EBK INTRODUCTION TO HEALTH PHYSICS, FIF
EBK INTRODUCTION TO HEALTH PHYSICS, FIF
5th Edition
ISBN: 9780071835268
Author: Johnson
Publisher: VST
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Chapter 5, Problem 5.21P

Calculate the probability that a 2-MeV photon in a narrow, collimated beam will be removed from the beam by each of the following shields:
(a) Lead, 1-cm thick
(b) Iron, 1-cm thick
(c) Lead, 1-g/cm 2 thick
(d) Iron, 1-g/cm 2 thick

(a)

Expert Solution
Check Mark
To determine

The probability of 2 MeV photon in a narrow collimated beam will be removal from the beam by 1 cm thick lead shield.

Answer to Problem 5.21P

The probability of 2 MeV photon in a narrow, collimated beam will be removal from the beam by 1 cm thick lead shield is, Pr=0.4066

Explanation of Solution

Given:

Energy, E=2MeV

The thickness of shield, x=1cm

Formula Used:

The intensity of gamma beam through the absorber curve is,

  I=I0eμx

Where,

  I = The intensity of gamma beam through the absorber curve

  I0 =The intensity of gamma beam through zero thickness

  μ =Linear attenuation

  x =Absorber thickness

Calculation:

The linear attenuation of Lead having energy 2 MeV is, μ=0.522cm1

The probability of transmission is given by,

  P=II0

So, the probability of removal is given by,

  Pr=1II0=1eμxPr=1e0.522×1Pr=10.5933Pr=0.4066

Conclusion:

The probability of 2 MeV photon in a narrow, collimated beam will be removal from the beam by 1 cm thick lead shield is, Pr=0.4066

(b)

Expert Solution
Check Mark
To determine

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 cm thick Iron shield.

Answer to Problem 5.21P

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 cm thick iron shield is, Pr=0.3064

Explanation of Solution

Given:

Energy, E=2MeV

The thickness of shield, x=1cm

Formula Used:

The intensity of gamma beam through the absorber curve is,

  I=I0eμx

Where,

  I = The intensity of gamma beam through the absorber curve

  I0 =The intensity of gamma beam through zero thickness

  μ =Linear attenuation

  x =Absorber thickness

Calculation:

The linear attenuation of Iron having energy 2 MeV is, μ=0.366cm1

The probability of transmission is given by,

  P=II0

So, the probability of removal is given by,

  Pr=1II0=1eμxPr=1e0.366×1Pr=10.6935Pr=0.3064

Conclusion:

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 cm thick iron shield is, Pr=0.3064

(c)

Expert Solution
Check Mark
To determine

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 g/cm2 thick lead shield.

Answer to Problem 5.21P

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 g/cm2thick lead shield is, Pr=0.0409

Explanation of Solution

Given:

Energy, E=2MeV

Thickness density of shield, td=1g/cm2

Formula Used:

The intensity of gamma beam through the absorber curve is,

  I=I0eμx

Where,

  I = The intensity of gamma beam through the absorber curve

  I0 =The intensity of gamma beam through zero thickness

  μ =Linear attenuation

  x =Absorber thickness

The thickness of the shield is,

  x=tdρ

Where,

  x =Thickness

  td =Thickness density

  ρ =Density

Calculation:

The thickness of the Lead shield is,

  x=tdρx=111.34x=0.088cm

The linear attenuation of Lead having energy 2 MeV is, μ=0.522cm1

The probability of transmission is given by,

  P=II0

So, the probability of removal is given by:

  Pr=1II0=1eμxPr=1e0.522×0.088Pr=10.9590Pr=0.0409

Conclusion:

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 g/cm2 thick lead shield is, Pr=0.0409

(d)

Expert Solution
Check Mark
To determine

The probability of 2 MeV photon in a narrow, collimated beam will be removed from the beam by 1 g/cm2thick iron shield.

Answer to Problem 5.21P

The probability of 2 MeV photon in a narrow, collimated beam will be removal from the beam by 1 g/cm2 thick Iron shield is, Pr=0.0454

Explanation of Solution

Given:

Energy, E=2MeV

Thickness density of shield, td=1g/cm2

Formula used:

The intensity of gamma beam through the absorber curve is,

  I=I0eμx

Where,

  I = The intensity of gamma beam through the absorber curve

  I0 =The intensity of gamma beam through zero thickness

  μ =Linear attenuation

  x =Absorber thickness

The thickness of the shield is,

  x=tdρ

where,

  x =Thickness

  td =Thickness density

  ρ =Density

Calculation:

The thickness of the Iron shield is,

  x=tdρx=17.87x=0.1270cm

The linear attenuation of Lead having energy 2 MeV is, μ=0.366cm1

The probability of transmission is given by:

  P=II0

So, the probability of removal is given by,

  Pr=1II0=1eμxPr=1e0.366×0.1270Pr=10.9545Pr=0.0454

Conclusion:

The probability of 2 MeV photon in a narrow, collimated beam will be removal from the beam by 1 g/cm2 thick lead shield is, Pr=0.0454

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