EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Question
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Chapter 40, Problem 64P

(a)

To determine

The binding energy of a pair of α - particles from the atomic masses of 4He and 16O .

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of one 4He nucleus is 4.002603u .

The mass of one 16O nucleus is 15.994915u .

The number of α - particles in 16O is 4 .

The number of α - particles at each vertex of the tetrahedron structure is 1 .

Formula used:

Write the expression for the mass difference between α - particles and 16O .

  Δm=4mαm16O

Here, Δm is the mass difference between α - particles and 16O , mα is the mass of α - particles and m16O is the mass of one 16O .

Write the expression for the binding energy of the pair of α - particles.

  Eb=(Δm)c2

Here, Eb is the binding energy and c is the speed of light.

Substitute 4mαm16O for Δm in the above expression.

  Eb=(4mαm16O)c2 …… (1)

Write the expression for the total binding energy for the regular tetrahedron structure.

  Ebt=Ebbond=Eb6 …… (2)

Here, Ebt is the total binding energy for the regular tetrahedron structure.

Calculation:

Substitute 4.002603u for mα and 15.994915u in equation (1).

  Eb=(4(4.002603u)(15.994915u))c2=(16.010412u15.994915u)c2=(0.015497u)c2

Substitute (0.015497u)c2 for Eb in equation (2).

  Ebt=(0.015497u)c26=(0.015497u)c26(931.5MeVc2u)=2.406MeV

Conclusion:

Thus, the binding energy of a pair of α - particles from the atomic masses of 4He and 16O is 2.406MeV .

(b)

To determine

The binding energy of 12C from the given model and from the atomic mass of 12C and comparison between the two results.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of one proton is 1.007825u .

The mass of one neutron is 1.008665u .

The mass of one 12C nucleus is 12.000000u .

The mass of one 4He nucleus is 4.002603u .

The mass of one 16O nucleus is 15.994915u .

The number of α - particles in 16O is 4 .

The number of α - particles at each vertex of the tetrahedron structure is 1 .

Formula used:

Write the expression for the mass difference between α - particles and 16O .

  Δm=4mαm16O

Here, Δm is the mass difference between α - particles and 16O , mα is the mass of α - particles and m16O is the mass of one 16O .

Write the expression for the binding energy of the pair of α - particles.

  Eb=(Δm)c2

Here, Eb is the binding energy and c is the speed of light.

Substitute 4mαm16O for Δm in the above expression.

  Eb=(4mαm16O)c2 …… (1)

Write the expression for the total binding energy for the regular tetrahedron structure.

  Ebt=Ebbond=Eb6 …… (2)

Here, Ebt is the total binding energy for the regular tetrahedron structure.

Write the expression for the masses of protons and neutrons in 4He .

  mp+a=2(mp+ma)

Here, mp is the mass of one proton, ma is the mass of one neutron and mp+a is the total mass of both proton and neutron in 4He .

Write the expression for the mass difference between total mass of both proton and neutron in 4He and mass of 4He alone.

  Δm=2(mp+ma)m4He

Here, Δm is the mass difference between total mass of both proton and neutron in 4He and m4He is the mass of 4He .

Write the expression for the binding energy of 4He .

  Eb(4He)=(Δm)c2

Here, Eb(4He) is the binding energy of 4He and c is the speed of light.

Substitute 2(mp+ma)m4He for Δm in the above expression.

  Eb(4He)=(2(mp+ma)m4He)c2 …… (3)

Write the expression for the binding energy of 12C from the given model.

  Eb(12C)=3(Eb(4He))+3(Ebt) …… (4)

Write the expression for the masses of protons and neutrons in 12C .

  mp+a=6(mp+ma)

Here, mp is the mass of one proton, ma is the mass of one neutron and mp+a is the total mass of both proton and neutron in 12C .

Write the expression for the mass difference between total mass of both proton and neutron in 12C and mass of 12C alone.

  Δm=6(mp+ma)m12C

Here, Δm is the mass difference between total mass of both proton and neutron in 4He and m12C

  m4He is the mass of 12C .

Write the expression for the binding energy of 12C from the atomic masses of 12C , proton and neutron.

  Eb(12C)=(Δm)c2

Here, Eb(12C) is the binding energy of 12C and c is the speed of light.

Substitute 6(mp+ma)m12C for Δm in the above expression.

  Eb(12C)=(6(mp+ma)m12C)c2 …… (5)

Calculation:

Substitute 4.002603u for mα and 15.994915u in equation (1).

  Eb=(4(4.002603u)(15.994915u))c2=(16.010412u15.994915u)c2=(0.015497u)c2

Substitute (0.015497u)c2 for Eb in equation (2).

  Ebt=(0.015497u)c26=(0.015497u)c26(931.5MeV/c2u)=2.406MeV

Substitute 1.007825u for mp , 1.008665u for ma and 4.002603u for m4He in equation (3).

  Eb(4He)=(2(1.007825u+1.008665u)4.002603u)c2=(2(1.007825u+1.008665u)4.002603u)c2(931.5MeV/c2u)=28.30MeV

Substitute 28.30MeV for Eb(4He) and 2.406MeV for Ebt in equation (4).

  Eb(12C)=3(28.30MeV)+3(2.406MeV)=92.1MeV

Substitute 1.007825u for mp , 1.008665u for ma and 12.000000u for m12C in equation (5).

  Eb(12C)=(6(1.007825u+1.008665u)12.000000u)c2=(6(1.007825u+1.008665u)12.000000u)c2(931.5MeV/c2u)=92.2MeV

Conclusion:

Thus, the binding energy of 12C from the given model is 92.1MeV and from the atomic mass of 12C is 92.2MeV . Hence the results are good agreement with the given model.

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