Introduction To Health Physics
Introduction To Health Physics
5th Edition
ISBN: 9780071835275
Author: Johnson, Thomas E. (thomas Edward), Cember, Herman.
Publisher: Mcgraw-hill Education,
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Chapter 5, Problem 5.35P

(a)

To determine

The relative proportions of the two groups after passing through 25 cm of water.

(a)

Expert Solution
Check Mark

Answer to Problem 5.35P

The relative proportions of the two groups after passing through 25 cm of water is, 1:1

Explanation of Solution

Given:

Energy, E1=1MeV

Energy, E2=10MeV

Thickness, x=25cm

Cross-section, σH,1MeV=3.4×105b=3.4×1029cm2

Cross-section, σH,10MeV=3.2×105b=3.2×1029cm2

Cross-section, σO,1MeV=1×104b=1×1028cm2

Cross-section, σO,10MeV=1×104b=1×1028cm2

Formula used:

The intensity of gamma beam through the absorber curve is,

  I=I0eNσx

Where,

  I = The intensity of gamma beam through the absorber curve

  I0 =The intensity of gamma beam through zero thickness

  N =Number of atoms

  x =Absorber thickness

x =Thickness

Calculation:

The number of Hydrogen atoms per cm3 is,

  NH=1 g H2O cm3×1 mol H2O18gH2O×1molH1molH2O×6.023× 10 23molHNH=6.692×1022atomsofHydrogenpercm3

The number of Oxygen atoms per cm3 is,

  NO=1 g H2O cm3×1 mol H2O18gH2O×1molO1molH2O×6.023× 10 23molHNO=3.346×1022atomsofOxygenpercm3

For Energy, E1=1MeV

The intensity is given by

  II0=eσ H,1MeVNHxσ O,1MeVNHxII0=e( 3.4× 10 29 ×6.692× 10 22 1.0× 10 28 ×3.346× 10 22 )×25II0=0.9998

For Energy, E1=10MeV

The intensity is given by

  II0=eσ H,1MeVNHxσ O,1MeVNHxII0=e( 3.2× 10 29 ×6.692× 10 22 1.0× 10 28 ×3.346× 10 22 )×25II0=0.9998

The relative proportions of the two groups after passing through 25 cm of water is,

  0.99980.9998=1:1

Conclusion:

The relative proportions of the two groups after passing through 25 cm of water is, 1:1

(b)

To determine

The relative proportions of the two groups after passing through a slab of lead of the same density thickness.

(b)

Expert Solution
Check Mark

Answer to Problem 5.35P

The relative proportions of the two groups after passing through 25 cm of water is 0.7344:1 .

Explanation of Solution

Given:

Energy, E1=1MeV

Energy, E2=10MeV

Density thickness, t=25g/cm2

Cross-section, σPb204,1MeV=0.24b=0.24×1024cm2

Cross-section, σPb204,10MeV=1.79b=1.79×1024cm2

Cross-section, σPb206,1MeV=0.36b=0.36×1024cm2

Cross-section, σPb206,10MeV=1.47b=1.47×1024cm2

Cross-section, σPb207,1MeV=0.55b=0.55×1024cm2

Cross-section, σPb207,10MeV=0.98b=0.98×1024cm2

Cross-section, σPb208,1MeV=2.4×104b=2.4×1028cm2

Cross-section, σPb208,10MeV=1.15b=1.15×1024cm2

Formula used:

The intensity of gamma beam through the absorber curve is,

  I=I0eNσx

Where,

  I = The intensity of gamma beam through the absorber curve

  I0 =The intensity of gamma beam through zero thickness

  N =Number of atoms

  x =Absorber thickness

x =Thickness

The thickness is given by,

  x=Density thicknessDensity

Calculation:

The thickness is given by,

  x=Density thicknessDensityx=2511.34x=2.20cm

For energy 1 MeV

Number of Pb-204 atoms are,

  NPb204=11.34204×6.023×1023NPb204=3.34×1022

So,

  σ204N204=0.24×1024×3.34×1022σ204N204=0.008016

Number of Pb-206 atoms are,

  NPb206=11.34206×6.023×1023NPb206=3.32×1022

So,

  σ206N206=0.36×1024×3.32×1022σ206N206=0.01195

Number of Pb-207 atoms are,

  NPb206=11.34207×6.023×1023NPb206=3.29×1022

So,

  σ207N207=0.55×1024×3.29×1022σ207N207=0.0180

Number of Pb-208 atoms are,

  NPb208=11.34208×6.023×1023NPb208=3.28×1022

So,

  σ208N208=2.4×1028×3.28×1022σ208N208=7.872×106

Thus,

  σN=(σN)204+(σN)206+(σN)207+(σN)208σN=0.008016+0.01195+0.0180+7.872×106σN=0.03797

The intensity is given by:

  II0=eσNxII0=e( 0.03797)×2.20II0=1.087

For energy 10 MeV

Number of Pb-204 atoms are,

  NPb204=11.34204×6.023×1023NPb204=3.34×1022

So,

  σ204N204=1.79×1024×3.34×1022σ204N204=0.05978

Number of Pb-206 atoms are,

  NPb206=11.34206×6.023×1023NPb206=3.32×1022

So,

  σ206N206=1.47×1024×3.32×1022σ206N206=0.0488

Number of Pb-207 atoms are,

  NPb206=11.34207×6.023×1023NPb206=3.29×1022

So,

  σ207N207=0.98×1024×3.29×1022σ207N207=0.0322

Number of Pb-208 atoms are:

  NPb208=11.34208×6.023×1023NPb208=3.28×1022

So,

  σ208N208=1.15×1024×3.28×1022σ208N208=0.03772

Thus,

  σN=(σN)204+(σN)206+(σN)207+(σN)208σN=0.05978+0.0488+0.0322+0.03772σN=0.1785

The intensity is given by

  II0=eσNxII0=e( 0.1785)×2.20II0=1.480

The relative proportions of the two groups after passing through 25 cm of water is,

  1.0871.480=0.7344:1

Conclusion:

The relative proportions of the two groups after passing through 25 cm of water is:

  0.7344:1

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