Сарычева Л.И. Физика фундаментальных взаимодействий
Подождите немного. Документ загружается.
Photon Energy
1 Mb
1 kb
1 b
10 mb
10 eV 1 keV 1 MeV 1 GeV 100 GeV
(b) Lead (Z = 82)
− experimental σ
tot
σ
p.e.
κ
e
Cross section (barns/atom)Cross section (barns/atom)
10 mb
1 b
1 kb
1 Mb
(a) Carbon (Z = 6)
σ
Rayleigh
σ
Rayleigh
σ
Compton
σ
Compton
κ
nuc
κ
nuc
κ
e
σ
p.e.
− experimental σ
tot
N(t) = 2
t
E(t) =
E
0
/2
t
E
c
t
max
= ln(E
0
/E
c
)/ ln 2
∼ 2E
0
/E
c
E
0
X
0
/E
c
E =
t
R
0
N(t) · dt
σ(E)/E ≃ 1/
√
E
~
E
~
B F = q
~
E +
1
c
~v ×
~
B
~
E
~
B
q m
q
m
v
10
−5
÷ 10
−6
E
2
= m
2
c
4
+ p
2
c
2
c = 1 m =
√
E
2
− p
2
m =
E
γ
γ =
1
√
1−β
2
β =
v
c
m =
~p
γ·v
~p = mγ ·~v
E p v
∆t v
τ > 10
−6
÷10
−7
v =
l
∆t
~p
ρ
~
B pc = 300B ·ρ s
1
s
2
l
m =
~p
γ·v
pc =
mβc
2
√
1−β
2
=
mβc
2
· γ
τ <
10
−16
m
x
τ
x
∼ 6 ·10
−24
β ∼ 1
l ∼ vt ∼ 1.5 1.5 ·1 0
−13
m
x
π
−
p →
π
+
π
−
n
π
−
p →
π
+
π
−
n
π
+
π
−
m
x
π
+
π
−
M
2
c
4
= (
X
i
E
i
)
2
− (
X
i
p
i
)
2
c
2
.
m
1,2
=
1
c
2
[(E
1
+ E
2
)
2
− (~p
1
+ ~p
2
)
2
c
2
]
1/2
.
~p
1
~p
2
E
1,2
= β
Z
dE
dx
dx
δ
π
−
p → nπ
+
π
−
π
+
π
−
m
1,2
J/ψ
J/ψ
J/ψ
ψ
1
ψ
2
m
ψ
=
1
c
2
[(E
1
+ E
2
)
2
− (p
1
+ p
2
)
2
c
2
]
1/2
,
µ
+
µ
−
e
+
e
−
J/ψ
J/ψ c¯c
m
J/ψ
= 3000 m
J
1
= 3770
m
J
2
= 4040
J/ψ J/ψ → µ
+
µ
−
→ e
+
e
−
∆t · ∆E ≥ ¯h,
∆E = Γ =
¯h
τ
τ =
¯h
Γ
τ ∼ 10
−19
p +
11
→ 3α
⌊
→
8
+ α
⌊
→ 2α
8
α τ ∼ 2 · 10
−16
π
0
→ 2γ
γ E m =
E
γ
γ
8
→ 2α
γ = 1/
√
1 − β
2
dE
dx
= 2π
e
4
Z
2
m
e
v
2
N
e
"
ln
2m
e
· v
2
· T
max
I
2
(1 −β
2
)
− 2β
2
− δ
#
.
γ =
1
√
1 − β
2
β =
v
c
γ
dE
dx
D
dE
dx
E
γ ∆
dE
dx
∼
1
√
N
N
dE
dx
N
dE
dx
D
dE
dx
E
γ
E
E = a
i
R
x
L
0
I(x)dx
N
dE
dx
X
L
E = a
i
R
x
L
0
I(x)dx
p + A → π
±,0
+ K
±,0
+ Λ
0
+ p¯p.
π
+
π
−
π
0
π
0
π
+
π
−
I
x
γ
E m =
E
γ
γ
ε
1
ε
2
Ze
ε
1
ε
2