jackson milo choate atlas em calorimeter presentation

Reading & Presentation Assignment

Jackson Milo Choate

Atlas EM Calorimeter

Presentation November 3

Harisankar Namasivayam

Vexter Reconstruction in Atlas

Presentation October 27

Susmita Jyotishmati The LHC: the energy, cooling, and operation Presentation October 20

Wei Cheng Wong The Atlas Pixel Detector Presentation October 20

Juan Navarro-Sorroche Neutron Detection Presentation October 13

Lionel Cohen Atlas Muon Detectors Presentation October 13

Yuen-Jung Chang Babar Cherenkov Detector Presentation October 6

Jackson Milo Choate Gamma Ray Detection for Inspection Presentation October 6

√√√√ Wei Cheng Wong The Atlas Transition Radiation Tracker sub-detector Presentation Sept. 22

√√√√ Harisankar Namasivayam Materials in the Atlas Detector Presentation Sept. 29

09/27/2010 1PHYS6314 Prof. Lou

Plan:

Chapter 2,3, 4 examples

Reading & study assignments

Units of radiations & SourcesUnits of radiations & Sources

Detector Chararteristics

09/27/2010 2PHYS6314 Prof. Lou

Chapter 2 Examples

Examples:6. The Cherenkov angle is normally derived to be related to the particle

velocity β and index of refraction n according to cosθ=1/nβ. This, however,

neglects the recoil of the emitted Cherenkov photon on the incident

particle. Determine the exact relation for the Cherenkov angle considering

the recoil effect.

source: Grupen09/27/2010 3PHYS6314 Prof. Lou

Chapter 2 Examples

Examples: 6. The Cherenkov angle is normally derived to be related to the

particle velocity β and index of refraction n according to cosθ=1/nβ. This,

however, neglects the recoil of the emitted Cherenkov photon on the

incident particle. Determine the exact relation for the Cherenkov angle

considering the recoil effect.

'

'

'

before Cherenkov emission: ,

after Cherenkov emission: q , q

Eq

p

hE

γ

ν

=

= =

��

��

source: Grupen09/27/2010

'

2 2 2

after Cherenkov emission: q , q

energy-momentum conservation:

q' = q-q

2Use E , E= m, , c/n= , =

It can be der

pp

m p p hkk

γ

γ

γ

γ

πνγ ν λ

= =

= + =

��

��

2

rived that

1 1 1cos = (1 )

2

(since the energy carried off by the Cherenkov light is very small)

hk

n p n nβ βΘ + − ≅

4PHYS6314 Prof. Lou

PART II Physics of Particle-Matter Interactions

Chapter 2 Energy Loss of Charged Particles

Chapter 3 Photons and Neutrons Chapter 3 Photons and Neutrons

Chapter 4 Radiation Measurements

units and sources of radiations

09/27/2010 5PHYS6314 Prof. Lou

Chapter 4 Measurements of Radiations

� Units of radiations

lifetime, dosage, weighting factors, equivalent dosage

equivalent whole body dose, radiation limit

� Sources of radiations

radioactive sources: β decays, electron capture,

09/27/2010

radioactive sources: β decays, electron capture,

annihilations, α decays, ….

radioactive sources: radio-isotope

� Additional Information

problems & examples

6PHYS6314 Prof. Lou6PHYS6314 Prof. Lou

Chapter 4 Examples

Examples: 4.1 In an accident in a nuclear lab a researcher has inhaled dust

containing the radioactive isotope 90Sr (Strontium), which led to a dose rate

of 1 µSv/h in his body. The physical half-life of 90Sr is 28.5 years, the

biological half-life is only 80 days. How long does it take this dose rate to

decay to a level of 0.1 µSV/h?


Chapter 4 Examples


containing the radioactive isotope 90Sr (Strontium), which led to a dose rate

of 1 µSv/h in his body. The physical half-life of 90Sr is 28.5 years, the

biological hal-life is only 80 days. How long does it take this dose rate to

decay to a level of 0.1 µSV/h?

because the biological half-life ( =80 days) is much smaller

than the physical hal-lifetime ( 28.5 ), the overal hal-life is then

b

pyrs

τ

τ =


taken as = =80 days.

( ) ( 0) , where D(t)/D(t=0)=0.1, we f

b

t

D t D t e τ

τ τ−

= = ind

D(t)t=- ln( ) 80( ) ln(0.1) 263 days

D(t=0)daysτ = − × =

8PHYS6314 Prof. Lou

Chapter 4 Examples


containing the radioactive isotope 90Sr, which led to a dose rate of 1 µSv/h in

his body. The physical half-life of 90Sr is 28.5 years, the biological hal-life is

only 80 days. How long does it take this dose rate to decay to a level of 0.1

µSV/h?

263263

During the period the total dose the researcher has received is

( 0) ( 0) (1 )t days

days

D D t e dt D t eτ ττ− −

= = = = × −∫


0( 0) ( 0) (1 )

note that = =55.5 days, we have ln 2

2.47 .

The maximum dose (t )=2.8 mSv

total

h

total

D D t e dt D t e

T

D mSv

τ ττ

τ

= = = = × −

=

→ ∞

∫

9PHYS6314 Prof. Lou

Chapter 4 Examples

Examples: 4.2 Consider a pocket dosimeter with a chamber volume of 2.5

cm2 and a capacitance of 7 pF. Originally it has been charged to a voltage 0f

200 V. After a visit in a nuclear power plant it only showed a voltage of 170

V. What is the received dose?


Chapter 4 Examples

Examples: 4.2 Consider a pocket dosimeter with a chamber volume of 2.5

cm2 and a capacitance of 7 pF. Originally it has been charged to a voltage 0f

200 V. After a visit in a nuclear power plant it only showed a voltage of 170

V. What is the received dose?

12 10

The recorded charge ( Q) is related to the voltage drop ( U)

Q U=7 10 ( ) 30( ) 2.1 10 ( )

The anount of air in the chamber is

C F V C− −

∆ ∆

∆ = ∆ × = ×i


3

5

The anount of air in the chamber is

m 3.23 10

Then the charge dose =

Q6.5 10 / 0.258

air air

air

V g

C kgm

ρ −

−

= = ×

∆= × = 0.258 (8.8 / ) 2.2R R mGy R mGy= × =

11PHYS6314 Prof. Lou

Chapter 4 Examples

Examples: how does a dosimeter work?

Luxel OSL Whole Body Dosimeter Luxel's optically stimulated luminescence (OSL)

dosimeter measures radiation exposure due to x-ray, beta, and gamma radiation

through a thin layer of aluminum oxide and different filters. The dosimeter is

enclosed in a water-resistant blister pack. After use, the RPO returns them to

Landauer for processing, where the aluminum oxide is stimulated with a blue-laser

causing it to become luminescent. This luminescence is proportional to the amount

of the radiation exposed to the dosimeter during use. This luminescence is measured

and a report of the exposure results is generated.

source: Harvard U09/27/2010 12PHYS6314 Prof. Lou

Chapter 4 Examples

Examples: how does a dosimeter work?

TLD Ring Thermoluminescent dosimeter (TLD) rings measure radiation exposure to your

extremities due to x-ray, beta, and gamma radiation with an encased lithium fluoride chip. The

TLD chip is sealed beneath the identification cover of the ring. After use, the RPO returns them

to Landauer, where the chip is removed and carefully heated causing the chip to become

luminescent. The luminescence is proportional to the amount of radiation exposure during use.

source: Harvard U09/27/2010 13PHYS6314 Prof. Lou

Chapter 4 Examples

Examples: 4.3 The absorption coefficient for 50 keV X-rays in aluminum is

µ=0.3 (g/cm2)-1. Work out the thickness of an aluminum shielding which

reduces the radiation level by a factor of 10,000.


Chapter 4 Examples

Examples: 4.3 The absorption coefficient for 50 keV X-rays in aluminum is

µ=0.3 (g/cm2)-1. Work out the thickness of an aluminum shielding which

reduces the radiation level by a factor of 10,000.

2

0

0

The attenuation of x-rays in material is given

1 1( / )( ) ln( ) ln(10000.0)

( ) 0.3

x I g cmI x I e x

I x

µ

µ−= ⇒ = =


0

2

3

*

( ) ln( ) ln(10000.0)( ) 0.3

30.7 /

density of Al is 2.7g/cm , we have

( ) / 11.4Al

I x I e xI x

g cm

The

x length x cm

µ

ρ

= ⇒ = =

=

= =


Chapter 4 Examples

Examples: 4.4 How does the radiation does received in a four-week

holiday in the high mountain (3000 m) compare ti the radiation load caused

by an X-ray of the human chest in an X-ray mass screening?


V.S.


Chapter 4 Examples

Examples: 4.4 How does the radiation does received in a four-week

holiday in the high mountain (3000 m) compare ti the radiation load caused

by an X-ray of the human chest in an X-ray mass screening?

The who-body dose a person gets from a modern x-rays tube

is ~ .

For a 4-week holiday, the average cosmic ray dose rate at 3000 m

0.1 mSv


For a 4-week holiday, the average cosmic ray dose rate at 3000 m

is about 0.1 67 Sv/h that amounts to . The radiation dose

du

Sv/hµ µe to terrestrial radiation is about . The s

comparabl

u

e

m

a

is ~

moun

107 Sv/

t of ra

40 S

diat

.v/ hh

. ions

µµ


Chapter 4 Examples

Examples: 4.5 A 10 Ci 60Co source falls down and is almost immediately

recovered by a technician with his naked, unprotected hand. Work out the

partical body dose and also estimate a value for the whole-body does

(exposure time ~60 seconds for the hands, and 5 minutes for the whole

body).


Database of nuclear accidents in the past 60 years –

http://pages.prodigy.net/wrjohnston/nuclear/radevents/index.html


Chapter 4 Examples



partical body dose and also estimate a value for the whole-body does


body).

The specific does constants for β and γ radiation of 60Co are

Γβ = 2.62×10-11 Svm2/Bqh, Γγ = 3,41×10-13 Svm2/Bqhβ γ

Since Γβ >> Γγ, the exposure of the hands is mainly due to β radiation. Assume hand-source

distance ~10 cm and the handling time is 60 s, the partial-body does is

The whole-body dose depends on the g radiation. Assume distance 0.5 m, exposure time 300 s.


11

11

2 2

3.7 10 12.62 10 16.1

0.1 60

AH t Sv Sv

rβ β

− ×= Γ × ∆ = × =

242 m

AH t Sv

rγ γ= Γ × ∆ =

Chapter 4 Examples



partial body dose and also estimate a value for the whole-body does


body).

A real accident in Saintes, France in 1981 –

Date: 2 April 1981

Location: Saintes, France

Type of event: accidental irradiation with teletherapy source

source: Wm. Robert Johnston09/27/2010

Type of event: accidental irradiation with teletherapy source

Description:

While a technician was changing the cobalt-60 source in a teletherapy machine,

the source fell to the ground. The technician picked up the source, touching it for

11 seconds. The dose to the hands was over 10,000 rad, and both hands had to be

amputated. A second operator had to have both hands amputated, and a third had

three fingers amputated. Eight other individuals in the room received doses of

1 rad to 100 rad.

Consequences: 3 injuries.


09/27/2010 21PHYS6314 Prof. Lou

Chapter 4 Examples

Examples: 4.6 A nuclear lab has been contaminated with a radioactive

isotope. The decontamination procedure had an efficiency of ε=80%. After

three DP a remaining surface contamination of 512Bq/cm2 was still

measured. What is the initial contamination? If the level of contamination

were to be lowered to 1 Bq/cm2, how many more DB procedures must be

performed?

09/27/2010 PHYS6314 Prof. Lou 22


Chapter 4 Examples

Examples: 4.6 A nuclear lab has been contaminated with a radioactive

isotope. The decontamination procedure had an efficiency of ε=80%. After

three DP a remaining surface contamination of 512Bq/cm2 was still

measured. What is the initial contamination? If the level of contamination

were to be lowered to 1 Bq/cm2, how many more DB procedures must be

performed?

Each DP procedure reduces the comtamination to 20% of the original levele.

09/27/2010 PHYS6314 Prof. Lou 23


2 3

0

0

After three DPs the level of contamination is 512 Qb/cm =N (1.0 0.8) ,

(N is the original levelof contamination), we find

−

2

0 3

2

2 0

0

2512 N = Qb/cm =

0.2

To reduce the level to 1 Qb/cm , we would need n DP procedures, such that

log( ) log(64000)1 Qb/cm =N (1.0 0.8) ,

64,000 Qb/cm

leading to n=log0.2 log0

n N− −− =

27 DP procedures are required to clean up the contamination to 1 QB/

6..

.

9

cm

2=


Chapter 4 Examples

Three Mile Island nuclear plant

09/27/2010 24source: Scientific American

Chapter 4 Examples

Environment impact

of nuclear activities

09/27/2010 25source: US Nuclear Regulatory Commission

Chapter 4 Examples

“The primary obstacle in disposing of nuclear waste and cleaning nuclear development

facilities is the duration of halflives of the elements that compose nuclear waste. For

example, Uranium 235 has a half-life of 703,800,000 years. There is no real way of

disposing of the waste, the only option is effectively manage the waste for the

thousands of years until it decays completely. ”

The methods used to clean contaminated soil include washing, extraction and

incineration. Extracting soil involves removing contaminated soil which is often

subsequently buried in lined landfills (Boudlen 1999). This method is not favored

09/27/2010 26source: U of Michigan

subsequently buried in lined landfills (Boudlen 1999). This method is not favored

because it does not eliminate existing waste. Incinerating techniques use a special

furnace that releases the contaminants in non-harmful forms (Boudlen 1999). Although

this is an effective solution, the method is expensive and there is a certain amount of

difficulty involved in the operation of the furnaces. Soil washing techniques extract

contaminated soils from their original sites, typically separate stones from finer soils,

mix the soils with a solvent that causes contaminants to turn into a liquid form and then

trap the newly contaminated solvent (Boulden 1999). Though the original sites can be

cleaned, the liquid waste still needs to be disposed of.

Chapter 4 Examples

Sites storing radioactive materials

09/27/2010 27source: US DOE

PART III Energy Physics Detectors: Principles

and Operations

Chapter 5 General Concepts

Chapter 6 Ionization Detectors

09/27/2010 PHYS6314 Prof. Lou 28

Chapter 6 Ionization Detectors

Chapter 7 Scintillation Detectors

Chapter 8 Semiconductor Detectors

Chapter 9 Calorimeters

Chapter 10 Other Detectors

Chapter 5 General Concepts

� Sensitivity, detector response, energy/position resolution

� Response time, detection efficiency

09/27/2010 PHYS6314 Prof. Lou 29

�Dead time, noise

Chapter 5 Detectors: General Concepts

Resolutions: measurement of spatial position and time of radiation and

high energy particles.

Position

Resolution

A wire chamber hit gives a position measurement that corresponds to a

( )z z PDF z dz< >= ×∫2 2( ) ( )z

z z PDF z dzσ = −< > ×∫

A wire chamber hit gives a position measurement that corresponds to a

rectangular box:

source: Grupen

2 2

2 2

2

2 2 2

2

1 / 1 0

1 12 12

( 0) 1 / z

z

z

z

z dz dz

dz

z

z dz

δ δ

δ δ

δ

δ

δσ

δσ

+ +

− −

+

−

× =

= ⇒

< >=

= − × =

∫ ∫

∫∫


Basic statistics: In many experiments the results are Gaussian distributed.

2

2

( )

21

2( )

z z

eG z σ

πσ

−< >−

=

source: Grupen

( ) ( the confidence level, 5-95% or 1-99%)1z

z

G z dzδ

δ

αα< >+

< >−

− = ∫

2ln 2Full Width at Half Maximum: 2 2.3548 FWHM σ σ× ==


Characteristic times: dead time, recover time, readout timeDead time – time period during which a detector can not ‘see’ a particle passing through it.

Recover time – the period of time between the detector just begin to see a passing particle

(although not with full sensitivity), and the time when the detector is fully recovered and is able

to see with full sensitivity the passing particle.

Readout time – is the time required to read the event to an electronic memory.

during the dead time events

or particles coming in are

source: Grupen

Dead time correction: N1

real count rate; N the observed count rate; deadtime

true

true D

N

N D

N

ττ

=− ×

or particles coming in are

not recorded. The apparent

count rate is not exactly the

true rate of event. A correction

is needed to determine the true

rate.

jackson milo choate atlas em calorimeter presentation

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