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Science Chemistry Chemistry: An Atoms-Focused Approach

To compare: The energy released during fusion and fission processes. Solution: The energy generated by fusion of Hydrogen is 9.9 × 10 - 13 J / n u c l e o n , and the energy generated by fission of Uranium-235 is 1.3617 × 10 - 13 J / n u c l e o n . Therefore, the energy generated by fusion is more as compared to the energy generated by fission ( 7.3 times.) Explanation: Concept: The energy is calculated using Einstein’s relation E = m c 2 where, E is the energy, m is the mass of the matter in k g , and, c is the velocity of light. In nuclear reactions, the energy released during the fission or fusion process is determined by the total mass change of the reactants and the products. The energy due to the fission of Uranium is given to us and we need to determine the energy due to the fusion of four Hydrogen atoms and compare them. Formulae: T o t a l m a s s c h a n g e = ∆ m = m p r o d u c t s - m r e a c t a n t s E = ∆ m c 2 Given: Energy released during fission of Uranium E f i s s i o n = 3.2 × 10 - 11 J / a t o m Mass of Hydrogen = 1.67262 × 10 - 27 k g Mass of Helium = 6.64465 × 10 - 27 k g Mass of positron = 9.10938 × 10 - 31 k g Velocity of light c = 2.998 × 10 8 m / s 2 Calculations: 4 H 1 1 → H e 2 4 + 2 β 1 0 Change in the mass of the reaction: ∆ m = m H e l i u m + 2 × m p o s i t r o n - 4 × m H y d r o g e n = 6.64465 × 10 - 27 k g + 2 × 9.10938 × 10 - 31 k g - 4 × 1.67262 × 10 - 27 k g = 6.64465 × 10 - 27 k g + 1.821876 × 10 - 30 k g - 6.69048 × 10 - 27 k g = 6.64647 × 10 - 27 k g - 6.69048 × 10 - 27 k g = - 4.401 × 10 - 29 k g The energy corresponding to this loss in mass is: E = ∆ m c 2 = - 4.401 × 10 - 29 k g × 2.998 × 10 8 m / s 2 2 = - 4.401 × 10 - 29 k g × 8.988004 × 10 16 m 2 s 4 = - 3.9556 × 10 - 12 J This is the energy per atom. Helium contains four neutrons in one atom. Thus, the energy generated by fusion in J / n u c l e o n is: E f u s i o n = E n e r g y i n J / a t o m 4 n u c l e o n s / a t o m = 3.9556 × 10 - 12 J / a t o m 4 n u c l e o n s / a t o m = 9.889 × 10 - 13 ~ 9.90 × 10 - 13 J / n u c l e o n For Uranium-235, the number of neutrons in one atom of Uranium is 235 . Thus, the energy generated by fission in J / n u c l e o n is: E f i s s i o n = E n e r g y i n J / a t o m 235 n u c l e o n s / a t o m = 3.2 × 10 - 11 J / a t o m 4 n u c l e o n s / a t o m = 1.3617 × 10 - 13 ~ 1.36 × 10 - 13 J / n u c l e o n The ratio of the energy generated by fusion of four hydrogen nuclie and that by fission of Uranium-235is: E f u s i o n E f i s s i o n = 9.889 × 10 - 13 J / n u c l e o n 1.3617 × 10 - 13 J / n u c l e o n = 7.2622 ~ 7.3 Therefore, the energy generated by fusion of four hydrogen atoms is 7.3 times greater than the energy generated by fission of Uranium-235. Conclusion: The energy generated by either fusion or fission is calculated by Einstein’s relation that says the mass and energy are related to each other.

To compare: The energy released during fusion and fission processes. Solution: The energy generated by fusion of Hydrogen is 9.9 × 10 - 13 J / n u c l e o n , and the energy generated by fission of Uranium-235 is 1.3617 × 10 - 13 J / n u c l e o n . Therefore, the energy generated by fusion is more as compared to the energy generated by fission ( 7.3 times.) Explanation: Concept: The energy is calculated using Einstein’s relation E = m c 2 where, E is the energy, m is the mass of the matter in k g , and, c is the velocity of light. In nuclear reactions, the energy released during the fission or fusion process is determined by the total mass change of the reactants and the products. The energy due to the fission of Uranium is given to us and we need to determine the energy due to the fusion of four Hydrogen atoms and compare them. Formulae: T o t a l m a s s c h a n g e = ∆ m = m p r o d u c t s - m r e a c t a n t s E = ∆ m c 2 Given: Energy released during fission of Uranium E f i s s i o n = 3.2 × 10 - 11 J / a t o m Mass of Hydrogen = 1.67262 × 10 - 27 k g Mass of Helium = 6.64465 × 10 - 27 k g Mass of positron = 9.10938 × 10 - 31 k g Velocity of light c = 2.998 × 10 8 m / s 2 Calculations: 4 H 1 1 → H e 2 4 + 2 β 1 0 Change in the mass of the reaction: ∆ m = m H e l i u m + 2 × m p o s i t r o n - 4 × m H y d r o g e n = 6.64465 × 10 - 27 k g + 2 × 9.10938 × 10 - 31 k g - 4 × 1.67262 × 10 - 27 k g = 6.64465 × 10 - 27 k g + 1.821876 × 10 - 30 k g - 6.69048 × 10 - 27 k g = 6.64647 × 10 - 27 k g - 6.69048 × 10 - 27 k g = - 4.401 × 10 - 29 k g The energy corresponding to this loss in mass is: E = ∆ m c 2 = - 4.401 × 10 - 29 k g × 2.998 × 10 8 m / s 2 2 = - 4.401 × 10 - 29 k g × 8.988004 × 10 16 m 2 s 4 = - 3.9556 × 10 - 12 J This is the energy per atom. Helium contains four neutrons in one atom. Thus, the energy generated by fusion in J / n u c l e o n is: E f u s i o n = E n e r g y i n J / a t o m 4 n u c l e o n s / a t o m = 3.9556 × 10 - 12 J / a t o m 4 n u c l e o n s / a t o m = 9.889 × 10 - 13 ~ 9.90 × 10 - 13 J / n u c l e o n For Uranium-235, the number of neutrons in one atom of Uranium is 235 . Thus, the energy generated by fission in J / n u c l e o n is: E f i s s i o n = E n e r g y i n J / a t o m 235 n u c l e o n s / a t o m = 3.2 × 10 - 11 J / a t o m 4 n u c l e o n s / a t o m = 1.3617 × 10 - 13 ~ 1.36 × 10 - 13 J / n u c l e o n The ratio of the energy generated by fusion of four hydrogen nuclie and that by fission of Uranium-235is: E f u s i o n E f i s s i o n = 9.889 × 10 - 13 J / n u c l e o n 1.3617 × 10 - 13 J / n u c l e o n = 7.2622 ~ 7.3 Therefore, the energy generated by fusion of four hydrogen atoms is 7.3 times greater than the energy generated by fission of Uranium-235. Conclusion: The energy generated by either fusion or fission is calculated by Einstein’s relation that says the mass and energy are related to each other.

Solution Summary: The author compares the energy generated during fusion and fission processes. The energy is calculated using Einstein's relation E=mc2.

Chemistry: An Atoms-Focused Approach

Chemistry: An Atoms-Focused Approach

14th Edition

ISBN: 9780393912340

Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster

Publisher: W. W. Norton & Company

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Writing and Balancing Nuclear Equations

A nuclear equation is a symbolic representation of a nuclear reaction using certain symbols. It is studied in nuclear chemistry and nuclear physics.

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Chapter 21, Problem 21.84QA

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Chemistry: An Atoms-Focused Approach

Ch. 21 - Prob. 21.1VP Ch. 21 - Prob. 21.2VP Ch. 21 - Prob. 21.4VP Ch. 21 - Prob. 21.5VP Ch. 21 - Prob. 21.6VP Ch. 21 - Prob. 21.7VP Ch. 21 - Prob. 21.8VP Ch. 21 - Prob. 21.9VP Ch. 21 - Prob. 21.10VP Ch. 21 - Prob. 21.11VP

Ch. 21 - Prob. 21.12VP Ch. 21 - Prob. 21.13QA Ch. 21 - Prob. 21.14QA Ch. 21 - Prob. 21.15QA Ch. 21 - Prob. 21.16QA Ch. 21 - Prob. 21.17QA Ch. 21 - Prob. 21.18QA Ch. 21 - Prob. 21.19QA Ch. 21 - Prob. 21.20QA Ch. 21 - Prob. 21.22QA Ch. 21 - Prob. 21.23QA Ch. 21 - Prob. 21.24QA Ch. 21 - Prob. 21.25QA Ch. 21 - Prob. 21.26QA Ch. 21 - Prob. 21.27QA Ch. 21 - Prob. 21.28QA Ch. 21 - Prob. 21.29QA Ch. 21 - Prob. 21.30QA Ch. 21 - Prob. 21.31QA Ch. 21 - Prob. 21.32QA Ch. 21 - Prob. 21.33QA Ch. 21 - Prob. 21.34QA Ch. 21 - Prob. 21.35QA Ch. 21 - Prob. 21.36QA Ch. 21 - Prob. 21.37QA Ch. 21 - Prob. 21.38QA Ch. 21 - Prob. 21.39QA Ch. 21 - Prob. 21.40QA Ch. 21 - Prob. 21.41QA Ch. 21 - Prob. 21.42QA Ch. 21 - Prob. 21.43QA Ch. 21 - Prob. 21.44QA Ch. 21 - Prob. 21.45QA Ch. 21 - Prob. 21.46QA Ch. 21 - Prob. 21.47QA Ch. 21 - Prob. 21.48QA Ch. 21 - Prob. 21.49QA Ch. 21 - Prob. 21.50QA Ch. 21 - Prob. 21.51QA Ch. 21 - Prob. 21.52QA Ch. 21 - Prob. 21.53QA Ch. 21 - Prob. 21.54QA Ch. 21 - Prob. 21.55QA Ch. 21 - Prob. 21.56QA Ch. 21 - Prob. 21.57QA Ch. 21 - Prob. 21.58QA Ch. 21 - Prob. 21.59QA Ch. 21 - Prob. 21.60QA Ch. 21 - Prob. 21.61QA Ch. 21 - Prob. 21.62QA Ch. 21 - Prob. 21.63QA Ch. 21 - Prob. 21.64QA Ch. 21 - Prob. 21.65QA Ch. 21 - Prob. 21.66QA Ch. 21 - Prob. 21.67QA Ch. 21 - Prob. 21.68QA Ch. 21 - Prob. 21.69QA Ch. 21 - Prob. 21.70QA Ch. 21 - Prob. 21.71QA Ch. 21 - Prob. 21.72QA Ch. 21 - Prob. 21.73QA Ch. 21 - Prob. 21.74QA Ch. 21 - Prob. 21.75QA Ch. 21 - Prob. 21.76QA Ch. 21 - Prob. 21.77QA Ch. 21 - Prob. 21.78QA Ch. 21 - Prob. 21.79QA Ch. 21 - Prob. 21.80QA Ch. 21 - Prob. 21.81QA Ch. 21 - Prob. 21.82QA Ch. 21 - Prob. 21.83QA Ch. 21 - Prob. 21.84QA Ch. 21 - Prob. 21.85QA Ch. 21 - Prob. 21.86QA Ch. 21 - Prob. 21.87QA Ch. 21 - Prob. 21.88QA Ch. 21 - Prob. 21.89QA Ch. 21 - Prob. 21.90QA Ch. 21 - Prob. 21.91QA Ch. 21 - Prob. 21.92QA Ch. 21 - Prob. 21.93QA Ch. 21 - Prob. 21.94QA Ch. 21 - Prob. 21.95QA Ch. 21 - Prob. 21.96QA Ch. 21 - Prob. 21.97QA Ch. 21 - Prob. 21.98QA Ch. 21 - Prob. 21.99QA

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