Chemistry Smartwork Access Code Fourth Edition
Chemistry Smartwork Access Code Fourth Edition
4th Edition
ISBN: 9780393521368
Author: Gilbert
Publisher: NORTON
Question
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Chapter 21, Problem 21.83QP

(a)

Interpretation Introduction

Interpretation: The energy released per nucleus of Tritium in the given reactions is to be calculated.

Concept introduction: The process of formation of a heavy nucleus from two lighter nuclei is known as nuclear fusion.

To determine: The energy released per nucleus of Tritium in this reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 21.83QP

Solution

The energy released per nucleus of Tritium in this reaction is -7.68×10-13J_ .

Explanation of Solution

Explanation

The given reaction is,

01n+36Li24He+13H

The mass of 1n is 1.00866amu .

The mass of 4He is 4.00260amu .

The mass of 3H is 3.0160492amu .

The mass of 6Li is 6.015122794amu .

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+3.0160492amu)(1.00866amu+6.015122794amu)=7.0186492amu7.02378794amu=0.00513874amu

The conversion of amu into kg is done as,

1amu=1.66054×1027kg

Hence, the conversion of 0.00513874amu into kg is done as,

0.00513874amu=0.00513874×1.66054×1027kg=0.0085330833×1027kg

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

ΔE=Δmc2=0.0085330833×1027kg×(3×108m/s)2=7.68×1013kgm2s2

The conversion of kgm2s2 to J is done as,

1kgm2s2=1J

Hence, the conversion of 7.68×1013kgm2s2 to J is done as,

7.68×1013kgm2s2=7.68×1013×1J=-7.68×10-13J_

Therefore, energy released per nucleus of Tritium in this reaction is -7.68×10-13J_ .

(b)

Interpretation Introduction

To determine: The energy released per nucleus of Tritium in this reaction.

(b)

Expert Solution
Check Mark

Answer to Problem 21.83QP

Solution

The energy released per nucleus of Tritium in this reaction is 3.95××10-13J_ .

Explanation of Solution

Explanation

The given reaction is,

01n+37Li24He+13H+01n

The mass of 1n is 1.00866amu .

The mass of 4He is 4.00260amu .

The mass of 3H is 3.0160492amu .

The mass of 7Li is 7.016004548amu .

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+3.0160492amu+1.00866amu)(1.00866amu+7.016004548amu)=8.0273092amu8.024664548amu=0.002664652amu

The conversion of amu into kg is done as,

1amu=1.66054×1027kg

Hence, the conversion of 0.002664652amu into kg is done as,

0.002664652amu=0.002664652×1.66054×1027kg=0.0043915504×1027kg

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

ΔE=Δmc2=0.0043915504×1027kg×(3×108m/s)2=3.95×1013kgm2s2

The conversion of kgm2s2 to J is done as,

1kgm2s2=1J

Hence, the conversion of 3.95×1013kgm2s2 to J is done as,

3.95×1013kgm2s2=3.95×1013×1013×1J=3.95××10-13J_

Therefore, energy released per nucleus of Tritium in this reaction is 3.95××10-13J_ .

Conclusion

  1. a. The energy released per nucleus of Tritium in this reaction is -7.68×10-13J_ .
  2. b. The energy released per nucleus of Tritium in this reaction is 3.95××10-13J_

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Chapter 21 Solutions

Chemistry Smartwork Access Code Fourth Edition

Ch. 21 - Prob. 21.5VPCh. 21 - Prob. 21.6VPCh. 21 - Prob. 21.7VPCh. 21 - Prob. 21.8VPCh. 21 - Prob. 21.9VPCh. 21 - Prob. 21.10VPCh. 21 - Prob. 21.11QPCh. 21 - Prob. 21.12QPCh. 21 - Prob. 21.13QPCh. 21 - Prob. 21.14QPCh. 21 - Prob. 21.15QPCh. 21 - Prob. 21.16QPCh. 21 - Prob. 21.17QPCh. 21 - Prob. 21.18QPCh. 21 - Prob. 21.19QPCh. 21 - Prob. 21.20QPCh. 21 - Prob. 21.21QPCh. 21 - Prob. 21.22QPCh. 21 - Prob. 21.23QPCh. 21 - Prob. 21.24QPCh. 21 - Prob. 21.25QPCh. 21 - Prob. 21.26QPCh. 21 - Prob. 21.27QPCh. 21 - Prob. 21.28QPCh. 21 - Prob. 21.29QPCh. 21 - Prob. 21.30QPCh. 21 - Prob. 21.31QPCh. 21 - Prob. 21.32QPCh. 21 - Prob. 21.33QPCh. 21 - Prob. 21.34QPCh. 21 - Prob. 21.35QPCh. 21 - Prob. 21.36QPCh. 21 - Prob. 21.37QPCh. 21 - Prob. 21.38QPCh. 21 - Prob. 21.39QPCh. 21 - Prob. 21.40QPCh. 21 - Prob. 21.41QPCh. 21 - Prob. 21.42QPCh. 21 - Prob. 21.43QPCh. 21 - Prob. 21.44QPCh. 21 - Prob. 21.45QPCh. 21 - Prob. 21.46QPCh. 21 - Prob. 21.47QPCh. 21 - Prob. 21.48QPCh. 21 - Prob. 21.49QPCh. 21 - Prob. 21.50QPCh. 21 - Prob. 21.51QPCh. 21 - Prob. 21.52QPCh. 21 - Prob. 21.53QPCh. 21 - Prob. 21.54QPCh. 21 - Prob. 21.55QPCh. 21 - Prob. 21.56QPCh. 21 - Prob. 21.57QPCh. 21 - Prob. 21.58QPCh. 21 - Prob. 21.59QPCh. 21 - Prob. 21.60QPCh. 21 - Prob. 21.61QPCh. 21 - Prob. 21.62QPCh. 21 - Prob. 21.63QPCh. 21 - Prob. 21.64QPCh. 21 - Prob. 21.65QPCh. 21 - Prob. 21.66QPCh. 21 - Prob. 21.67QPCh. 21 - Prob. 21.68QPCh. 21 - Prob. 21.69QPCh. 21 - Prob. 21.70QPCh. 21 - Prob. 21.71QPCh. 21 - Prob. 21.72QPCh. 21 - Prob. 21.73QPCh. 21 - Prob. 21.74QPCh. 21 - Prob. 21.75QPCh. 21 - Prob. 21.76QPCh. 21 - Prob. 21.77QPCh. 21 - Prob. 21.78QPCh. 21 - Prob. 21.79QPCh. 21 - Prob. 21.80QPCh. 21 - Prob. 21.81QPCh. 21 - Prob. 21.82QPCh. 21 - Prob. 21.83QPCh. 21 - Prob. 21.84QPCh. 21 - Prob. 21.85APCh. 21 - Prob. 21.86APCh. 21 - Prob. 21.87APCh. 21 - Prob. 21.88APCh. 21 - Prob. 21.89APCh. 21 - Prob. 21.90APCh. 21 - Prob. 21.91APCh. 21 - Prob. 21.92APCh. 21 - Prob. 21.93APCh. 21 - Prob. 21.94APCh. 21 - Prob. 21.95APCh. 21 - Prob. 21.96APCh. 21 - Prob. 21.97APCh. 21 - Prob. 21.98APCh. 21 - Prob. 21.99APCh. 21 - Prob. 21.100AP
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