Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Question
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Chapter 40, Problem 63P

(a)

To determine

The lowest energy of a nucleon in the well.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of the nucleon is 1.0u .

The length of the square well is 3.0fm .

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 1.0u for m , 3.0fm for L and 6.626×1034Js for h in equation (1).

  E1= ( 6.626× 10 34 Js )28( 1.0u)( 1.661× 10 27 Kg/u) ( 3.0fm( 1m 10 15 fm ) )2=3.678×1012J( 1MeV 1.602× 10 13 J)=23MeV

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of the nucleon is 1.0u .

The length of the square well is 3.0fm .

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 1.0u for m , 3.0fm for L and 6.626×1034Js for h in equation (1).

  E1= ( 6.626× 10 34 Js )28( 1.0u)( 1.661× 10 27 Kg/u) ( 3.0fm( 1m 10 15 fm ) )2=3.678×1012J( 1MeV 1.602× 10 13 J)=23MeV

Substitute 23MeV for E1 in equation (2).

  E=182(23MeV)=4186MeV( 1GeV 10 3 MeV)=4.186GeV4.2GeV

Conclusion:

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

(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
Check Mark

Explanation of Solution

Given:

The mass of the nucleon is 1.0u .

The length of the square well is 3.0fm .

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 1.0u for m , 3.0fm for L and 6.626×1034Js for h in equation (1).

  E1= ( 6.626× 10 34 Js )28( 1.0u)( 1.661× 10 27 Kg/u) ( 3.0fm( 1m 10 15 fm ) )2=3.678×1012J( 1MeV 1.602× 10 13 J)=23MeV

Substitute 23MeV for E1 in equation (3).

  E=56(23MeV)=1288MeV( 1GeV 10 3 MeV)=1.288GeV1.3GeV

Conclusion:

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

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