Nuclear Forces and Models Questions and Answers

Q: If nucleons in a nucleus are considered to be confined in a three-dimensional cubical box, then the first four magic numbers are

The first four magic numbers are 2, 8, 14, 20.In a cubical box the levels are \(E \propto n_{x}^{2} + n_{y}^{2} + n_{z}^{2}\) with each \(n \ge 1\). Count the distinct arrangements (degeneracy), and double for the two spin states:(1,1,1), sum 3: 1 state, 2 nucleons, total 2.(2,1,1) and permutations, sum 6: 3 states, 6 nucleons, total 8.(2,2,1), sum 9: 3 states, 6 nucleons, total 14.(3,1,1), sum 11: 3 states, 6 nucleons, total 20.Shell closures, the magic numbers in this model, are therefore 2, 8, 14, 20. The sequence 2, 8, 20, 28 is the observed set from the real nuclear shell model with spin-orbit coupling, which a simple box does not reproduce.

Q: The nuclear energy levels of mirror nuclei are similar. Using this empirical fact alone, the nuclear force can be said to be independent of which one of the following properties of the nucleons?

The nuclear force is independent of the nucleons' charge.Mirror nuclei, such as \(^{3}\)H and \(^{3}\)He or \(^{13}\)C and \(^{13}\)N, have their proton and neutron numbers interchanged. Once the Coulomb energy is allowed for, their energy levels are nearly identical, which shows that the nn and pp interactions are the same: the strong force does not distinguish a proton from a neutron. This is charge symmetry, part of the wider idea of charge independence.Swapping protons and neutrons leaves spins and parities of the nucleons unchanged, so mirror nuclei say nothing about dependence on spin or parity; and the tiny proton-neutron mass difference is not what the comparison tests.

Q: In the energy levels predicted by shell model, the labelled energy states and the corresponding nucleon number, starting from the lowest to higher energy levels are as follows :(i) 1s1/2 − 2 Nucleons(ii) 1p3/2 − 4 Nucleons(iii) 1p1/2 − 2 Nucleons(iv) 1d5/2 − 6 NucleonsThe estimated groundstate spins of \( {}^{27}_{13}\text{Al} \) and \( {}^{11}_{5}\text{B} \) nuclei are :

The spins are \( \frac{3}{2} \) for \( ^{11}_{5}\text{B} \) and \( \frac{5}{2} \) for \( ^{27}_{13}\text{Al} \).In the shell model, paired nucleons contribute nothing, so an odd-A nucleus takes the spin of its single unpaired nucleon.\( ^{11}\text{B} \): 5 protons and 6 neutrons. The neutrons fill 1s1/2 (2) and 1p3/2 (4) completely. The protons put 2 in 1s1/2 and 3 in 1p3/2, so the odd proton is in 1p3/2: spin \( \frac{3}{2} \).\( ^{27}\text{Al} \): 13 protons and 14 neutrons. The neutrons fill up to 1d5/2 exactly (2 + 4 + 2 + 6). The protons put 8 in the first three levels and 5 in 1d5/2, so the odd proton is in 1d5/2: spin \( \frac{5}{2} \).The other options give spins of levels the odd proton does not occupy, or, like \( \frac{11}{2} \), values not available in these shells.

Q: Nuclear forces are :

Nuclear forces are spin dependent and have a non-central part.The deuteron is bound only in the spin-1 state, and neutron-proton scattering differs for singlet and triplet spins, which shows spin dependence. The deuteron also has an electric quadrupole moment, meaning it is not spherical, which requires a non-central tensor force that depends on the orientation of the spins relative to the line joining the nucleons.Each of the other options denies one of these two experimentally established features.

Q: To obtain splitting in the nuclear energy levels, the concept of spin-orbit interaction was used in the nuclear :

Spin-orbit interaction was introduced in the nuclear shell model.Maria Goeppert Mayer and, independently, Haxel, Jensen and Suess (1949) added a strong spin-orbit term to the shell model potential. It splits each level of orbital angular momentum l into \(j = l + \frac{1}{2}\) and \(j = l - \frac{1}{2}\), with the higher j pushed down, and this reproduces the observed magic numbers 2, 8, 20, 28, 50, 82 and 126. Mayer and Jensen shared the 1963 Nobel Prize for it.The liquid drop model treats the nucleus as a charged fluid with no single-particle levels, the collective model adds rotations and vibrations, and the optical model describes nuclear scattering with a complex potential.

Q: Experimentally it is observed that neutron (n)-proton(p) system (deuteron) is bound, but the proton-proton and neutron-neutron systems are not bound. This is due to :

The difference is due to Pauli's principle.The nuclear force is spin dependent: it is strong enough to bind two nucleons only in the spin-triplet state (S = 1). The deuteron's ground state is exactly this, an L = 0, S = 1 state of a neutron and a proton, which are different particles and face no restriction.Two protons or two neutrons are identical fermions, so their total wave function must be antisymmetric. In the lowest, L = 0 state the space part is symmetric, which forces the spin part to be the antisymmetric singlet (S = 0), and the singlet force is too weak to bind. So the Pauli principle keeps p-p and n-n out of the one state that binds.The nuclear force itself is charge independent to a good approximation, so 'unequal nuclear force' is not the reason, and charge only adds Coulomb repulsion for p-p, which cannot explain why n-n is unbound. Angular momentum is conserved throughout.

Q: In the nuclear shell model, orbitals are filled in the following order \({}^{1}s_{1/2}\), \({}^{1}p_{3/2}\), \({}^{1}p_{1/2}\), \({}^{1}d_{5/2}\), \({}^{2}s_{1/2}\), \({}^{1}d_{3/2}\), .......... etc. The splitting between the \(p_{3/2}\) and \(p_{1/2}\) orbitals is caused by :

The splitting is caused by spin-orbit coupling.Maria Goeppert Mayer and Hans Jensen (1949) showed that a strong spin-orbit term in the nuclear potential, proportional to \(\vec{l}\cdot\vec{s}\), splits each level of orbital angular momentum l into two with \(j = l \pm \frac{1}{2}\). For nuclei the \(j = l + \frac{1}{2}\) level lies lower, which is why \(p_{3/2}\) fills before \(p_{1/2}\) and why the model reproduces the magic numbers 2, 8, 20, 28, 50, 82 and 126.Orbital angular momentum alone, or spin alone, cannot split the \(p_{3/2}\) and \(p_{1/2}\) levels, which have the same l and the same s; the splitting comes from their coupling, which is what 'spin-orbit' names and what the general phrase 'angular momentum and spin' leaves out.

Q: The only di-nucleon n-p bound system is observed in nature, but not the n-n or p-p di-nucleon bound system, this is because :

The reason is that nuclear forces are strongly spin-dependent.The deuteron (n-p) is bound only in the spin-triplet state (\(S = 1\), \(L = 0\)); the n-p singlet state is not bound. For two identical nucleons in the lowest state, \(L = 0\) is symmetric in space, so the Pauli principle forces the spins into the antisymmetric singlet. The singlet nuclear force is too weak to bind, so neither n-n nor p-p forms a bound pair.Charge independence actually says the n-n, p-p and n-p forces in the same state are equal, which is why it cannot explain the difference; charge dependence is a small effect that does not decide binding. A purely central force would not depend on spin orientation, so it could not single out the n-p triplet either.

If nucleons in a nucleus are considered to be confined in a three-dimensional cubical box, then the first four magic numbers are

A.2, 8, 20, 28
B.2, 8, 16, 24
C.2, 8, 14, 20
D.2, 10, 16, 28

Exam Year(s)

GATE Physics - 2022,

Answer: C

The first four magic numbers are 2, 8, 14, 20.

In a cubical box the levels are \(E \propto n_{x}^{2} + n_{y}^{2} + n_{z}^{2}\) with each \(n \ge 1\). Count the distinct arrangements (degeneracy), and double for the two spin states:

(1,1,1), sum 3: 1 state, 2 nucleons, total 2.
(2,1,1) and permutations, sum 6: 3 states, 6 nucleons, total 8.
(2,2,1), sum 9: 3 states, 6 nucleons, total 14.
(3,1,1), sum 11: 3 states, 6 nucleons, total 20.

Shell closures, the magic numbers in this model, are therefore 2, 8, 14, 20. The sequence 2, 8, 20, 28 is the observed set from the real nuclear shell model with spin-orbit coupling, which a simple box does not reproduce.

Discuss in Forum

The nuclear energy levels of mirror nuclei are similar. Using this empirical fact alone, the nuclear force can be said to be independent of which one of the following properties of the nucleons?

A.Mass
B.Spin
C.Charge
D.Parity

Exam Year(s)

GATE Physics - 2025,

Answer: C

The nuclear force is independent of the nucleons' charge.

Mirror nuclei, such as \(^{3}\)H and \(^{3}\)He or \(^{13}\)C and \(^{13}\)N, have their proton and neutron numbers interchanged. Once the Coulomb energy is allowed for, their energy levels are nearly identical, which shows that the nn and pp interactions are the same: the strong force does not distinguish a proton from a neutron. This is charge symmetry, part of the wider idea of charge independence.

Swapping protons and neutrons leaves spins and parities of the nucleons unchanged, so mirror nuclei say nothing about dependence on spin or parity; and the tiny proton-neutron mass difference is not what the comparison tests.

Discuss in Forum

In the energy levels predicted by shell model, the labelled energy states and the corresponding nucleon number, starting from the lowest to higher energy levels are as follows :

(i) 1s1/2 − 2 Nucleons
(ii) 1p3/2 − 4 Nucleons
(iii) 1p1/2 − 2 Nucleons
(iv) 1d5/2 − 6 Nucleons

The estimated groundstate spins of \( {}^{27}_{13}\text{Al} \) and \( {}^{11}_{5}\text{B} \) nuclei are :

A.\( {}^{11}_{5}\text{B} \rightarrow \left(\frac{1}{2}\right) \) and \( {}^{27}_{13}\text{Al} \rightarrow \left(\frac{3}{2}\right) \)
B.\( {}^{11}_{5}\text{B} \rightarrow \left(\frac{3}{2}\right) \) and \( {}^{27}_{13}\text{Al} \rightarrow \left(\frac{5}{2}\right) \)
C.\( {}^{11}_{5}\text{B} \rightarrow \left(\frac{5}{2}\right) \) and \( {}^{27}_{13}\text{Al} \rightarrow \left(\frac{1}{2}\right) \)
D.\( {}^{11}_{5}\text{B} \rightarrow \left(\frac{11}{2}\right) \) and \( {}^{27}_{13}\text{Al} \rightarrow \left(\frac{7}{2}\right) \)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The spins are \( \frac{3}{2} \) for \( ^{11}_{5}\text{B} \) and \( \frac{5}{2} \) for \( ^{27}_{13}\text{Al} \).

In the shell model, paired nucleons contribute nothing, so an odd-A nucleus takes the spin of its single unpaired nucleon.

\( ^{11}\text{B} \): 5 protons and 6 neutrons. The neutrons fill 1s_1/2 (2) and 1p_3/2 (4) completely. The protons put 2 in 1s_1/2 and 3 in 1p_3/2, so the odd proton is in 1p_3/2: spin \( \frac{3}{2} \).
\( ^{27}\text{Al} \): 13 protons and 14 neutrons. The neutrons fill up to 1d_5/2 exactly (2 + 4 + 2 + 6). The protons put 8 in the first three levels and 5 in 1d_5/2, so the odd proton is in 1d_5/2: spin \( \frac{5}{2} \).

The other options give spins of levels the odd proton does not occupy, or, like \( \frac{11}{2} \), values not available in these shells.

Discuss in Forum

Nuclear forces are :

A.Spin dependent and have no non-central part
B.Spin dependent and have a non-central part
C.Spin independent and have no non-central part
D.Spin independent and have a non-central part

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

Nuclear forces are spin dependent and have a non-central part.

The deuteron is bound only in the spin-1 state, and neutron-proton scattering differs for singlet and triplet spins, which shows spin dependence. The deuteron also has an electric quadrupole moment, meaning it is not spherical, which requires a non-central tensor force that depends on the orientation of the spins relative to the line joining the nucleons.

Each of the other options denies one of these two experimentally established features.

Discuss in Forum

To obtain splitting in the nuclear energy levels, the concept of spin-orbit interaction was used in the nuclear :

A.Liquid drop model
B.Collective model
C.Optical model
D.Shell model

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

Spin-orbit interaction was introduced in the nuclear shell model.

Maria Goeppert Mayer and, independently, Haxel, Jensen and Suess (1949) added a strong spin-orbit term to the shell model potential. It splits each level of orbital angular momentum l into \(j = l + \frac{1}{2}\) and \(j = l - \frac{1}{2}\), with the higher j pushed down, and this reproduces the observed magic numbers 2, 8, 20, 28, 50, 82 and 126. Mayer and Jensen shared the 1963 Nobel Prize for it.

The liquid drop model treats the nucleus as a charged fluid with no single-particle levels, the collective model adds rotations and vibrations, and the optical model describes nuclear scattering with a complex potential.

Discuss in Forum

Experimentally it is observed that neutron (n)-proton(p) system (deuteron) is bound, but the proton-proton and neutron-neutron systems are not bound. This is due to :

A.unequal nuclear force between n-p, n-n and p-p systems
B.unequal total charges of n-p, n-n and p-p systems
C.Pauli's principle
D.non-conservation of angular momentum

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The difference is due to Pauli's principle.

The nuclear force is spin dependent: it is strong enough to bind two nucleons only in the spin-triplet state (S = 1). The deuteron's ground state is exactly this, an L = 0, S = 1 state of a neutron and a proton, which are different particles and face no restriction.

Two protons or two neutrons are identical fermions, so their total wave function must be antisymmetric. In the lowest, L = 0 state the space part is symmetric, which forces the spin part to be the antisymmetric singlet (S = 0), and the singlet force is too weak to bind. So the Pauli principle keeps p-p and n-n out of the one state that binds.

The nuclear force itself is charge independent to a good approximation, so 'unequal nuclear force' is not the reason, and charge only adds Coulomb repulsion for p-p, which cannot explain why n-n is unbound. Angular momentum is conserved throughout.

Discuss in Forum

In the nuclear shell model, orbitals are filled in the following order \({}^{1}s_{1/2}\), \({}^{1}p_{3/2}\), \({}^{1}p_{1/2}\), \({}^{1}d_{5/2}\), \({}^{2}s_{1/2}\), \({}^{1}d_{3/2}\), .......... etc. The splitting between the \(p_{3/2}\) and \(p_{1/2}\) orbitals is caused by :

A.angular momentum of the nucleus
B.only the spins of the nucleus
C.spin-orbit coupling of the nucleus
D.angular momentum and spin of the nucleus

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The splitting is caused by spin-orbit coupling.

Maria Goeppert Mayer and Hans Jensen (1949) showed that a strong spin-orbit term in the nuclear potential, proportional to \(\vec{l}\cdot\vec{s}\), splits each level of orbital angular momentum l into two with \(j = l \pm \frac{1}{2}\). For nuclei the \(j = l + \frac{1}{2}\) level lies lower, which is why \(p_{3/2}\) fills before \(p_{1/2}\) and why the model reproduces the magic numbers 2, 8, 20, 28, 50, 82 and 126.

Orbital angular momentum alone, or spin alone, cannot split the \(p_{3/2}\) and \(p_{1/2}\) levels, which have the same l and the same s; the splitting comes from their coupling, which is what 'spin-orbit' names and what the general phrase 'angular momentum and spin' leaves out.

Discuss in Forum

The only di-nucleon n-p bound system is observed in nature, but not the n-n or p-p di-nucleon bound system, this is because :

A.Nuclear forces are mainly central
B.Nuclear forces are strongly spin-dependent
C.Nuclear forces are charge independent
D.Nuclear forces are charge dependent

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The reason is that nuclear forces are strongly spin-dependent.

The deuteron (n-p) is bound only in the spin-triplet state (\(S = 1\), \(L = 0\)); the n-p singlet state is not bound. For two identical nucleons in the lowest state, \(L = 0\) is symmetric in space, so the Pauli principle forces the spins into the antisymmetric singlet. The singlet nuclear force is too weak to bind, so neither n-n nor p-p forms a bound pair.

Charge independence actually says the n-n, p-p and n-p forces in the same state are equal, which is why it cannot explain the difference; charge dependence is a small effect that does not decide binding. A purely central force would not depend on spin orientation, so it could not single out the n-p triplet either.

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