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 10, Problem 10.1P

(a)

To determine

The minimum thickness of the paraffin shield.

(a)

Expert Solution
Check Mark

Answer to Problem 10.1P

The minimum thickness of the paraffin shield is 44 cm.

Explanation of Solution

Given:

Source emits, S=107neutrons

The fast neutron flux, ϕ=10neutronscm2.s

Attenuation coefficient, μ=0.126cm-1

Build up factor for hydrogenous shields, B = 5

Formula used:

  ϕ=BS4πt2eμt

Where,

t = thickness of paraffin shield

Calculation:

The shield's surface

  ϕ=BS4πt2eμt

10ncm2sec=5×107neutronsec4πt2e0.126t

Using the method of iteration,

First, we assume a small value of t, and then calculating the resulting value ϕ , and then we repeated the process until the value for t is found to be nearest to the desired value of ϕ .

Conclusion:

Att = 44 cm results in ϕ=8neutroncm2.sec which is close to the desired maximum of ϕ=10neutroncm2.sec .

(b)

To determine

Thermal neutron leakage atthe surface of the field.

(b)

Expert Solution
Check Mark

Answer to Problem 10.1P

The thermal neutron leakage is 0.08ncm2sec

Explanation of Solution

Given:

Point source emit, S = 107 n/sec

Radius, R = 44 cm

Diffusion coefficient, D = 0.381 cm

thermal diffusion length, L = 3cm

Formula used:

The thermal neutron leakage

  ϕ=SeR/L4πRD

Calculation:

The thermal neutron leakage:

  ϕ=SeR/L4πRD

ϕth=107nsec4×π×44cm×0.381cme44cm3cm

  ϕth=0.08ncm2sec

Conclusion:

The thermal neutron leakage is 0.08ncm2sec

(c)

To determine

The gamma ray dose rate, due to the hydrogen capture gammas, atthe surface of the shield.

(c)

Expert Solution
Check Mark

Answer to Problem 10.1P

  1.4mradshr

Explanation of Solution

Given:

Point source emit, S = 107 n/sec

Radius, R = 44 cm

The flux of fast neutron, ϕfast=8ncm2sec

The flux of thermal neutron, ϕthermal=0.08ncm2sec

Formula used:

The concentration of gamma emitter

C=S4πR2(ϕ fast+ϕ temal)43πR3

  ϕfast = flux of fast neutron

  ϕthermal = flux of thermal neutron

The source strength

  Γ=3.69×109fiEi(C/kg)m2MBqh×34GyC/kg

Where,

fi = fraction of the transformations that yield a photon of the ith energy, and

Ei = energy of the ith photon, MeV

  μ1=μm×ρ

Where,

  μl=linearenergyabsoptioncoefficientofparaffin.μm=linearenergyabsoptioncoefficientofwater.

The dose rate :

  D˙=12CΓ4πμ(1eμR)

Calculation:

Concentration of gamma emitter

C=S4πr2(ϕ fast+ϕ temal)43πr3

  C=107ns4π(44cm)2(8+0.08)n cm2s43π(44)3=27.5γ/seccm3=27.5Bqcm3

  C=27.5×106MBqcm3

The source strength

  Γ=3.69×109fiEi(C/kg)m2MBqh×34GyC/kg

  Γ=3.69×109×2.26MeV(C/kg)m2MBqh×104cm2m2×34GyC/kg

  Γ=2.8×103Gycm2MBqh=2.8mGycm2MBqh

  μm (2.26MeV, H20) = 0.0252 cm2/g. Since H2CH20 for gamma energy absorption, we

Linear energy absorption coefficient for paraffin

  μ1=μm×ρμ1=0.252cm2g×0.89gcm3=0.022cm1

Inserting these values into the dose rate equation, we have

  D˙=12CΓ4πμ(1eμr)

  D˙=12×27.5×106MBqcm3×2.8mGycm2hrMBq×4×π0.022cm1(1e0.022×(44))

  D˙=0.014mGyhr=1.4mradshr

Conclusion:

Gamma ray dose rate is 1.4mradshr

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