Explanation Given info: The half-life of C 137 s for β decay is 30.07 y , The energy released per decay is 1.17 MeV , the energy of the emitted electron is 0.51 MeV , the energy of the γ ray is 0.66 MeV , the equivalent dose is 3.5 Sv , the mass of the tissue is 1.0 kg and RBE value for electron is 1.5 . Let Δ N represent the number of cerium atoms that decay in one week. Write the expression for the equivalent dose. equivalent dose = RBE × absorbed dose Here, RBE is the relative biological effectiveness Write the expression for the absorbed dose. absoebed dose = Δ N × E m (1) Here, E is the energy delivered from the radiation m is the mass of the tissue Write the expression for the equivalent dose in terms of Δ N , m and the RBE values of both electron and γ ray. equivalent dose = Δ N m ( ( RBE ) e E e + ( RBE ) γ E γ ) Here, E γ is the energy delivered by the γ ray E e is the energy delivered by the electron ( RBE ) γ is the RBE value of γ ray ( RBE ) e is the RBE value of the electron Substitute 3.5 Sv for equivalent dose, 1.0 kg for m , 1.5 for ( RBE ) e , 1.0 for ( RBE ) γ , 0.51 MeV for E e and 0.66 MeV for E γ to find Δ N . Δ N = ( 3.5 Sv ) ( 1.0 kg ) ( 1.5 × 0.51 MeV + 1.0 × 0.66 MeV ) = ( 3.5 Sv ) ( 1.0 kg ) ( 1.425 MeV ) ( 1.602 × 10 − 13 J 1 MeV ) = 1

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Question

Chapter 43, Problem 43.66P

To determine

The number of C137s atoms would need to be present in each kilogram of body tissue to deliver the given dose.

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Chapter 43, Problem 43.65P

Chapter 43, Problem 43.67P

Chapter 43 Solutions

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Ch. 43.1 - Prob. 43.1TYU Ch. 43.2 - Rank the following nuclei in order from largest to...Ch. 43.3 - Prob. 43.3TYU Ch. 43.4 - Prob. 43.4TYU Ch. 43.5 - Prob. 43.5TYU Ch. 43.6 - Prob. 43.6TYU Ch. 43.7 - Prob. 43.7TYU Ch. 43.8 - Prob. 43.8TYU Ch. 43 - Prob. 43.1DQ Ch. 43 - Prob. 43.2DQ

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