NEEP 408
Assignment 4
1. (3pts) Turner 8.4
The threshold wavelength for producing photoelectrons from sodium is 5650Å.
a) What is the work function?
b) What is the maximum kinetic energy of photoelectrons produced by light with a wavelength of
4000Å?
Solution:
a)
Φ
Te
Φ
Φ
hc
+ Te
λ
= 0
hc
=
λ
= 2.19 eV
=
b)
Tmax
Tmax
Tmax
hc
−Φ
λ
hc
=
− 2.19
4E −7
= 0.91 eV
=
2. (4pts) A 2.754 MeV gamma from the decay of 24 Na is scattered two times; first through an angle of 45◦
and then through an angle of 135◦ . (a) What is the energy of the photon after the second scattering?
(b) What is the energy of the scattered photon if the angular scattering sequence is reversed? (c) What
are the energies of the scattered electrons for each case.
Solution: For the compton scattering interaction, the energy loss relationship is
Ef =
Ei
Ei
1+
(1 − cos(θ))
me c2
In terms of collisions, we write:
Ei0
First collison: Ef1 =
1+
Ei0
(1 − cos(θ1 ))
me c2
and
Ef1
the second collision: Ef2 =
1+
Ef1
(1 − cos(θ2 ))
me c2
Case(1): θ1 = 45◦ and θ2 = 135◦
=⇒ Ef1 = 1.068 MeV and Ef2 = 0.234 MeV
Case(2): θ1 = 135◦ and θ2 = 45◦
=⇒ Ef1 = 0.270 MeV and Ef2 = 0.234 MeV
We can use the conservation of energy expression to compute the energy of the electron after the
scattering event: Te = Ei − Ef . In terms of collisions, we write:
First collison: Te1 = Ei0 − Ef1
and
the second collision: Te2 = Ef1 − Ef2
Case(1): θ1 = 45◦ and θ2 = 135◦
=⇒ Te1 = 1.686 MeV and Te2 = 0.834 MeV
Case(2): θ1 = 135◦ and θ2 = 45◦
=⇒ Te1 = 2.484 MeV and Te2 = 0.036 MeV
1
3. (6 pts) Turner 8.20
In a Compton scattering experiment a photon is observed to be scattered at an angle of 122◦ while the
electron recoils at an angle of 17◦ with respect to the incident photon direction.
a) What is the incident photon energy?
b) What is the frequency of the scattered photons?
c) How much energy does the electron receive?
d) What is the recoil momentum of the electron?
Solution:
a) The energy can be found using Eq. 8.24 with θ = 122o , φ = 17o
θ
2
122
cot
2
E
cot
=
=
=
E
tan φ
1+
me c2
E
tan 17
1+
me c2
0.415 MeV
b) Eq. 8.12
hν 0
=
0
=
ν
E
1+
E
me c2 (1 −
19
4.476 ∗ 10
cos θ)
Hz
c)
Te
=
E − hν 0
Te
=
0.415 − h ∗ 4.476 ∗ 1019
Te
=
0.23 MeV
d)
E
= Te + me c2
p
p2 c2 + m2e c4
=
E
=
0.549 MeV
p
=
2.87 ∗ 10−22
kg m
s
4. (5 pts) Turner 8.25 (use iron (Fe) instead of aluminum given in the problem).
The Klein-Nishina cross section for the collision of a 1-MeV photon with an electron is 2.11 × 10−25
cm2 . Calculate, for Compton scattering on aluminum,
a) the energy-transfer cross section (per electron cm−2 )
b) the energy-scattering cross section (per electron cm−2 )
c) the atomic cross section
c) the linear attenuation coefficient.
Solution:
Note e σc = 2.11 ∗ 10−25 cm2 for 1 MeV photons.
a)
e σtr
=
Tev
e σc
hν
0.44 e σc
=
9.28 ∗ 10−26 cm2
=
2
b)
e σs
−e σtr
=
e σc
=
1.18 ∗ 10−25 cm2
=
26 ∗ 2.11 ∗ 10−25
=
5.486 ∗ 10−24 cm2
c)
Z e σc
d)
3
Data: ρ = 7.87 g/cm , M = 55.847 g/mole
N
=
ρNA /M
N
=
8.488 ∗ 1022 cm−3
σ
=
N Z e σc
=
0.466 cm−1
5. (4 pts) Calculate the mass attenuation coefficient of silica glass (SiO2 , ρ = 2.21 g/cm3 ) for 2 MeV
photons. (use the attached Tables).
Solution:
Beginning from the basic definition:
µSiO2
µSiO2
ρSiO
2
µ
ρ SiO
2
µ
ρ SiO
2
µ
ρ SiO2
= µSi + µO
µSi
µO
=
+
ρSiO2
ρSiO2
µSi
µO
ρSi
ρO
=
+
ρSi
ρSiO2
ρO
ρSiO2
µ
µ
=
(weight fraction Si) +
(weight fraction O)
ρ Si
ρ O
µ
µ
=
(WSi ) +
(WO )
ρ Si
ρ O
The weight fractions of O and Si are WSi = 28/60 and WO = 32/60. Thus
µ
µ
µ
=
(WSi ) +
(WO )
ρ SiO2
ρ Si
ρ O
µ
= (0.0447) (28/60) + (0.0445) (32/60)
ρ SiO2
=
0.04459 cm2 /g
6. (5 pts) The absorption of radiation is often measured in units called Grays, where one Gray (abbreviated Gy) is equal to the absorption of 1 J/kg. What intensity of 1 MeV photons incident on a thin
slab of water is required to give an absorption rate of 0.01 Gy per second?
Solution:
3
From the Tables provided,
µ
ρ |H2 O
= 0.0311 cm2 /g.
I
=
absorption rate
Eγ µρ |H2 O
I
=
0.01
(1.6 ∗ 10−13 )(0.031)(1000)
I
=
2.01 ∗ 109 photons/cm2 -s
7. (12pts) The compton photon collision rate at a point ~r in a medium is given by
R(~r, E) = (N Z)m
d e σc
(~r) Φ(~r, E),
d Ef
where
d e σc
(~r)
d Ef
is the differential energy compton interaction cross section per electron for an incoming photon with
energy E at ~r and (N Z)m is the electron density in the medium. The photon flux at ~r is Φ(~r, E) =
φ0 (~r) δ(E − E0 ). The energy distribution of scattered photons is computed from the integral
Z
E0
S(~r, Ef ) =
R(~r, E) dE.
0
a) Determine and plot the energy distribution of the scattered photons S(~r, Ef ) at ~r for E0 = 1 MeV
photons for a Pb medium and φ0 = 1010 photons/cm2 -s.
b) Determine and plot the energy distribution of the emitted electrons at this point for the same initial
photon energy, medium and flux level.
The Klein-Nishina angular differential collision cross-section is
de σc
= r02
dΩ
1 + cos2 θ
2
1
1 + α (1 − cos θ)
and the energy loss relationships are
Eγ0 = Eγ
2 1
1 + α(1 − cos θ)
α2 (1 − cos θ)2
1+
(1 + cos2 θ)[1 + α (1 − cos θ)]
and Te = Eγ − Eγ0
and where Eγ0 is the final energy of the gamma after the collision (Ef ) and Eγ is the initial energy Ei .
Solution:
The key point for this problem is the construction of the energy differential cross sections for photon
and electron scattering. Both can be obtained from the Klein-Nishina angular differential collision
cross-section. Once they are constructed we can generate the collision rate and then compute the
integral. For both parts a and b we need to know (N Z)m for Pb. The Z for lead is 82, its density is
ρ = 11.34 g/cm3 and its molar mass is 207.19 grams/mole.
(N Z)m
=
(82)
(11.34)(6.023 × 1023 )
= 2.703 × 1024 electrons/cm3 .
207.19
4
a)
d e σc
=
d Eγ0
Now
Eγ0 =
d Ω̂ d Ω̂ d Eγ0 d e σc
Eγ
1 + α (1 − cos θ)
dθ Thus
d E0 2 π sin θ d θ =
d Eγ0
d Ω̂ dθ d e σc
=
(2 π sin θ) d Eγ0 d Ω̂
2
1
=⇒ d Eγ0 = −Eγ
α sin θ d θ
1 + α (1 − cos θ)
=
γ
d e σc
d Eγ0
Recall
d e σc
(1 + α (1 − cos θ ))
Eγ α sin θ
2
dθ (2 π sin θ) d Eγ0 d Ω̂
2 1 + cos2 θ
1
α2 (1 − cos θ)2
2
= r0
1+
2
1 + α (1 − cos θ)
(1 + cos2 θ)[1 + α (1 − cos θ)]
=
d e σc
2
(1 + α (1 − cos θ))
×(2 π sin(θ))
Eγ α sin θ
2
π r0
α2 (1 − cos θ)2
2
=
1 + cos θ +
Eγ α
1 + α (1 − cos θ)
From
Eγ0 =
Eγ
1 + α (1 − cos θ)
we obtain
and
also
(1 + α (1 − cos θ)) =
Eγ
Eγ0
Eγ
−1
Eγ0
2
1
Eγ
2
cos θ = 1 +
1− 0
α
Eγ
α (1 − cos θ) =
Substituting these relations into the above equation gives
d e σc
=
d Eγ0
π r02
Eγ α
2 !
2 0 Eγ
1
Eγ
Eγ
1+ 1+
1− 0
−1
+
α
Eγ
Eγ
Eγ0
5
for
Eγ
≤ Eγ0 ≤ Eγ
1 + 2α
Now
R(~r, Eγ )
=
=
d e σc
(~r) φ0 (~r) δ(Eγ − E0 )
d Eγ0
2 0 2 !
Eγ
π r02
1
Eγ
Eγ
+
−1
1+ 1+
1− 0
(N Z)m φ0 (~r) δ(Eγ − E0 )
Eγ α
α
Eγ
Eγ
Eγ0
(N Z)m
E0
Z
S(~r, Eγ0 )
=
R(~r, Eγ ) dEγ
0
2 0 2 !
Eγ
1
Eγ
Eγ
= (N Z)m φ0 (~r)
δ(Eγ − E0 )
1+ 1+
+
−1
dEγ
1− 0
α
Eγ
Eγ
Eγ0
0
2 0 2 !
Eγ
1
E0
E0
π r02
0
+
S(~r, Eγ ) = (N Z)m φ0 (~r)
−1
1+ 1+
1− 0
E0 α
α
Eγ
E0
Eγ0
Z
E0
π r02
Eγ α
for
E0
≤ Eγ0 ≤ E0 .
1 + 2α
We can plot this expression against all final energies Eγ0 .
b) The procedure begins the same as in part a) by constructing differentials:
d e σc
=
d Te
Now
0 d Ω̂ d Eγ d Ω̂ d Eγ0 d Te d e σc
Eγ
1 + α (1 − cos θ)
and
Eγ = Eγ0 + Te
dθ Thus
d E0 γ
d Eγ0 and
d Te Eγ0 =
2 π sin θ d θ d Eγ0 d Te d Eγ0
d Ω̂ d θ d Eγ0 d e σc
(2 π sin θ) =
d Eγ0 d Te d Ω̂
2
1
=⇒ d Eγ0 = −Eγ
α sin θ d θ
1 + α (1 − cos θ)
=⇒ d Te = −d Eγ0
=
d e σc
=
(1 + α (1 − cos θ ))
Eγ α sin θ
=
1
2
0
dE Note that d Teγ = 1, hence
d e σc
=
d Te
0 d Ω̂ d Eγ d Ω̂ d Eγ0 d Te d e σc
2 π sin θ d θ (1)
=
d Eγ0
d Ω̂ dθ d e σc
=
(2 π sin θ) d Eγ0 d Ω̂
d e σc
This gives
d e σc
=
d Te
π r02
Eγ α
1
1+ 1+
α
6
Eγ
1− 0
Eγ
2
+
Eγ0
Eγ
Eγ
−1
Eγ0
2 !
.
where Eγ0 = Eγ − Te . Replacing Eγ0 with Eγ − Te we have
d e σc
d Te
=
π r02
Eγ α
2 2 !
1
Eγ − Te
Eγ
Eγ
1+ 1+
+
1−
−1
α
(Eγ − Te )
Eγ
(Eγ − Te )
for
0 ≤ Te ≤
2 α E0
.
1 + 2α
Now
R(~r, Eγ )
=
=
d e σc
(~r) φ0 (~r) δ(Eγ − E0 )
d Te
π r02
×
(N Z)m φ0 (~r) δ(Eγ − E0 )
Eγ α
2 2 !
1
Eγ − Te
Eγ
Eγ
1+ 1+
+
1−
−1
α
(Eγ − Te )
Eγ
(Eγ − Te )
(N Z)m
Z
E0
R(~r, Eγ ) dEγ
2
π r0
= (N Z)m φ0 (~r)
×
α
2 2 !
Z E0
1
Eγ
Eγ − Te
Eγ
δ(Eγ − E0 )
1+ 1+
1−
+
−1
dEγ
Eγ
α
(Eγ − Te )
Eγ
(Eγ − Te )
0
2 !
2 π r02
1
E0
E0 − Te
E0
S(~r, Te ) = (N Z)m φ0 (~r)
−1
1+ 1+
1−
+
E0 α
α
(E0 − Te )
E0
(E0 − Te )
S(~r, Te )
=
0
for
We can plot this expression against all final energies Te .
7
0 ≤ Te ≤
2 α E0
.
1 + 2α