PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 29, Problem 64P
To determine

The comparison of the amount of energy released when 1.0 kg of 235U undergoes the fission reaction with the energy released when 1.0 kg of hydrogen undergoes the fusion reaction.

Expert Solution & Answer
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Answer to Problem 64P

The fission reaction of 235U releases 7.1×1013 J/kg whereas the fusion reaction of hydrogen releases 34×1013 J/kg .

Explanation of Solution

Write the equation for the energy released.

  E=|Δm|c2        (I)

Here, E is the energy released, Δm is the change in mass during the reaction and c is the velocity of light.

Write the chemical equation for the fission reaction of 235U nucleus.

  01n+92235U56141Ba+3692Kr+301n

Write the equation for Δm using the chemical equation of the fission reaction of 235U .

  Δm=mn+mU(mBa+mKr+3mn)        (II)

Here, mn is the mass of neutron, mU is the mass of 235U nucleus, mBa is the mass of 56141Ba nucleus and mKr is the mass of 3692Kr nucleus.

Write the chemical equation for the fusion reaction of hydrogen.

  12H+13H24He+01n

Write the equation for Δm using the chemical equation of the fusion reaction of hydrogen.

  Δm=m2H+m3H(mHe+mn)        (III)

Here, m2H is the mass of 12H nucleus, m3H is the mass of 13H nucleus and mHe is the mass of 24He nucleus.

Write the equation for the total energy released from 1.0 kg of 235U .

Conclusion:

The value of mn is 1.0086649 u , that of mU is 235.0439282 u , that of mBa is 140.9144035 u , that of mKr is 91.9261731 u , that of m2H is 2.0141018 u , that of m3H is 3.0160493 u , that of mHe is 4.0026033 u and that of c2 is 931.494 MeV/u .

Substitute 1.0086649 u for mn , 235.0439282 u for mU , 140.9144035 u for mBa and 91.9261731 u for mKr in equation (II) to find Δm of the fission reaction of 235U .

  Δm=1.0086649 u+235.0439282 u(140.9144035 u+91.9261731 u+3(1.0086649 u))=0.1860218 u

Substitute 0.1860218 u for Δm and 931.494 MeV/u for c2 in equation (I) to find E for the fission reaction of 235U.

  Efission=|0.1860218 u|(931.494 MeV/u)=173.278 MeV

Write the equation for the total energy released for 1.0 kg of 235U .

  Efission,tot=Efission1.0 kgmU

Here, Efission,tot is total energy released for 1.0 kg of 235U .

Substitute 173.278 MeV for Efission and 235.0439282 u for mU in the above equation to find Efission,tot .

  Efission,tot=(173.278 MeV106 eV1 MeV)1.0 kg(235.0439282 u1.6605×1027 kg1 u)=4.44×1032 eV1.602×1019 J1 eV=7.1×1013 J

Substitute 2.0141018 u for m2H , 3.0160493 u for m3H , 4.0026033 u for mHe and 1.0086649 u for mn in equation (III) to find Δm of the fusion reaction of hydrogen.

  Δm=2.0141018 u+3.0160493 u(4.0026033 u+1.0086649 u)=0.0188829 u

Substitute 0.0188829 u for Δm and 931.494 MeV/u for c2 in equation (I) to find E for the fusion reaction of hydrogen.

  Efusion=|0.0188829 u|(931.494 MeV/u)=17.5893 MeV

Write the equation for the total energy released from 1.0 kg of hydrogen.

  Efusion,tot=Efusion1.0 kgm2H+m3H

Here, Efusion,tot is total energy released from 1.0 kg of hydrogen.

Substitute 17.5893 MeV for Efusion , 2.0141018 u for m2H , 3.0160493 u for m3H in the above equation to find Efission,tot .

Efusion,tot=(17.5893 MeV106 eV1 MeV)1.0 kg(2.0141018 u1.6605×1027 kg1 u+3.0160493 u1.6605×1027 kg1 u)=2.11×1033 eV1.602×1019 J1 eV=34×1013 J

Therefore, the fission reaction of 235U releases 7.1×1013 J/kg whereas the fusion reaction of hydrogen releases 34×1013 J/kg .

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Chapter 29 Solutions

PHYSICS

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