COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
Question
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Chapter 27, Problem 58QAP
To determine

(a)

The energy released in fusion reaction, H2+H3H4e+nx

Expert Solution
Check Mark

Explanation of Solution

To calculate the no of neutrons, release from the above equation. The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,

  2+3=4+xx=1

So, 1 neutron will be released from the above equation

Hence, the complete reaction would be

  H2+H3H4e+n1

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of H2=2.01355321270u

Atomic mass of H3=3.01604928199u

Atomic mass of H4e=4.002602u

  H2+H3H4e+n1

Mass defect

  Δm=(2.01355321270u+3.01604928199u)(4.002602u+1.008664917u)=0.018335577u

Converting this into energy by below equation

  E=mc2

  E=0.018335577u×931.494061MeVu=17.07948MeV

Conclusion:

The energy released in fusion reaction is 17.07948MeV.

To determine

(b)

The energy released in fusion reaction, H4+H4eB7e+nx

Expert Solution
Check Mark

Explanation of Solution

To calculate the no of neutrons, release from the above equation. The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,

  4+4=7+xx=1

So, 1 neutron will be released from the above equation

Hence, the complete reaction would be

  H4+H4eB7e+n1

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of B7e=7.0169u

Atomic mass of H4=4.03188u

Atomic mass of H4e=4.002602u

  H4+H4eB7e+n1

Mass defect

  Δm=(4.03188u+4.002602u)(7.0169u+1.008664917u)=0.008917083u

Converting this into energy by below equation

  E=mc2

  E=0.008917083u×931.494061MeVu=8.306209MeV

Conclusion:

The energy released in fusion reaction is 8.306209MeV.

To determine

(c)

The energy released in fusion reaction, H2+H2H3e+nx

Expert Solution
Check Mark

Explanation of Solution

To calculate the no of neutrons, release from the above equation. The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,

  2+2=3+xx=1

So, 1 neutron will be released from the above equation

Hence, the complete reaction would be

  H2+H2H3e+n1

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of H2=2.01355321270u

Atomic mass of H3e=3.0160293u

  H2+H2H3e+n1

Mass defect

  Δm=(2.01355321270u+2.01355321270u)(3.0160293u+1.008664917u)=0.0024122084u

Converting this into energy by below equation

  E=mc2

  E=0.0024122084u×931.494061MeVu=2.246958MeV

Conclusion:

The energy released in fusion reaction is 2.246958MeV.

To determine

(d)

The energy released in fusion reaction, H12+H11γ+Xyx

Expert Solution
Check Mark

Explanation of Solution

To calculate the no of neutrons, release from the above equation. The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,

  2+1=0+xx=3

And

  1+1=0+yx=2

So, the atomic number is 2.

Hence, the complete reaction would be

  H12+H11γ+H23e

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of H2=2.01355321270u

Atomic mass of H3e=3.0160293u

  H12+H11γ+H23e

Mass defect

  Δm=(2.01355321270u+1.007825u)(3.0160293u)=0.0053489127u

Converting this into energy by below equation

  E=mc2

  E=0.0053489127u×931.494061MeVu=4.9825MeV

Conclusion:

The energy released in fusion reaction is 4.9825MeV

To determine

(e)

The energy released in fusion reaction, H2+H2H3+nx

Expert Solution
Check Mark

Explanation of Solution

To calculate the no of neutrons, release from the above equation. The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,

  2+2=3+xx=1

So, 1 neutron will be released from the above equation

Hence, the complete reaction would be

  H2+H2H3+n1

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of H2=2.01355321270u

Atomic mass of H3e=3.0160293u

  H2+H2H3+nx

Mass defect

  Δm=(2.01355321270u+2.01355321270u)(3.0160293u+1.008664917u)=0.0024122084u

Converting this into energy by below equation

  E=mc2

  E=0.0024122084u×931.494061MeVu=2.246958MeV

Conclusion:

The energy released in fusion reaction is 2.246958MeV

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Chapter 27 Solutions

COLLEGE PHYSICS

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