The lowest energy of a nucleon in the well.

Question

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Answer 1

Question

Chapter 40, Problem 63P

(a)

To determine

The lowest energy of a nucleon in the well.

(a)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Conclusion:

Thus, the lowest energy of a nucleon in the well is 23 MeV .

(b)

To determine

The ground state energy of 12 nucleons in the well, all are neutrons.

(b)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons which are fermions.

E=2(E1+E2+E3+E4+E5+E6)

Write the above expression again for less than two neutrons per state.

E=2(E1+22E1+32E1+42E1+52E1+62E1)=182E1 ........ (2)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (2).

E=182(23 MeV)=4186 MeV( 1 GeV 10 3 MeV)=4.186 GeV≈4.2 GeV

Conclusion:

Thus, the ground state energy of 12 nucleons in the well 4.2 GeV .

(c)

To determine

The ground state energy of 12 nucleons in the well, in which 6 are neutrons and 6 are protons.

(c)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons out of which 6 are protons and 6 are neutrons.

E=4(E1+E2+E3)

Write the above expression again for

4 nucleons per state.

E=4(E1+22E1+32E1)=56E1 ........ (3)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (3).

E=56(23 MeV)=1288 MeV( 1 GeV 10 3 MeV)=1.288 GeV≈1.3 GeV

Conclusion:

Thus, the ground state energy for 4 protons and 4 neutrons in the well is 1.3 GeV .

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Answer 2

Question

Chapter 40, Problem 63P

(a)

To determine

The lowest energy of a nucleon in the well.

(a)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Conclusion:

Thus, the lowest energy of a nucleon in the well is 23 MeV .

(b)

To determine

The ground state energy of 12 nucleons in the well, all are neutrons.

(b)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons which are fermions.

E=2(E1+E2+E3+E4+E5+E6)

Write the above expression again for less than two neutrons per state.

E=2(E1+22E1+32E1+42E1+52E1+62E1)=182E1 ........ (2)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (2).

E=182(23 MeV)=4186 MeV( 1 GeV 10 3 MeV)=4.186 GeV≈4.2 GeV

Conclusion:

Thus, the ground state energy of 12 nucleons in the well 4.2 GeV .

(c)

To determine

The ground state energy of 12 nucleons in the well, in which 6 are neutrons and 6 are protons.

(c)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons out of which 6 are protons and 6 are neutrons.

E=4(E1+E2+E3)

Write the above expression again for

4 nucleons per state.

E=4(E1+22E1+32E1)=56E1 ........ (3)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (3).

E=56(23 MeV)=1288 MeV( 1 GeV 10 3 MeV)=1.288 GeV≈1.3 GeV

Conclusion:

Thus, the ground state energy for 4 protons and 4 neutrons in the well is 1.3 GeV .

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Answer 3

Question

Chapter 40, Problem 63P

(a)

To determine

The lowest energy of a nucleon in the well.

(a)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Conclusion:

Thus, the lowest energy of a nucleon in the well is 23 MeV .

(b)

To determine

The ground state energy of 12 nucleons in the well, all are neutrons.

(b)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons which are fermions.

E=2(E1+E2+E3+E4+E5+E6)

Write the above expression again for less than two neutrons per state.

E=2(E1+22E1+32E1+42E1+52E1+62E1)=182E1 ........ (2)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (2).

E=182(23 MeV)=4186 MeV( 1 GeV 10 3 MeV)=4.186 GeV≈4.2 GeV

Conclusion:

Thus, the ground state energy of 12 nucleons in the well 4.2 GeV .

(c)

To determine

The ground state energy of 12 nucleons in the well, in which 6 are neutrons and 6 are protons.

(c)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons out of which 6 are protons and 6 are neutrons.

E=4(E1+E2+E3)

Write the above expression again for

4 nucleons per state.

E=4(E1+22E1+32E1)=56E1 ........ (3)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (3).

E=56(23 MeV)=1288 MeV( 1 GeV 10 3 MeV)=1.288 GeV≈1.3 GeV

Conclusion:

Thus, the ground state energy for 4 protons and 4 neutrons in the well is 1.3 GeV .

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Answer 4

Question

Chapter 40, Problem 63P

(a)

To determine

The lowest energy of a nucleon in the well.

(a)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Conclusion:

Thus, the lowest energy of a nucleon in the well is 23 MeV .

(b)

To determine

The ground state energy of 12 nucleons in the well, all are neutrons.

(b)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons which are fermions.

E=2(E1+E2+E3+E4+E5+E6)

Write the above expression again for less than two neutrons per state.

E=2(E1+22E1+32E1+42E1+52E1+62E1)=182E1 ........ (2)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (2).

E=182(23 MeV)=4186 MeV( 1 GeV 10 3 MeV)=4.186 GeV≈4.2 GeV

Conclusion:

Thus, the ground state energy of 12 nucleons in the well 4.2 GeV .

(c)

To determine

The ground state energy of 12 nucleons in the well, in which 6 are neutrons and 6 are protons.

(c)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons out of which 6 are protons and 6 are neutrons.

E=4(E1+E2+E3)

Write the above expression again for

4 nucleons per state.

E=4(E1+22E1+32E1)=56E1 ........ (3)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (3).

E=56(23 MeV)=1288 MeV( 1 GeV 10 3 MeV)=1.288 GeV≈1.3 GeV

Conclusion:

Thus, the ground state energy for 4 protons and 4 neutrons in the well is 1.3 GeV .

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Answer 5

Chapter 40, Problem 63P

(a)

To determine

The lowest energy of a nucleon in the well.

(a)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Conclusion:

Thus, the lowest energy of a nucleon in the well is 23 MeV .

(b)

To determine

The ground state energy of 12 nucleons in the well, all are neutrons.

(b)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons which are fermions.

E=2(E1+E2+E3+E4+E5+E6)

Write the above expression again for less than two neutrons per state.

E=2(E1+22E1+32E1+42E1+52E1+62E1)=182E1 ........ (2)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (2).

E=182(23 MeV)=4186 MeV( 1 GeV 10 3 MeV)=4.186 GeV≈4.2 GeV

Conclusion:

Thus, the ground state energy of 12 nucleons in the well 4.2 GeV .

(c)

To determine

The ground state energy of 12 nucleons in the well, in which 6 are neutrons and 6 are protons.

(c)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons out of which 6 are protons and 6 are neutrons.

E=4(E1+E2+E3)

Write the above expression again for

4 nucleons per state.

E=4(E1+22E1+32E1)=56E1 ........ (3)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (3).

E=56(23 MeV)=1288 MeV( 1 GeV 10 3 MeV)=1.288 GeV≈1.3 GeV

Conclusion:

Thus, the ground state energy for 4 protons and 4 neutrons in the well is 1.3 GeV .

Answer 6

Chapter 40, Problem 63P

(a)

To determine

The lowest energy of a nucleon in the well.

(a)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Conclusion:

Thus, the lowest energy of a nucleon in the well is 23 MeV .

Answer 7

Chapter 40, Problem 63P

(a)

To determine

The lowest energy of a nucleon in the well.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Conclusion:

Thus, the lowest energy of a nucleon in the well is 23 MeV .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

The ground state energy of 12 nucleons in the well, all are neutrons.

(b)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons which are fermions.

E=2(E1+E2+E3+E4+E5+E6)

Write the above expression again for less than two neutrons per state.

E=2(E1+22E1+32E1+42E1+52E1+62E1)=182E1 ........ (2)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (2).

E=182(23 MeV)=4186 MeV( 1 GeV 10 3 MeV)=4.186 GeV≈4.2 GeV

Conclusion:

Thus, the ground state energy of 12 nucleons in the well 4.2 GeV .

Answer 13

(b)

To determine

The ground state energy of 12 nucleons in the well, all are neutrons.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons which are fermions.

E=2(E1+E2+E3+E4+E5+E6)

Write the above expression again for less than two neutrons per state.

E=2(E1+22E1+32E1+42E1+52E1+62E1)=182E1 ........ (2)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (2).

E=182(23 MeV)=4186 MeV( 1 GeV 10 3 MeV)=4.186 GeV≈4.2 GeV

Conclusion:

Thus, the ground state energy of 12 nucleons in the well 4.2 GeV .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

The ground state energy of 12 nucleons in the well, in which 6 are neutrons and 6 are protons.

(c)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)

Write the expression for 12 neutrons out of which 6 are protons and 6 are neutrons.

E=4(E1+E2+E3)

Write the above expression again for

4 nucleons per state.

E=4(E1+22E1+32E1)=56E1 ........ (3)

Calculation:

Substitute 1.0 u for m , 3.0 fm for L and 6.626 × 10−34 J⋅s for h in equation (1).

E1= ( 6.626 × 10 −34 J⋅s )28( 1.0 u)( 1.661 × 10 −27 Kg/u) ( 3.0 fm( 1 m 10 15 fm ) )2=3.678 × 10−12 J( 1 MeV 1.602 × 10 −13 J)=23 MeV

Substitute 23 MeV for E1 in equation (3).

E=56(23 MeV)=1288 MeV( 1 GeV 10 3 MeV)=1.288 GeV≈1.3 GeV

Conclusion:

Thus, the ground state energy for 4 protons and 4 neutrons in the well is 1.3 GeV .

Answer 19

(c)

To determine

The ground state energy of 12 nucleons in the well, in which 6 are neutrons and 6 are protons.

Answer 20

(c)

Expert Solution

Explanation of Solution

Given:

The mass of the nucleon is 1.0 u .

The length of the square well is 3.0 fm .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

En=n2(h28mL2)

Here, En is the allowed energy for n energy levels, n are the allowed energy levels, h is the Plank’s constant, m is the mass of the particle and L is the length of the square well.

For lowest energy level, n is 1 .

Substitute 1 for n in the above expression.

E1=(1)2( h 2 8m L 2 )=h28mL2 ........ (1)