΩ = 2 = 2 r r = 2π sin θ d θ
rdθ • • • •
θ θ = 4π sin cos d θ 2 2
∴ α θ θ+dθ
dw D 2 1 = ( ) ⋅n ⋅t ⋅ 4 dΩ 4 θ sin 2
4u
10-3 10-5
§2
F
F r
R
r
R
1871— — 1937
1902
α 4.0034 u ),
(1) 2°. (2) 1/8000 90°.
α R
∆p
2 Ze 2R ∆p = F∆t = 4πε 0 R 2 v
F
2
P +Ze v m
θ
θ max
∆p 2 Ze 2 1 / mv 2 = = p 4πε 0 R 2
2
∴D
v v v v m r0 × v0 = m r × v
dϕ ⇒ mbv 0 = mr ( − ) dt
2
⇒
v0
1 1 dϕ =− 2 r v0 b dt
( 1)
v //
α
v0
2
dv // Z ' Ze m cos ϕ =− 2 4 πε 0 r dt
(2)
1
2
2
t
dv// 2 Ze dϕ m∫ dt = cos ϕ dt 2 ∫ dt 4πε 0 r dt
(D D mv02/2
) α
10-14m
)
(M
M
§5
1° 2° 3° (1967 1° 2° 10-8s) ) RBS) )
∴
2 Ze dv z Fz = sin ϕ = m 2 4 πε 0 r dt
2
(2)
(1)
( 2) 2 2 Ze sin ϕ − dv z = dϕ 4 πε 0 mv 0 b
( 3)
α
( 3)
A→C
v 0 sin θ
v z : 0 → v0 sin θ ϕ :π → θ
2
−∫ ⇒
0
) 1994 1982
—— 1991
.
,
§1 §2 §3 §4 §5
§1
u
12
1/12
12 ( g ) 1 1( g ) - 27 1u = = = 1.660 × 10 kg = 931.5MeV/c N 12 N A A
2
X
A(g ) M (X) = = Au NA
4 3 A( g ) π r NA ≡ , 3 ρ
θ
∫
v 0 cos θ
v0
θ 2 Ze dv // = cos ϕ d ϕ ∫ 4πε 0 mv 0 b π
2
1 − cos θ 2 θ = sin 2 2
⇒
θ D θ 2 Ze b= cot = cot 2 2 2 2 4 πε 0 mv 0
2
1 2 Ze D= 2 4 πε 0 mv 0
2
2
∴D
α a (2πbdb
S
α b b-db b b-db 2πbdb/S α
S b db t
b
b-db
2π bdb dw = ⋅ nSt = 2π ntbdb S
θ cos 1 D 2 2 b db = ( ) dθ 2 2 sin 3 θ 2
α θ θ + dθ θ θ+dθ
θ cos D 2 2 ∴ dw = π ( ) ⋅ n ⋅ t ⋅ dθ 2 3 θ sin 2
4 3 πr 3
ρ(g/cm )
3
3A r = 4πρ N A
Li Pb A=7 A=207
1 3
=0.7 rLi =0.16nm =11.34 rPb=0.19nm
_
1897
E B +
e/me
1910 e =1.602×10
19
c
me=9.109×10
31kg=0.511MeV/c2 =5.487×10
)Байду номын сангаас2000( 2006( 1992( 1989 1979 2002 2000 2001 1983 ) ) )
) )H.C.Wolf (200306 ) Fundamentals of physics, 6th ed., D. Halliday, R. Resnick & J. Walker, John Wiley & Sons (Asia), 2001 Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particles R.Eisberg, R.Resnick, 1985 )H.Haken,(
2 Z × 1 . 44 fm ⋅ MeV = 0 . 1nm E K ( MeV ) = 3 × 10
−5
Z E K ( MeV )
V0 ≈ 2 × 10 m
7
s Z = 79( !!!) )
1 mV0 ~ 8MeV 2 −4 θ ≈ 1.4 × 10 (rad ) (
θ >3
o o
< 1% < 1 3500 10
1 10 1 3500 10
θ > 90
θ ≥ 90
o
. Ze 10-10m 10-10m
§3
F z z
A db
b
v F
C
α
ϕ
θ
Ze
EK
b
(1)
(2)
M>>m m
,
Fz z
A db
b
v F
C
α
ϕ
θ
Ze
F z z
α Ze θ b
2
v F
C
A db
b
α
ϕ
θ
Ze
2 Ze θ 1 θ b= cot = D cot 2 2 4 πε 0 mv 0 2 2