Nuclear methods in the inspection of macroscopical samplesof material

Angel Humberto CruzMartha Liliana CortesNancy ForeroWilson Javier BlancoZandra Omaira SilvaLiliana Maritza MeloMarıa Fernanda NaderEdana MerchanFernando CristanchoUniversidad Nacional de Colombia

Eduardo FajardoJose Enrique GarcıaCentro Internacional de Fısica

Jurgen GerlFrederic AmeilGesellschaft furSchwerionenforschungDarmstadt, Alemania

November 12, 2009

The problem

Explosives with high content of organic material and low content in metal→ does not work the metal detector.

Alternative: methods that

• “see” the organic material , and...

• distinguish it from the soil.

1

The methods

1. Compton γ-ray Backscattering

2. Thermal Neutron Backscatering

The colombian case:

A. Wide range of humidity valuesB. Non-standard antipersonnel mines

2

The method

d

s

252Cfdetector A detector B

suelo

mina

neutrones

coordenate x0 of the mine

A(x)−B(x)

x0

0

B(x)A(x)signal

D(x) = D0

[

exp

(

x− x0 − s√2σ0

)2

− exp

(

x− x0 + s√2σ0

)2]

F. D. Brooks, M. Drosg, The HYDAD-D antipersonnel landmine detector.Appl. Rad. Isot. 63, 565 (2005).

3

Experimental set-up

Electronics + spectra

4

The dummy landmine (DLM2)

Acrylic→[H:C:N:O]=[4:2:0:1]

3.4

3.5

2.2

4 cm

TNT Simulant→[H:C:N:O]=[5:7:3:6]

simulates 100 g explosive

5

Simulation (Geant4) and experimental results

Soil = dry sand

6

7

8

9

10

11

B(x)(Counts×10

3) (a)

d = 0 cm

Sim

Exp

-2

-1

0

1

2

-30 -20 -10 0 10 20 30

D(x)(Counts×10

3)

x (cm)

(b)

6

The dependence on depth

-4

-2

0

2

4

-30 -20 -10 0 10 20 30

D(x)(Counts×10

3)

x(cm)

d(cm)

θ = 0%

051015

humidity = θ = 0%

7

The dependence on soil humidity

252Cfdetector A detector B

wminehumid soil

dry soil

w = 5 cm

-8-6-4-20246810

-30 -20 -10 0 10 20 30

D(x)(Counts×10

3)

x (cm)

d = 0 cm

Moisture (weight %)0510

8

7.55.02.5

θ(%) = 0w = 18 cm

depth (cm)

D0(103

counts)

1614121086420

15

10

5

0

d = 0 cm

D0/D

0(θ

=0%)

2

1.5

1

0.5

0

-0.5

-1

-1.5

-2

-2.5

Sim.

Exp

0 10 20 40 50 60

θ(%)

30

9

Conclusions neutrons

1. A mine of around 100 g can be located with certainty up to a depth of 10 cm in drysand.

2. Humidity has an amplyfying effect → pre-thermalization.

3. Swamp-humidity produces negative signals, i.e., mines are detectable.

4. It is important to characterize the humidity region around D0 ≈ 0.

10

Towards a Colombian prototype: the preamp is ready

11

II. Gamma-ray Compton backscattering

Goal: To obtain images of objects buried in soil.

Physical principle: γ-rays detect spatial variations of the electronic density.

backscatteringdetector

shielding

detectorposition

22Na source

landmine soil

γ2

γ1 s

d

99.962%

0.038%

2211Na

2.60 a

β+

3.63 ps

2210Ne

1274.5 keV

A game of probabilities:

Pbackscattering ∝ σC ∝ Z

Pabsorption ∝ σph ∝ Z3.5

J. Gerl et al., Nucl. Instr. Meth. Phys. Res. A 525, 328 (2004).

12

The Compton camera

13

R&D program

1. Independent research of γ-soil.

• Problem: existing data and theory describe just one event...

• ... whereas γ-soil:

(a) multiple colisions

(b) target: multielemental

2. Imaging: DAQ, pattern recognition, ...

3. Characterize the prototype

(a) Geometry: view of field, relative sizes, ...

(b) Operation values

(c) ...

4. Geant4 wherever it may seem to help.

14

The γ-soil experiments: Ge-NaI coincidences

polystyrene

polypropylenesoil

lead

NaI detector Gedetector

22Na

Two set-up modes:

1. Transmission (in the figure).

2. Backscattering

Spectroscopy as a function of

1. thickness of sample

2. humidity,

3. type of soil

15

backscattering:

transmision:

16

Transmitted intensity at low angle (∼ 400− 500 keV)

• At small depth: intensity becomes larger with thickness. For thickness > 10 cmintensity decreases.

• No clear effect with humidity.

17

Imaging

Simulated plate, gradient in 3D and 2D:

X (pixels)

Y (

pixe

ls)

10 20 30 40 50 60 70 80

10

20

30

40

50

60

70

80 400

300

200

100

0

Intensity (counts)

X (pixels)

Y (pixels)

Intensity (counts)

X (pixels)

Y (

pixe

ls)

10 20 30 40 50 60 70 80

10

20

30

40

50

60

70

80

Thresholding, k = 0.8, k = 0.9:

0

10

20

30

40

50

60

70

80

0 10 20 30 40 50 60 70 80

Y (

pixe

ls)

X (pixels)

0

10

20

30

40

50

60

70

80

0 10 20 30 40 50 60 70 80

Y (

pixe

ls)

X (pixels)

18

Image (with late Minen-Verifier):

X (pixels)

Y (

pixe

ls)

0 10 20 30 40 50 60 70 80 0

10

20

30

40

50

60

70

80

Gradient:

X (pixels)

Y (

pixe

ls)

10 20 30 40 50 60 70 80

10

20

30

40

50

60

70

80

Thresholding an fitting:

k = 0.5

0

10

20

30

40

50

60

70

80

0 10 20 30 40 50 60 70 80

Y (

pixe

ls)

X (pixels)

k = 0.8

0

10

20

30

40

50

60

70

80

0 10 20 30 40 50 60 70 80

Y (

pixe

ls)

X (pixels)

19

The response of the Compton Camera

What does the camera “see”?

Case of study: massive objects of specific material in front of its field of view:

0

5

10

15

20

25

30

35

40

(a) Air

0

5

10

15

20

25

30

35

(b) Polystyrene

0

5

10

15

20

25

30

35

40

(c) Paper

0 5 10 15 20 25 30 35 40 45

(d) Water

0 5 10 15 20 25 30 35 40 45

(e) Sand

0

5

10

15

20

25

30

35

40

(f) Pb

20

s = 10 cm

counts/pixel18016014012010080604020

100

80

60

40

20

0

watersand

paraffinpaperleadair s = 5 cm

frequency

100

80

60

40

20

0

s = 0 cm100

80

60

40

20

0

Frequency analysis

• s = 0: The less reflective ma-terials

– air

– lead

group together. The others:

– paper

– paraffin

– water, sand

• The distinction of the materialsdecreases with larger stand-offdistance. At s = 10 cm there isno distinction.

21

The Compton camera according to Geant4

Backscattering

detector

detectorPosition

Source

Object Soil

0 10 20 30 40 50 60 70 80 90

X (pixels)

Y (

pixe

ls)

15 20 25 30 35 40 45 50

15

20

25

30

35

40

45

50

According to Geant4 the camera should see a lead plate r = 5 cm, s = 2 cm...

22

Experimental images

No object

0

20

40

60

80

100

120

140

160

d = 0, lead brick 2× 2× 5 cm3

0

20

40

60

80

100

120

140

d = 1.8 cm. Object on the letf

0

50

100

150

200

250

300

350

d = 1.8 cm. Object on the right

0

50

100

150

200

250

300

350

23

THANKS

24

La interaccion γ-suelo

1. Efecto Compton: seccion eficaz de Klein-Nishina:

dσ

dΩ(θ, ǫ) =

r2e2

1

[1 + ǫ(1− cos θ)]2

[

1 + cos2 θ +ǫ2(1− cos θ)2

1 + ǫ(1− cos θ)

]

, ǫ =Eγ

mec2

1010.50.10.01 MeV

~pγ

90

60

30

γ − edσdΩ(θ)

σC ∝ Z

2. Efecto fotoelectrico:σph ∝ Z3.5

25

totalproduccion de pares

fotoelectricoCompton

1.2750.511

γ-SiO2

Eγ(MeV)

µ/ρ

(gm/cm

−2)

1010.10.01

101

100

10−1

10−2

10−3

10−4

http://www.physics.nist.gov/PhysRefData/Xcom/Text/XCOM.html

26

NaI

Ge

fuente de 22Na

Pb (blindaje)

muestra

t

Retrodispersion

NaI

muestra

t

Ge

fuente de 22Na

Transmision

27