Explanation Given: Thickness of shield, t = 15 cm Energy, E 1 = 1 MeV Energy, E 2 = 0.1 MeV Formula Used: The intensity of gamma beam through the absorber curve is, I = I 0 e − μ x Where, I = The intensity of gamma beam through the absorber curve I 0 = The intensity of gamma beam through zero thickness μ =Linear attenuation x =Absorber thickness Calculation: The intensity of gamma beam through the absorber curve is, I = I 0 e − μ x For 0.1 MeV we have, I 1 I 01 = e − μ 1 x ∵ μ 1 = 0.400 I 1 I 01 = 2.47 × 10 − 3 ..

5th Edition

ISBN: 9780071835275

Author: Johnson, Thomas E. (thomas Edward), Cember, Herman.

Publisher: Mcgraw-hill Education,

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Chapter 5, Problem 5.15P

To determine

The relative portion of 1 MeV photons to 0.1 MeV photons in emergent beam.

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Chapter 5, Problem 5.14P

Chapter 5, Problem 5.16P

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The point source Co-60 emits gamma with energies of 1.17 MeV and 1.33 MeV. Each of these energies emits the same number of photon particles. At a certain location, the flux that is read is 5.7 x 109 photons/cm2.s, what is the intensity or energy flux with units of J/m^2.min

The point source Co-60 emits gamma with energies of 1.17 MeV and 1.33 MeV. Each of these energies emits the same number of photon particles. At a certain location, the flux read is 5.7 x 109 photons/cm2.s, what is the flux energy for 1.17 MeV for 24 hours of radiation exposure?

Chapter 5 Solutions

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