1.1 A 0.5 MeV neutron strikes a target atom with mass A, which is initially at rest. Calculate the
velocity and energy of both particles in the laboratory reference frame after a head-on collision for
A=27 (Al) and A=207.2 (Pb)
Before Collision After Collision
m1 ,v1
m1 ,v i m2
θ
m2 ,v 2
For a head-on collision: θ=π
Ei = 0.5 MeV mn = 1.675×10-27 kg m1 = mn m2 = Amn
Necessary Equations:
1 2 2E
(1) E = mv → v =
2 m
(2) Conservation of Energy: Ei = T + E f
γ
(3) T = Ei (1 − cos θ )
2
4A
(4) γ =
(1 + A)2
i) A = 27
Using (4): γ = 0.137
Using (3): T = 68.9keV
Using (2): E f = 431.1keV
Using (1): v1 = 9.08×10 m , v 2 = 7.02 ×10 m
6 5
s s
,ii) A = 207.2
Using (4): γ = 0.019
Using (3): T = 9.56 keV
Using (2): E f = 490.4 keV
Using (1): v1 = 9.69 ×10 m , v 2 = 9.44 ×10 m
6 4
s s
,1.2 A detector of 100% efficiency (i.e., every particle entering the detector is registered) and area of
1cm2 is placed a distance r from a target (taken to be of zero dimension, i.e. a point). The target is
bombarded with neutrons. Assuming that only elastic scattering occurs, that scattering is
azimuthally symmetric and that the scattering cross section is isotropic:
Position 2, θ2
Position 1, θ1
dθ = Increment of scattering angle
dΩ = Increment of solid angle about θ
θ = Scattering angle in the center-of-mass system
a) What is the ratio of the number of particles detected by the detector at positions 1 and 2
shown in the figure?
b) What is the ratio of the number of particles scattered through an angular increment of 10°
about θ1 = 5° and θ2 = 85° ?
c) Repeat parts (a) and (b) assuming that instead of being isotropic the differential scattering
cross section varies as σ s ( Ei ,θ ) = cos θ .
a) Since 𝜎! is isotropic, and for both positions the distance to the detector is the same, the fluxes and cross
sections cancel yielding:
N1 φ1σ s (E i , θ )1 d Ω1
= =1
N 2 φ2σ s (E i ,θ ) 2 d Ω 2
b)
10 10 10
∫ φ1σ s (Ei ,θ )1 d Ω1 ∫φσ 1 s (E i ,θ )1 2π sin θ1dθ1 ∫ sin θ dθ
1 1
N1
= 900 = 0
90
= 0
90
N2
∫φ σ 2 s (Ei ,θ ) 2 d Ω 2 ∫φ σ 2 s (E i , θ ) 2 2π sin θ 2 dθ 2 ∫ sin θ dθ
2 2
80 80 80
N1 − cos θ1 |100
= 90
= 0.087
N 2 − cos θ 2 |80
,c) For part (a) repeat as before, except this time σ s ( Ei ,θ ) = cos θ .
N1 φ1σ s (Ei ,θ )1 d Ω1 φ1σ s (E i ,θ )1 d Ω1 cos θ1
= = =
N 2 φ2σ s (Ei ,θ ) 2 d Ω2 φ2σ s (E i , θ ) 2 d Ω 2 cos θ 2
N1 cos(5°)
= = 11.43
N 2 cos(85°)
For repeating part (b), the integrals are modified as follows:
10 10 10
φ1σ s (Ei ,θ )1 d Ω1 ∫φσ (Ei ,θ )1 2π sin θ1dθ1 ∫ cos θ sin θ dθ
N1 ∫0
1 s
0
= = = 900
N 2 90 90
∫φ σ 2 s (Ei ,θ )2 d Ω2 ∫φ σ 2 s (Ei ,θ )2 2π sin θ 2 dθ 2 ∫ cosθ sin θ dθ
80 80 80
Continue integration by substitution:
u = sin θ
du = cos θ dθ
sin(10)
udu 1 2 sin(10)
N1 sin(0)
∫ u |sin(0)
= =2 =1
N 2 sin(90) 1 2 sin(90)
udu u |sin(80)
∫
sin(80)
2
,1.3 A Ti plate is bombarded with 1014 neutrons per cm2 per second at perpendicular incidence.
The entire plate is hit by the beam.
a) Calculate the number of particles scattered per second at
i) 85˚ ≤ θ ≤ 86˚
ii) 5˚ ≤ θ ≤ 6˚.
The plate size is 1 cm2 by 0.6 mm. Scattering is isotropic with a total scattering cross section
of 2.87 barns (1 barn = 10-24 cm2).
b) The same target is bombarded with particles such that the differential angular scattering
cross section is proportional to θ2. Calculate the ratio of the atomic flux in interval (i) to
that in interval (ii). In both cases, perform full integration of the differential cross section.
c) Approximate the integrals in (b) by assuming the differential angular scattering cross
section to be constant in each integration interval and equal to the value at the interval’s
center.
a) Converting angles to radians:
85° ≤ θ ≤ 86° → 1.48 ≤ θ ≤ 1.50
5° ≤ θ ≤ 6° → 0.087 ≤ θ ≤ 0.105
Rate of scattering events: Rs = ∫∫ N σ s ( Ei , θ )φ dxd Ω
x Ω
ρ NA
Number Density: N = = 5.71×1022 at
M cm3
σ s ( Ei ) 2.87
Scattering is isotropic: σ s ( Ei ,θ ) = =
4π 4π
Fluence: φ = 1014 n
s
Thickness of Plate: dx = 0.06 cm
Solid Angle: d Ω = 2π sin θ d θ
2π Nφσ s ( Ei )dx θ2 1 θ
Rs = ∫ sin θ dθ = Nφσ s ( Ei )dx [− cos θ ]θ2
4π θ1 2 1
i) 85˚ ≤ θ ≤ 86˚
Rs = 8.55 ×109 scatters
s
ii) 5˚ ≤ θ ≤ 6˚
Rs = 8.22 ×109 scatters
s
, b) Scattering is proportional to r2: σ s ( Ei , θ ) = kθ 2 where k is an arbitrary constant
1.50 1.50
2π k ∫ σ s (E i ,θ )d Ω ∫ θ 2 sin θ dθ ⎡⎣ 2θ sin θ + (2 − θ 2 ) cos θ ⎤⎦
(i ) 1.48 1.48
Rs = = 0.105
= 0.105
= 2493
2π k ∫ σ s (E , θ )d Ω ∫
i θ 2 sin θ dθ ⎡⎣ 2θ sin θ + (2 − θ 2 ) cos θ ⎤⎦
( ii ) 0.087 0.087
c) σ s ( Ei ,θ ) ; σ s ( Ei ,1.49) = 1.492 = 2.227
σ s ( Ei ,θ ) ; σ s ( Ei , 0.096) = 0.0962 = 0.009
1.50
2π k ∫ σ s (E i ,θ )d Ω ∫ 2.227 sin θ dθ [−2.227 cos θ ]1.48
1.50
(i ) 1.48
Rs = = 0.105
= 0.105
= 2513
2π k ∫ σ s
( ii )
(E ,θ )d Ω ∫
i 0.087
0.009sin θ dθ [−0.009 cos θ ]0.087
velocity and energy of both particles in the laboratory reference frame after a head-on collision for
A=27 (Al) and A=207.2 (Pb)
Before Collision After Collision
m1 ,v1
m1 ,v i m2
θ
m2 ,v 2
For a head-on collision: θ=π
Ei = 0.5 MeV mn = 1.675×10-27 kg m1 = mn m2 = Amn
Necessary Equations:
1 2 2E
(1) E = mv → v =
2 m
(2) Conservation of Energy: Ei = T + E f
γ
(3) T = Ei (1 − cos θ )
2
4A
(4) γ =
(1 + A)2
i) A = 27
Using (4): γ = 0.137
Using (3): T = 68.9keV
Using (2): E f = 431.1keV
Using (1): v1 = 9.08×10 m , v 2 = 7.02 ×10 m
6 5
s s
,ii) A = 207.2
Using (4): γ = 0.019
Using (3): T = 9.56 keV
Using (2): E f = 490.4 keV
Using (1): v1 = 9.69 ×10 m , v 2 = 9.44 ×10 m
6 4
s s
,1.2 A detector of 100% efficiency (i.e., every particle entering the detector is registered) and area of
1cm2 is placed a distance r from a target (taken to be of zero dimension, i.e. a point). The target is
bombarded with neutrons. Assuming that only elastic scattering occurs, that scattering is
azimuthally symmetric and that the scattering cross section is isotropic:
Position 2, θ2
Position 1, θ1
dθ = Increment of scattering angle
dΩ = Increment of solid angle about θ
θ = Scattering angle in the center-of-mass system
a) What is the ratio of the number of particles detected by the detector at positions 1 and 2
shown in the figure?
b) What is the ratio of the number of particles scattered through an angular increment of 10°
about θ1 = 5° and θ2 = 85° ?
c) Repeat parts (a) and (b) assuming that instead of being isotropic the differential scattering
cross section varies as σ s ( Ei ,θ ) = cos θ .
a) Since 𝜎! is isotropic, and for both positions the distance to the detector is the same, the fluxes and cross
sections cancel yielding:
N1 φ1σ s (E i , θ )1 d Ω1
= =1
N 2 φ2σ s (E i ,θ ) 2 d Ω 2
b)
10 10 10
∫ φ1σ s (Ei ,θ )1 d Ω1 ∫φσ 1 s (E i ,θ )1 2π sin θ1dθ1 ∫ sin θ dθ
1 1
N1
= 900 = 0
90
= 0
90
N2
∫φ σ 2 s (Ei ,θ ) 2 d Ω 2 ∫φ σ 2 s (E i , θ ) 2 2π sin θ 2 dθ 2 ∫ sin θ dθ
2 2
80 80 80
N1 − cos θ1 |100
= 90
= 0.087
N 2 − cos θ 2 |80
,c) For part (a) repeat as before, except this time σ s ( Ei ,θ ) = cos θ .
N1 φ1σ s (Ei ,θ )1 d Ω1 φ1σ s (E i ,θ )1 d Ω1 cos θ1
= = =
N 2 φ2σ s (Ei ,θ ) 2 d Ω2 φ2σ s (E i , θ ) 2 d Ω 2 cos θ 2
N1 cos(5°)
= = 11.43
N 2 cos(85°)
For repeating part (b), the integrals are modified as follows:
10 10 10
φ1σ s (Ei ,θ )1 d Ω1 ∫φσ (Ei ,θ )1 2π sin θ1dθ1 ∫ cos θ sin θ dθ
N1 ∫0
1 s
0
= = = 900
N 2 90 90
∫φ σ 2 s (Ei ,θ )2 d Ω2 ∫φ σ 2 s (Ei ,θ )2 2π sin θ 2 dθ 2 ∫ cosθ sin θ dθ
80 80 80
Continue integration by substitution:
u = sin θ
du = cos θ dθ
sin(10)
udu 1 2 sin(10)
N1 sin(0)
∫ u |sin(0)
= =2 =1
N 2 sin(90) 1 2 sin(90)
udu u |sin(80)
∫
sin(80)
2
,1.3 A Ti plate is bombarded with 1014 neutrons per cm2 per second at perpendicular incidence.
The entire plate is hit by the beam.
a) Calculate the number of particles scattered per second at
i) 85˚ ≤ θ ≤ 86˚
ii) 5˚ ≤ θ ≤ 6˚.
The plate size is 1 cm2 by 0.6 mm. Scattering is isotropic with a total scattering cross section
of 2.87 barns (1 barn = 10-24 cm2).
b) The same target is bombarded with particles such that the differential angular scattering
cross section is proportional to θ2. Calculate the ratio of the atomic flux in interval (i) to
that in interval (ii). In both cases, perform full integration of the differential cross section.
c) Approximate the integrals in (b) by assuming the differential angular scattering cross
section to be constant in each integration interval and equal to the value at the interval’s
center.
a) Converting angles to radians:
85° ≤ θ ≤ 86° → 1.48 ≤ θ ≤ 1.50
5° ≤ θ ≤ 6° → 0.087 ≤ θ ≤ 0.105
Rate of scattering events: Rs = ∫∫ N σ s ( Ei , θ )φ dxd Ω
x Ω
ρ NA
Number Density: N = = 5.71×1022 at
M cm3
σ s ( Ei ) 2.87
Scattering is isotropic: σ s ( Ei ,θ ) = =
4π 4π
Fluence: φ = 1014 n
s
Thickness of Plate: dx = 0.06 cm
Solid Angle: d Ω = 2π sin θ d θ
2π Nφσ s ( Ei )dx θ2 1 θ
Rs = ∫ sin θ dθ = Nφσ s ( Ei )dx [− cos θ ]θ2
4π θ1 2 1
i) 85˚ ≤ θ ≤ 86˚
Rs = 8.55 ×109 scatters
s
ii) 5˚ ≤ θ ≤ 6˚
Rs = 8.22 ×109 scatters
s
, b) Scattering is proportional to r2: σ s ( Ei , θ ) = kθ 2 where k is an arbitrary constant
1.50 1.50
2π k ∫ σ s (E i ,θ )d Ω ∫ θ 2 sin θ dθ ⎡⎣ 2θ sin θ + (2 − θ 2 ) cos θ ⎤⎦
(i ) 1.48 1.48
Rs = = 0.105
= 0.105
= 2493
2π k ∫ σ s (E , θ )d Ω ∫
i θ 2 sin θ dθ ⎡⎣ 2θ sin θ + (2 − θ 2 ) cos θ ⎤⎦
( ii ) 0.087 0.087
c) σ s ( Ei ,θ ) ; σ s ( Ei ,1.49) = 1.492 = 2.227
σ s ( Ei ,θ ) ; σ s ( Ei , 0.096) = 0.0962 = 0.009
1.50
2π k ∫ σ s (E i ,θ )d Ω ∫ 2.227 sin θ dθ [−2.227 cos θ ]1.48
1.50
(i ) 1.48
Rs = = 0.105
= 0.105
= 2513
2π k ∫ σ s
( ii )
(E ,θ )d Ω ∫
i 0.087
0.009sin θ dθ [−0.009 cos θ ]0.087
