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Q1. (a) The mass defect for ²⁴⁰Pu is 50.23 MeV and ²⁴Na is — 8.48 MeV. Calculate the corresponding atomic mass. 4 (b) Write the Weiszacker semi-empirical mass formula and explain its various terms. Draw the contribution of various terms of this formula on B.E/A vs. mass number(A) curve. 6

Ans. (a) The term “mass defect” is likely used incorrectly here. The binding energy (B.E.) of ²⁴⁰Pu is about 1813 MeV and for ²⁴Na is about 192 MeV. The given values are extremely small in comparison. It is more plausible that these values represent the mass excess (or mass surplus), defined as Δ = (M – A)c², where M is the atomic mass and A is the mass number. We will proceed with this assumption. The atomic mass M (in amu) can be calculated from the mass excess Δ (in MeV) using the formula: M (in amu) = A + Δ (in MeV) / 931.5 MeV/amu 1. For ²⁴⁰Pu: A = 240, Δ = 50.23 MeV Atomic Mass M = 240 + (50.23 / 931.5) amu M = 240 + 0.05392 amu M ≈ 240.05392 amu 2. For ²⁴Na: A = 24, Δ = -8.48 MeV Atomic Mass M = 24 + (-8.48 / 931.5) amu M = 24 – 0.00910 amu M ≈ 23.99090 amu This interpretation yields physically consistent results, as the actual mass of ²⁴Na is very close to 23.99096 amu. (b) Weizsacker Semi-Empirical Mass Formula: This formula provides an approximation for the binding energy (B.E.) of a nucleus based on the liquid drop model. The formula is: B.E.(A, Z) = aᵥA – aₛA²/³ – a_c Z(Z-1)/A¹/³ – aₐ(A-2Z)²/A – δ(A, Z) Explanation of the Terms:

• Volume Energy Term (aᵥA): This term arises from the fact that each nucleon interacts only with its nearest neighbors. As the nuclear force is saturated, the binding energy is proportional to the number of nucleons, A. This is a positive and dominant contribution.
• Surface Energy Term (-aₛA²/³): Nucleons on the surface are surrounded by fewer nucleons and are therefore less tightly bound. This reduces the binding energy. This term is proportional to the surface area of the nucleus (R² ∝ A²/³).
• Coulomb Energy Term (-a_c Z(Z-1)/A¹/³): The electrostatic repulsion between protons destabilizes the nucleus and reduces the binding energy. This term is proportional to the number of proton pairs [Z(Z-1)/2] and inversely proportional to the radius of the nucleus (R ∝ A¹/³).
• Asymmetry Energy Term (-aₐ(A-2Z)²/A): Due to the Pauli exclusion principle, the most stable nuclei have approximately equal numbers of neutrons and protons (N ≈ Z). When there is a difference between N and Z, energy levels are filled to higher states, reducing stability. This term is proportional to the square of the N-Z difference.
• Pairing Energy Term (δ(A, Z)): This term accounts for the tendency of nucleons to form pairs. Even-even nuclei are the most stable, while odd-odd nuclei are the least stable. δ = +aₚA⁻¹/² for even-even nuclei δ = 0 for odd-even nuclei δ = -aₚA⁻¹/² for odd-odd nuclei

Contribution of terms to the B.E/A vs. A curve:

In a graph, the overall B.E/A curve can be seen as a combination of the contributions from the different terms:

1. The

volume term

gives a constant positive value (aᵥ) to B.E/A.

2. The

surface term

‘s contribution is negative and decreases in magnitude as A increases, lowering B.E/A more significantly for low A.

3. The

Coulomb term

‘s negative contribution increases with A (as Z²/A⁴/³), causing the B.E/A to droop at high A.

4. The

asymmetry term

becomes more significant for heavy nuclei where N becomes much larger than Z.

The combination of these effects explains the characteristic shape of the B.E/A vs. A curve, which peaks at A ≈ 56 (iron) and then slowly decreases.

Q2. (a) Obtain the charge form factor and <r²> for the nuclear charge density ρ(r) = ρ₀ e⁻ʳ/ᵃ / r, where a and ρ₀ are constants. You may use : ∫₀^∞ xⁿe⁻ᵃˣ dx = n! / aⁿ⁺¹. (b) Show that : [L_y, L_z] = iħL_y [Note: This seems to be a typo, likely intended to be [L_y, L_z] = iħL_x]. 4

Ans. (a) Charge Form Factor F(q): The charge form factor is the Fourier transform of the charge density ρ(r). For a spherically symmetric charge distribution, it is given by: F(q) = (4π/q) ∫₀^∞ r sin(qr) ρ(r) dr Substituting the given charge density ρ(r) = ρ₀ e⁻ʳ/ᵃ / r: F(q) = (4πρ₀/q) ∫₀^∞ r sin(qr) (e⁻ʳ/ᵃ / r) dr F(q) = (4πρ₀/q) ∫₀^∞ e⁻ʳ/ᵃ sin(qr) dr We use the standard integral: ∫₀^∞ e⁻ᵇˣ sin(ax) dx = a / (a² + b²) Here, b = 1/a and a = q. ∫₀^∞ e⁻ʳ/ᵃ sin(qr) dr = q / (q² + (1/a)²) = q / (q² + 1/a²) = qa² / (1 + q²a²) Plugging this into the expression for F(q): F(q) = (4πρ₀/q) * [qa² / (1 + q²a²)] F(q) = 4πρ₀a² / (1 + q²a²)

Mean Square Radius <r²>: The mean square radius can be calculated using the formula: <r²> = ∫ r² ρ(r) d³r / ∫ ρ(r) d³r Numerator integral: ∫ r² ρ(r) d³r = ∫₀^∞ r² (ρ₀ e⁻ʳ/ᵃ / r) 4πr² dr = 4πρ₀ ∫₀^∞ r³ e⁻ʳ/ᵃ dr Using the given formula ∫₀^∞ xⁿe⁻ᵇˣ dx = n! / bⁿ⁺¹ (here n=3, b=1/a): = 4πρ₀ [3! / (1/a)⁴] = 4πρ₀ [6a⁴] = 24πρ₀a⁴ Denominator integral: ∫ ρ(r) d³r = ∫₀^∞ (ρ₀ e⁻ʳ/ᵃ / r) 4πr² dr = 4πρ₀ ∫₀^∞ r e⁻ʳ/ᵃ dr Using the given formula (here n=1, b=1/a): = 4πρ₀ [1! / (1/a)²] = 4πρ₀ [a²] = 4πρ₀a² Now, calculate <r²>: <r²> = (24πρ₀a⁴) / (4πρ₀a²) = 6a² <r²> = 6a²

(b) The question contains a typo: [L_y, L_z] = iħL_y. The correct relation in angular momentum algebra is [L_y, L_z] = iħL_x . We will prove this correct relation. The angular momentum operators are defined in terms of position (r) and momentum (p) operators: L_x = y p_z – z p_y L_y = z p_x – x p_z L_z = x p_y – y p_x The canonical commutation relations are: [x_i, p_j] = iħδ_ij, [x_i, x_j] = 0, [p_i, p_j] = 0. Let’s compute the commutator [L_y, L_z]: [L_y, L_z] = [z p_x – x p_z, x p_y – y p_x] Using the commutator identity [A-B, C-D] = [A,C] – [A,D] – [B,C] + [B,D]: [L_y, L_z] = [z p_x, x p_y] – [z p_x, y p_x] – [x p_z, x p_y] + [x p_z, y p_x] Evaluate the four terms using the identity [AB, C] = A[B,C] + [A,C]B: 1. [z p_x, x p_y] = z[p_x, x p_y] + [z, x p_y]p_x = z( [p_x,x]p_y + x[p_x,p_y] ) + 0 = z(-iħ)p_y = -iħ z p_y 2. [z p_x, y p_x] = z[p_x, y p_x] + [z, y p_x]p_x = z( y[p_x, p_x] + [p_x, y]p_x ) + 0 = 0 3. [x p_z, x p_y] = x[p_z, x p_y] + [x, x p_y]p_z = x( x[p_z, p_y] + [p_z, x]p_y ) + 0 = 0 4. [x p_z, y p_x] = x[p_z, y p_x] + [x, y p_x]p_z = x( y[p_z, p_x] + [p_z, y]p_x ) + 0 = x(0)p_x = 0 Wait, the expansion was not quite right. Let’s use [A, BC] = [A,B]C + B[A,C]. 1. [z p_x, x p_y] = z [p_x, x p_y] = z( [p_x, x]p_y + x[p_x, p_y] ) = z(-iħ)p_y + 0 = -iħ z p_y This is also incorrect. The simplest way is to expand fully. [L_y, L_z] = (z p_x – x p_z)(x p_y – y p_x) – (x p_y – y p_x)(z p_x – x p_z) = (z p_x x p_y – z p_x y p_x – x p_z x p_y + x p_z y p_x) – (x p_y z p_x – x p_y x p_z – y p_x z p_x + y p_x x p_z) Using p_x x = x p_x – iħ, etc., and that operators for different axes commute (e.g., p_x y = y p_x). = z(x p_x – iħ)p_y – zy p_x² – x² p_z p_y + xy p_z p_x – (xz p_y p_x – x² p_y p_z – yz p_x² + y(x p_x – iħ)p_z) = xz p_x p_y – iħz p_y – zy p_x² – x² p_z p_y + xy p_z p_x – xz p_y p_x + x² p_y p_z + yz p_x² – yx p_x p_z + iħy p_z The momentum components commute, so p_x p_y = p_y p_x, etc. Canceling terms: = – iħ z p_y + iħ y p_z = iħ (y p_z – z p_y) Since L_x = y p_z – z p_y, we have: [L_y, L_z] = iħL_x

Q3. Assuming the following square well potential for the deuteron : V(r) = -V₀, r < r₀ V(r) = 0, r > r₀ derive the relationship among V₀, r₀ and E_B. Estimate the potential depth assuming the deuteron to be a just bound system (E_B ≈ 0). Also, estimate the radius (decay length, R_D) of deuteron. Given : The binding energy (E_B) of deuteron is 2.225 MeV and r₀ = 2 fm. 6+3+1

Ans. For the deuteron, we consider the radial Schrödinger equation for the l=0 (s-state). Letting u(r) = rψ(r), the equation becomes: -ħ²/2μ d²u/dr² + V(r)u = -E_B u where μ ≈ M_p/2 is the reduced mass of the neutron-proton system, and E_B is the binding energy. Region I: r < r₀ (inside the well) V(r) = -V₀ -ħ²/2μ d²u/dr² – V₀ u = -E_B u d²u/dr² + (2μ/ħ²)(V₀ – E_B)u = 0 d²u/dr² + k²u = 0, where k = √[2μ(V₀ – E_B)]/ħ The solution is u₁(r) = A sin(kr) + B cos(kr). Since u(0) = 0, we must have B=0. Thus, u₁(r) = A sin(kr) Region II: r > r₀ (outside the well) V(r) = 0 -ħ²/2μ d²u/dr² = -E_B u d²u/dr² – (2μ/ħ²)E_B u = 0 d²u/dr² – κ²u = 0, where κ = √[2μE_B]/ħ The solution is u₂(r) = Ce⁻ᵏʳ + De⁺ᵏʳ. Since the wavefunction must vanish at r → ∞, D=0. Thus, u₂(r) = Ce⁻ᵏʳ Boundary Conditions: The wavefunction and its derivative must be continuous at r = r₀. 1) u₁(r₀) = u₂(r₀) ⇒ A sin(kr₀) = Ce⁻ᵏʳ₀ 2) u₁'(r₀) = u₂'(r₀) ⇒ Ak cos(kr₀) = -Cκe⁻ᵏʳ₀ Dividing equation (2) by (1) gives: k cot(kr₀) = -κ Substituting the values of k and κ gives the relationship among V₀, r₀, and E_B: √[2μ(V₀ – E_B)]/ħ cot(√[2μ(V₀ – E_B)]/ħ r₀) = -√[2μE_B]/ħ

Estimate of potential depth (V₀) for E_B ≈ 0: When the deuteron is considered a “just bound system”, E_B → 0. This means κ → 0. The condition k cot(kr₀) = -κ becomes k cot(kr₀) ≈ 0. This implies cot(kr₀) ≈ 0, which means kr₀ must be an odd multiple of π/2. For the lowest energy state: kr₀ = π/2 With E_B ≈ 0, k ≈ √[2μV₀]/ħ. (√[2μV₀]/ħ) * r₀ = π/2 V₀ = (π²/4) * (ħ² / 2μr₀²) We use ħc ≈ 197 MeV·fm, 2μc² ≈ M_N c² ≈ 938 MeV, and r₀ = 2 fm. V₀ = (π²/4) ( (ħc)² / (2μc² r₀²) ) V₀ ≈ (π²/4) ( (197)² / (938 2²) ) MeV V₀ ≈ 2.467 (38809 / 3752) ≈ 2.467 10.34 ≈ 25.5 MeV The estimated potential depth for the just-bound case is approximately 25.5 MeV . Using the actual E_B gives a more accurate value of V₀ ≈ 35 MeV.

Estimate of the deuteron radius (decay length, R_D): The “radius” or decay length of the deuteron is the distance over which the wavefunction in the outside region decays to 1/e. It is given by 1/κ. R_D = 1/κ = ħ / √[2μE_B] Using ħc ≈ 197 MeV·fm, 2μc² ≈ 938 MeV, E_B = 2.225 MeV: R_D = ħc / √[2μc²E_B] = 197 / √[938 * 2.225] R_D = 197 / √[2087.05] = 197 / 45.68 ≈ 4.31 fm The estimated radius (decay length) of the deuteron is 4.31 fm . This is significantly larger than the potential range r₀ = 2 fm, which implies that the neutron and proton spend a large fraction of their time in the classically forbidden region, outside the range of the nuclear force. This indicates that the deuteron is a very loosely bound system.

Q4. (a) Explain the significance of phase shift (δ₀) in scattering. How does attractive and repulsive potential affect the phase of an incident plane wave ? 3+2 (b) Using the effective range formula, evaluate the total n-p scattering cross-section for a neutron interacting with a free proton in center of mass frame at a relative energy of 2.5 MeV. 5 Given : Scattering lengths : a_t = 5.38 fm, a_s =— 23.7 fm. Effective range : r_ot = 1.78 fm, r_os = 2.40 fm.

Ans. (a) Significance of Phase Shift (δ₀): In scattering theory, an incident plane wave interacts with a potential and emerges as a scattered spherical wave. Far from the potential, the total wave function is a superposition of the incident and scattered waves. The phase shift (δ_l) measures how much the phase of the l-th partial wave (corresponding to angular momentum quantum number l) has been shifted by the interaction with the potential, compared to the case with no potential. Essentially, it quantifies the “pulling in” or “pushing out” of the wavefunction due to the presence of the potential. The phase shift holds crucial information about the nature and strength of the scattering potential. The scattering cross-section can be expressed directly in terms of the phase shifts: σ_total = (4π/k²) Σ (2l+1) sin²(δ_l) Thus, the experimental determination of phase shifts is the key to understanding the underlying potential responsible for the scattering process. Effect of Attractive and Repulsive Potentials:

• Attractive Potential: An attractive potential “pulls” the incident particle in, causing the wavelength to decrease and the phase of the wave function to be advanced. This results in a positive phase shift (δ_l > 0) . The wavefunction behaves as if it has traveled a longer path within the potential region.
• Repulsive Potential: A repulsive potential “pushes” the incident particle away, causing the wave function to stay away from the potential. This effectively retards the phase of the wave function. This results in a negative phase shift (δ_l < 0) . The wavefunction behaves as if it has traveled a shorter path.

(b) The total cross-section for low-energy n-p scattering is the sum of contributions from the spin-singlet (s) and spin-triplet (t) states:

σ_total = (π/k²) [sin²(δ_s) + 3sin²(δ_t)]

The factor of 3 comes from the three-fold degeneracy of the triplet state.

The effective range formula relates the phase shift to the energy (via k):

k cot(δ) = -1/a + (1/2)r₀k²

where ‘a’ is the scattering length and ‘r₀’ is the effective range.

1. Calculate the wave number k:

The relative energy in the CM frame is E = 2.5 MeV.

E = ħ²k² / M_N (since reduced mass μ ≈ M_N/2, and E_lab = 2E_cm, so E_cm = ħ²k²/M_N)

Given M_N c² ≈ 938 MeV, ħc ≈ 197 MeV·fm.

k² = E

M_N / ħ² = E

M_N c² / (ħc)²

k² = 2.5 MeV * 938 MeV / (197 MeV·fm)²

k² = 2345 / 38809 fm⁻² ≈ 0.06042 fm⁻²

k ≈ 0.2458 fm⁻¹

2. Calculate for the singlet state (δ_s):

a_s = -23.7 fm, r_os = 2.40 fm

k cot(δ_s) = -1/a_s + (1/2)r_os k²

k cot(δ_s) = -1/(-23.7) + (1/2)(2.40)(0.06042)

k cot(δ_s) = 0.04219 + 0.07250 = 0.11469 fm⁻¹

cot(δ_s) = 0.11469 / 0.2458 ≈ 0.4666

cot²(δ_s) ≈ 0.2177

sin²(δ_s) = 1 / (1 + cot²(δ_s)) = 1 / (1 + 0.2177) = 1 / 1.2177 ≈ 0.8212

3. Calculate for the triplet state (δ_t):

a_t = 5.38 fm, r_ot = 1.78 fm

k cot(δ_t) = -1/a_t + (1/2)r_ot k²

k cot(δ_t) = -1/(5.38) + (1/2)(1.78)(0.06042)

k cot(δ_t) = -0.18587 + 0.05377 = -0.1321 fm⁻¹

cot(δ_t) = -0.1321 / 0.2458 ≈ -0.5374

cot²(δ_t) ≈ 0.2888

sin²(δ_t) = 1 / (1 + cot²(δ_t)) = 1 / (1 + 0.2888) = 1 / 1.2888 ≈ 0.7760

4. Calculate the total cross-section (σ_total):

σ_total = (π/k²) [sin²(δ_s) + 3sin²(δ_t)]

σ_total = (π / 0.06042) [0.8212 + 3 * 0.7760]

σ_total = 51.99 * [0.8212 + 2.328]

σ_total = 51.99 * 3.1492 ≈ 163.73 fm²

Since 1 barn = 100 fm².

σ_total ≈ 1.64 barns

The total n-p scattering cross-section is approximately

1.64 barns

.

Q5. (a) Discuss the shell model and its limitations. 3+2 (b) The ground state energy and spin of a nucleus is 0.025 MeV and [5/2]⁺ respectively. Estimate the spin and energy of the first three rotational bands. 5

Ans. (a) The Shell Model: The nuclear shell model is analogous to the shell model of the atom. Its central idea is that each nucleon (proton or neutron) in a nucleus moves independently in an average, central potential created by all the other nucleons. This independent particle motion results in discrete energy levels or “shells”. Key Features:

1. Independent Particle Motion: It assumes that the complex interactions between nucleons can be replaced by an average central potential.
2. Spin-Orbit Coupling: A crucial breakthrough of the model was the inclusion of a strong spin-orbit interaction term. This term describes a coupling between the orbital angular momentum (l) and the intrinsic spin (s) of a nucleon. This l.s coupling splits levels that would otherwise be degenerate, and this splitting is critical for correctly reproducing the experimentally observed “magic numbers” (2, 8, 20, 28, 50, 82, 126). These magic numbers correspond to nuclei that are particularly stable.

Successes:

The shell model successfully explains the magic numbers, the ground state spins and parities of nuclei, magnetic dipole moments, and the phenomenon of nuclear isomerism.

Limitations:

1. Neglect of Collective Behavior: The shell model is an independent-particle model and fails to explain phenomena that arise from the correlated, collective motion of many nucleons.
2. Static Deformations: It cannot account for the large measured electric quadrupole moments of nuclei far from magic numbers. These large moments indicate a permanently deformed (non-spherical) nuclear shape, which contradicts the assumption of a simple central potential.
3. Rotational Spectra: The model is unable to predict the rotational energy bands observed in deformed nuclei, whose energy levels follow a J(J+1) rule.

To overcome these limitations, the shell model is often combined with other models, such as the Collective Model.

(b) The phrasing “The ground state energy… is 0.025 MeV” is contradictory, as the ground state energy is 0 by definition. We will assume the question means either that the band head (J=5/2 state) of the K=5/2 band is at an excitation energy of 0.025 MeV, or more likely, that the first rotational excited state (J=7/2) in the K=5/2 ground state band has an energy 0.025 MeV above the ground state (J=5/2). We will proceed with this second, more standard interpretation.

The energy levels for a rotational band are given by the formula:

E(J) = E₀ + A[J(J+1) – K(K+1)]

where K is constant for the band (K=J_min), J is the spin, E₀ is the energy of the band head, and A = ħ²/2I is the rotational constant.

Given: Ground state band with K=5/2, Jπ = 5/2⁺.

We assume E(J=5/2) = 0 (as reference).

The energy of the first excited state is E(J=7/2) = 0.025 MeV.

Calculate the rotational constant A:

E(7/2) – E(5/2) = A [ (7/2)(9/2) – K(K+1) ] – A [ (5/2)(7/2) – K(K+1) ]

0.025 MeV = A [ (63/4) – (35/4) ] = A [28/4] = 7A

A = 0.025 / 7 MeV ≈ 0.00357 MeV

We can now estimate the spin and energy of the “first three rotational bands” (which likely means the first three states of the band: J=7/2, 9/2, 11/2).

1. First rotational state:

• Spin: J = K+1 = 5/2 + 1 = 7/2 (parity remains +)
• Energy: E(7/2) = E(5/2) + 0.025 MeV = 0.025 MeV (as given/assumed)

2. Second rotational state:

• Spin: J = K+2 = 5/2 + 2 = 9/2 (parity remains +)
• Energy: E(9/2) = E(5/2) + A [ (9/2)(11/2) – (5/2)(7/2) ] = 0 + A [ (99/4) – (35/4) ] E(9/2) = A [64/4] = 16A E(9/2) = 16 * (0.00357 MeV) ≈ 0.0571 MeV

3. Third rotational state:

• Spin: J = K+3 = 5/2 + 3 = 11/2 (parity remains +)
• Energy: E(11/2) = E(5/2) + A [ (11/2)(13/2) – (5/2)(7/2) ] = 0 + A [ (143/4) – (35/4) ] E(11/2) = A [108/4] = 27A E(11/2) = 27 * (0.00357 MeV) ≈ 0.0964 MeV

Thus, the first three rotational states have spins (7/2)⁺, (9/2)⁺, (11/2)⁺ and energies of approximately 0.025 MeV, 0.057 MeV, and 0.096 MeV, respectively, relative to the J=5/2 ground state.

Q6. (a) Describe the experiment which confirmed parity violation in beta decay. 5 (b) Write short notes on the following : (i) Nuclear isomerism (ii) Internal conversion 2+2

Ans. (a) The Experiment Confirming Parity Violation in Beta Decay (The Wu Experiment): The violation of parity conservation in beta decay was definitively proven in a landmark experiment conducted in 1956 by Chien-Shiung Wu and her collaborators. The experiment tested the theoretical proposal by T. D. Lee and C. N. Yang that the weak interaction (responsible for beta decay) might not conserve parity. Experimental Setup:

1. Nuclear Sample: The experiment used Cobalt-60 (⁶⁰Co), which decays via beta decay to Nickel-60 (⁶⁰Ni). ⁶⁰Co → ⁶⁰Ni + e⁻ + ν̅ₑ
2. Polarization: The nuclei of ⁶⁰Co have a spin. By applying a strong external magnetic field and cooling the sample to a very low temperature (around 0.01 K), the nuclear spins were aligned (polarized) in the direction of the magnetic field. The low temperature was necessary to reduce thermal motion that would disrupt the alignment.
3. Electron Detection: Detectors were placed to count the electrons (e⁻) emitted from the beta decay. These detectors were positioned to measure the number of electrons emitted both “along” (parallel) and “opposite” (anti-parallel) to the direction of the nuclear spin.

Observation and Conclusion:

If parity were conserved, the electron emission should be isotropic, meaning there would be an equal probability of emission in any direction relative to the nuclear spin direction. The process viewed in a mirror would be indistinguishable from the original process.

However, Wu’s experiment showed a striking result:

more electrons were emitted in the direction opposite to the nuclear spin.

This asymmetry was clear proof that the process is not invariant under parity. In a mirror reflection, the spin of the nucleus (being an axial vector) would reverse, while the momentum of the electron (a polar vector) would also reverse. If in the original experiment electrons are emitted opposite to the spin, in the mirror world, they would appear to be emitted along with the

new

direction of the spin. Since the results of the original and mirror-image experiments were different, it was concluded that

the weak interaction maximally violates parity.

(b)

Short Notes:

(i) Nuclear Isomerism:

Nuclear isomerism is a phenomenon in which a single nucleus (same Z and A) can exist in one or more metastable excited states with measurable lifetimes (typically from nanoseconds to years). These metastable states are called

isomeric states

, and the nucleus in that state is a

nuclear isomer

.

This occurs when there is a large difference in angular momentum (spin) between an excited nuclear state and a lower-energy state (often the ground state). A normal decay via the emission of a gamma-ray requires a small spin change (typically ΔJ ≤ 2). When a large spin change (ΔJ > 3) is required, gamma emission is highly “forbidden” or suppressed, resulting in a very long half-life. These isomeric states eventually decay, either through the delayed gamma emission (known as an isomeric transition), internal conversion, or beta decay. Regions of the chart of nuclides where isomers are common are known as “islands of isomerism,” which are successfully explained by the Shell Model.

(ii) Internal Conversion:

Internal conversion is a non-radiative process by which an excited nucleus can release its energy and de-excite to a lower energy state. Instead of emitting a gamma-ray, the nucleus transfers its excitation energy directly to one of its own orbital electrons (usually from the K, L, or M shell).

This energy transfer is sufficient to eject the electron from the atom. This emitted electron is called a

conversion electron

. The kinetic energy of the conversion electron is given by:

T_e = E_γ – B.E._electron

where E_γ is the nuclear transition energy (which would be the energy of the gamma-ray if it were emitted) and B.E._electron is the binding energy of the shell from which the electron was removed.

Internal conversion is always in competition with gamma emission. The probability of internal conversion (measured by the internal conversion coefficient, α) increases for heavy elements (high Z) and low transition energies (low E_γ). It is the only possible de-excitation mode for 0 → 0 transitions, where single-photon gamma emission is absolutely forbidden by angular momentum conservation.

Q7. (a) What are different types of lepton flavor quantum numbers ? State whether lepton numbers are conserved in strong, weak and electromagnetic interactions. Hence, explain whether the following reaction is allowed : 5 p → e⁺ + π⁰ (b) Describe strong interactions and the quantities conserved. in strong interactions. Explain whether the following reaction is allowed : 5 K⁺ + n → K⁰ + p

Ans. (a) Lepton Flavor Quantum Numbers: Leptons are elementary particles that do not participate in the strong interaction. In the Standard Model, there are three generations or “flavors” of leptons. A separate lepton number quantum number exists for each flavor:

1. Electron Lepton Number (Lₑ): The electron (e⁻) and the electron neutrino (νₑ) are assigned Lₑ = +1. Their antiparticles, the positron (e⁺) and electron antineutrino (ν̅ₑ), are assigned Lₑ = -1. All other particles have Lₑ = 0.
2. Muon Lepton Number (L_μ): The muon (μ⁻) and the muon neutrino (ν_μ) have L_μ = +1. Their antiparticles (μ⁺, ν̅_μ) have L_μ = -1. All other particles have L_μ = 0.
3. Tau Lepton Number (L_τ): The tau (τ⁻) and the tau neutrino (ν_τ) have L_τ = +1. Their antiparticles (τ⁺, ν̅_τ) have L_τ = -1. All other particles have L_τ = 0.

The total lepton number is L = Lₑ + L_μ + L_τ.

Conservation in Interactions:

• Strong and Electromagnetic Interactions: In these interactions, each of the lepton flavor numbers (Lₑ, L_μ, L_τ) is individually conserved. This is trivially true as leptons do not participate in strong interactions, and electromagnetic interactions do not change flavor.
• Weak Interactions: Initially, it was believed that individual lepton numbers were also conserved in weak interactions. However, the discovery of neutrino oscillations showed that neutrinos can change flavor in flight (e.g., a νₑ can turn into a ν_μ). This indicates that the individual lepton flavor numbers (Lₑ, L_μ, L_τ) are not conserved in weak interactions. However, within the Standard Model, the total lepton number (L) is still considered to be conserved.

Given Reaction: p → e⁺ + π⁰

We check the conservation laws for this reaction:

• Baryon Number (B): Initial state: p (B = +1) Final state: e⁺ (B = 0), π⁰ (B = 0). Total B = 0. Baryon number is not conserved (1 → 0).
• Lepton Number (Lₑ): Initial state: p (Lₑ = 0) Final state: e⁺ (Lₑ = -1), π⁰ (Lₑ = 0). Total Lₑ = -1. Electron lepton number is not conserved (0 → -1).

Since both baryon number and lepton number are violated, this reaction is

not allowed

in the Standard Model. It is an example of proton decay, which is predicted by some Grand Unified Theories (GUTs) but has never been experimentally observed.

(b)

Strong Interactions:

The strong interaction, or strong nuclear force, is the strongest of the four fundamental forces of nature. It is responsible for:

1. Binding quarks together to form hadrons, such as protons and neutrons.
2. Binding protons and neutrons together to form atomic nuclei (this is a residual effect of the more fundamental interaction between quarks).

It is mediated by the exchange of particles called

gluons

. Its characteristics are:

• Extreme Strength: Its coupling constant (α_s) is ~1, making it about 100 times stronger than electromagnetism.
• Short Range: Its influence is confined to a distance of about 10⁻¹⁵ meters (1 femtometer), the size of a nucleus.
• Color Charge: It acts on a property called “color” carried by quarks and gluons.

Quantities Conserved in Strong Interactions:

Strong interactions obey almost all known conservation laws, making them the most “symmetric” interactions. These include:

• Energy, momentum, angular momentum
• Electric charge (Q)
• Baryon number (B)
• Lepton number (L) (as leptons don’t participate)
• Parity (P)
• Charge conjugation (C)
• CP (the combination of C and P)
• Time reversal (T)
• Isospin (I) and its third component (I₃)
• Quark flavor numbers such as Strangeness (S), Charm (C), Bottomness (B’), Topness (T’)

Given Reaction: K⁺ + n → K⁰ + p

Let us check the conservation laws for this reaction, which proceeds via the strong force:

Quark content: K⁺(us̄), n(udd) → K⁰(ds̄), p(uud)

• Electric Charge (Q): (+1) + (0) = +1 → (0) + (+1) = +1. (Conserved)
• Baryon Number (B): (0) + (+1) = +1 → (0) + (+1) = +1. (Conserved)
• Strangeness (S): (+1) + (0) = +1 → (+1) + (0) = +1. (Conserved)
• Isospin (I₃): For K⁺, I₃ = +1/2; for n, I₃ = -1/2. Total initial I₃ = 0. For K⁰, I₃ = -1/2; for p, I₃ = +1/2. Total final I₃ = 0. (Conserved)

Since all relevant quantum numbers, including charge, baryon number, strangeness, and isospin are conserved, this reaction can occur via the strong interaction. Therefore, the reaction is

allowed

. It is an example of charge-exchange scattering.

Q8. (a) What is charge conjugation ? Derive the eigenvalues of the charge conjugation operator. 5 (b) What is the quark hypothesis ? Write down the quark composition of the n, n°, m mesons. 5

Ans. (a) Charge Conjugation: Charge conjugation, also called C-symmetry, is a discrete symmetry operation that transforms a particle into its corresponding antiparticle. The operation reverses the sign of all internal additive quantum numbers, including:

• Electric charge (Q → -Q)
• Baryon number (B → -B)
• Lepton number (L → -L)
• Flavor quantum numbers like strangeness, charm, etc.

However, it leaves the kinematic quantities of the particle, such as its momentum, position, and spin (angular momentum), unchanged. The operation performed by the C-operator is:

C |ψ(q, B, L, …)> = |ψ(-q, -B, -L, …)>

Eigenvalues of the Charge Conjugation Operator:

A state |ψ> will be an eigenstate of the charge conjugation operator C if it is its own antiparticle. Such particles can be used to test C-symmetry. Examples include the photon (γ), the pion-zero (π⁰), and the eta meson (η).

If |ψ> is an eigenstate, the eigenvalue equation is:

C |ψ> = η_C |ψ>

where η_C is the eigenvalue.

Consider applying the C-operator twice. The first operation changes the particle to its antiparticle, and the second operation changes the antiparticle back to the original particle. Thus, applying C twice is equivalent to an identity operation.

C² |ψ> = C (C |ψ>) = C (η_C |ψ>)

Since η_C is a number, it can pass through the operator:

C² |ψ> = η_C (C |ψ>) = η_C (η_C |ψ>) = η_C² |ψ>

Since C² = 1 (the identity operator), we can write:

1 |ψ> = η_C² |ψ>

From this it follows that:

η_C² = 1

Therefore, the possible eigenvalues of the charge conjugation operator are

η_C = +1

and

η_C = -1

.

• States with η_C = +1 are called C-even (e.g., π⁰, η).
• States with η_C = -1 are called C-odd (e.g., photon, J/ψ).

The strong and electromagnetic interactions conserve C-symmetry, while the weak interaction violates it.

(b)

The Quark Hypothesis:

The quark hypothesis, proposed independently by Murray Gell-Mann and George Zweig in 1964, is a fundamental theory in particle physics. It proposes that hadrons—the particles that participate in the strong interaction—are not elementary, but are instead composed of smaller, more fundamental constituents called

quarks

.

The main points are:

1. Fundamental Constituents: Hadrons (like protons, neutrons, pions) are made of quarks and antiquarks.
2. Types (Flavors) of Quarks: Three flavors were originally proposed: up (u), down (d), and strange (s). Later, three more were discovered: charm (c), bottom (b), and top (t).
3. Fractional Charge: Quarks have fractional electric charges of the elementary charge e. The up-type quarks (u, c, t) have a charge of +2/3 e, while the down-type quarks (d, s, b) have a charge of -1/3 e.
4. Hadron Composition:
   • Baryons are made of three quarks (qqq). Ex: proton (uud), neutron (udd).
   • Mesons are made of a quark and an antiquark pair (q q̄). Ex: pions.
5. Confinement: Quarks have never been observed in isolation. They are always “confined” inside hadrons.

Quark Composition of Mesons:

The mesons “n, n°, m” in the question are very likely typos for π⁻, π⁰, and π⁺, which are the most common mesons. Under this assumption, their quark compositions are:

• π⁺ (likely written as “m”): The pi-plus meson consists of an up quark and a down antiquark . π⁺ = u d̄ (Charge: (+2/3) + (+1/3) = +1)
• π⁰ (written as “n°”): The pi-zero or neutral pion is a quantum mechanical superposition of two quark-antiquark pairs. π⁰ = (u ū – d d̄) / √2 (This is a specific combination of u ū and d d̄ states that makes it its own antiparticle.)
• π⁻ (likely written as “n”): The pi-minus meson consists of a down quark and an up antiquark . π⁻ = d ū (Charge: (-1/3) + (-2/3) = -1)