Explanation Given Info: The incomplete fission reaction is, 0 1 n + 92 235 U → 38 88 Sr + 54 136 Xe + ( ? ) neutrons Formula to calculate the number of neutrons in 92 235 U is, 92 235 U = A − Z (1) Here, A is the mass number. Z is an atomic number. Substitute 235 for A and 92 for Z to find the number of neutrons in 92 235 U . 92 235 U = 235 − 92 = 143 Thus, the number of neutrons in 92 235 U is 143 . Therefore, the number of neutrons in the system before the reaction takes place is 143 + 1 = 144 . Similarly, formula to calculate the number of neutrons in 38 88 Sr is, 38 88 Sr = A − Z Substitute 88 for A and 38 for Z to find the number of neutrons in 38 88 Sr . 38 88 Sr = 88 − 38 = 50 Thus, the number of neutrons in 38 88 Sr is 50 . Similarly, formula to calculate the number of neutrons in 54 136 Xe is, 54 136 Xe = A − Z Substitute 136 for A and 54 for Z to find the number of neutrons in 54 136 Xe . 54 136 Xe = 136 − 54 = 82 Thus, the number of neutrons in 54 136 Xe is 82 . Therefore, the number of neutrons in the system after the reaction is 82 + 50 + x = 132 + x . Here, x is the number of neutrons released in the given fission reaction. Therefore, the number of neutrons released in the given fission reaction is, 144 = 132 + x x = 144 − 132 x = 12 Thus, the number of neutrons released in the given fission reaction is 12 . Hence, the complete fission reaction is, 0 1 n + 92 235 U → 38 88 Sr + 54 136 Xe + 12 0 1 n Formula to calculate the amount of energy released is, E = | Δ m | c 2 (3) Here, Δ m is the difference in mass. c is the speed of light . Formula to calculate the difference in mass ( Δ m ) is, Δ m = m f − m i Here, m f is the final mass of the products. m i is the initial mass of the reactants. Formula to calculate the final mass m f is, m f = m ( 38 88 Sr ) + m ( 54 136 Xe ) + 5 × m ( n ) Atomic mass of 38 88 Sr is 87

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ISBN: 9780321976444

Author: James S. Walker

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Chapter 32, Problem 52PCE

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The complete fission reaction and the amount of energy it releases.

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Chapter 32, Problem 51PCE

Chapter 32, Problem 53PCE

Chapter 32 Solutions

Physics (5th Edition)

Ch. 32.1 - Prob. 1EYU Ch. 32.2 - A given nucleus can decay by alpha decay, beta...Ch. 32.3 - Prob. 3EYU Ch. 32.4 - Prob. 4EYU Ch. 32.5 - Prob. 5EYU Ch. 32.6 - Prob. 6EYU Ch. 32.7 - Prob. 7EYU Ch. 32.8 - Prob. 8EYU Ch. 32.9 - Prob. 9EYU Ch. 32 - Prob. 1CQ

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