General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Question
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Chapter 30, Problem 47E

(a)

To determine

The energy released in the given process.

(a)

Expert Solution
Check Mark

Answer to Problem 47E

The energy released in the process is 24.7MeV.

Explanation of Solution

Write expression for mass defect.

    Δm=4mpmα        (1)

Here, Δm is mass defect, mp is mass of proton and mα is the mass of alpha particle.

Write expression for energy released.

    E=Δmc2        (2)

Here, c is speed of light.

Conclusion:

Substitute 1.00728u for mp and 4.0026u for mα in equation (1).

    Δm=4(1.00728u)4.0026uΔm=0.02652u

Substitute 0.02652u for Δm in equation (2).

    E=(0.02652uc2)(931MeV1uc2)E=24.7MeV

Thus, the energy released in the process is 24.7MeV.

(b)

To determine

The mass of hydrogen burned.

(b)

Expert Solution
Check Mark

Answer to Problem 47E

Themass of hydrogen burned is 1.99×106kg.

Explanation of Solution

Write expression for equivalent energy released by one hydrogen molecule.

    Eeq=Enαnh        (1)

Here, nα is number of alpha particles formed and nh is number of hydrogen atoms.

Write expression for energy relareased by 1kg hydrogen.

    EH=EeqNA2        (2)

Here, NA is avogradro’s number.

Write expression for number of hydrogen atoms.

    N=EHEH2O        (3)

Here, EH2O is energy released when one hydrogen molecule is used to produce water.

Write expression for mass of hydrogen.

    m=2NNA        (4)

Conclusion:

Substitute 24.7MeV for E, 1 for nα and 2 for nh in equation (1).

    Eeq=(24.7MeV)12Eeq=12.35MeV

Substitute 12.35MeV for Eeq and 6.02×1026 for NA in equation (2).

    EH=12.35MeV(6.02×10262)=3.7174×1027MeV

Substitute 3.7174×1027MeV for EH and 6.2eV for EH2O in equation (3).

    N=3.7174×1027MeV6.2eV(106eV1MeV)N=5.9958×1032

Substitute 5.9958×1032 for N and 6.02×1026 for NA in equation (4).

    m=2(5.9958×1032)6.02×1026m=1.99×106kg

Thus, mass of hydrogen burned is 1.99×106kg.

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