Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 19, Problem 19.87AP
Interpretation Introduction

Interpretation: The energy released during the fission process of Uranium 235 and fusion reaction involving the fusion of four nuclei of Hydrogen to Helium is given. By finding the energy released in J/nucleon the comparison between the energy released during the fission process of Uranium 235 and during the given fusion reaction is to be stated.

Concept introduction: The combining of two or more lighter nuclei to form heavy nuclei is known as nuclear fusion reaction while the splitting of heavy nuclei into smaller nuclei is known as nuclear fission.

To determine: The comparison between the energy released during the fission process of Uranium 235 and during the given fusion reaction by finding the energy released in J/nucleon .

Expert Solution & Answer
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Answer to Problem 19.87AP

Solution

The energy released per nucleon in fusion process is -9.01×10-13J_ and the value of energy released per nucleon during fission is -1.36×10-13J/nucleon_ .

The energy released is more during the fusion process rather than the fission process.

Explanation of Solution

Explanation

The given reaction is,

411H24He+210β

The mass of 4He is 4.00260amu .

The mass of 1H is 1.00794amu .

The mass of 10β is 0.000548756amu .

The formula for the amount of energy released during fusion is given as,

ΔE=Δmc2 (1)

Where,

  • Δm is the change in the mass.
  • c is the speed of the light (3.0×108m/s) .

The change in the mass is calculated as,

Δm=Mass of productsMass of reactants

Substitute the mass of reactants and products in the above equation as,

Δm=Mass of productsMass of reactants=(4.00260amu+2×0.000548756amu)(4×1.00794amu)=4.003697512amu4.03176amu=0.028062488amu

The conversion of amu into kg is done as,

1amu=1.66054×1027kg

Hence, the conversion of 0.028062488amu into kg is done as,

0.028062488amu=0.028062488×1.66054×1027kg=0.0450515182×1027kg

Substitute the value of Δm and c in the equation (1),

ΔE=Δmc2=0.0450515182×1027kg×(3×108m/s)2=9.01×1013kgm2s2

The conversion of kgm2s2 to J is done as,

1kgm2s2=1J

Hence, the conversion of 9.01×1013kgm2s2 to J is done as,

9.01×1013kgm2s2=9.01×1013×1J=-9.01×10-13J_

Therefore, energy released per nucleon in fusion process is -9.01×10-13J_ .

The number of nucleons present in Uranium 235 are 235 . Therefore, on dividing the energy released during the fission of Uranium 235 , the value of energy released per nucleon is =3.2×1011235=1.36×1013J/nucleon

As the energy is released, therefore the obtained energy is assigned negative sign. Therefore, the value of energy released per nucleon during fission is -1.36×10-13J/nucleon_ .

The energy released is more during the fusion process rather than the fission process.

Conclusion

The energy released per nucleon in fusion process is -9.01×10-13J_ and the value of energy released per nucleon during fission is -1.36×10-13J/nucleon_ .

The energy released is more during the fusion process rather than the fission process

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Chapter 19 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 19 - Prob. 19.5VPCh. 19 - Prob. 19.6VPCh. 19 - Prob. 19.7VPCh. 19 - Prob. 19.8VPCh. 19 - Prob. 19.9VPCh. 19 - Prob. 19.10VPCh. 19 - Prob. 19.11QPCh. 19 - Prob. 19.12QPCh. 19 - Prob. 19.13QPCh. 19 - Prob. 19.14QPCh. 19 - Prob. 19.15QPCh. 19 - Prob. 19.16QPCh. 19 - Prob. 19.17QPCh. 19 - Prob. 19.18QPCh. 19 - Prob. 19.19QPCh. 19 - Prob. 19.20QPCh. 19 - Prob. 19.21QPCh. 19 - Prob. 19.22QPCh. 19 - Prob. 19.23QPCh. 19 - Prob. 19.24QPCh. 19 - Prob. 19.25QPCh. 19 - Prob. 19.26QPCh. 19 - Prob. 19.27QPCh. 19 - Prob. 19.28QPCh. 19 - Prob. 19.29QPCh. 19 - Prob. 19.30QPCh. 19 - Prob. 19.31QPCh. 19 - Prob. 19.32QPCh. 19 - Prob. 19.33QPCh. 19 - Prob. 19.34QPCh. 19 - Prob. 19.35QPCh. 19 - Prob. 19.36QPCh. 19 - Prob. 19.37QPCh. 19 - Prob. 19.38QPCh. 19 - Prob. 19.39QPCh. 19 - Prob. 19.40QPCh. 19 - Prob. 19.41QPCh. 19 - Prob. 19.42QPCh. 19 - Prob. 19.43QPCh. 19 - Prob. 19.44QPCh. 19 - Prob. 19.45QPCh. 19 - Prob. 19.46QPCh. 19 - Prob. 19.47QPCh. 19 - Prob. 19.48QPCh. 19 - Prob. 19.49QPCh. 19 - Prob. 19.50QPCh. 19 - Prob. 19.51QPCh. 19 - Prob. 19.52QPCh. 19 - Prob. 19.53QPCh. 19 - Prob. 19.54QPCh. 19 - Prob. 19.55QPCh. 19 - Prob. 19.56QPCh. 19 - Prob. 19.57QPCh. 19 - Prob. 19.58QPCh. 19 - Prob. 19.59QPCh. 19 - Prob. 19.60QPCh. 19 - Prob. 19.61QPCh. 19 - Prob. 19.62QPCh. 19 - Prob. 19.63QPCh. 19 - Prob. 19.64QPCh. 19 - Prob. 19.65QPCh. 19 - Prob. 19.66QPCh. 19 - Prob. 19.67QPCh. 19 - Prob. 19.68QPCh. 19 - Prob. 19.69QPCh. 19 - Prob. 19.70QPCh. 19 - Prob. 19.71QPCh. 19 - Prob. 19.72QPCh. 19 - Prob. 19.73QPCh. 19 - Prob. 19.74QPCh. 19 - Prob. 19.75QPCh. 19 - Prob. 19.76QPCh. 19 - Prob. 19.77QPCh. 19 - Prob. 19.78QPCh. 19 - Prob. 19.79QPCh. 19 - Prob. 19.80QPCh. 19 - Prob. 19.81QPCh. 19 - Prob. 19.82QPCh. 19 - Prob. 19.83QPCh. 19 - Prob. 19.84QPCh. 19 - Prob. 19.85APCh. 19 - Prob. 19.86APCh. 19 - Prob. 19.87APCh. 19 - Prob. 19.88APCh. 19 - Prob. 19.89APCh. 19 - Prob. 19.90APCh. 19 - Prob. 19.91APCh. 19 - Prob. 19.92APCh. 19 - Prob. 19.93APCh. 19 - Prob. 19.94APCh. 19 - Prob. 19.95APCh. 19 - Prob. 19.96APCh. 19 - Prob. 19.97APCh. 19 - Prob. 19.98APCh. 19 - Prob. 19.99APCh. 19 - Prob. 19.100AP
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