CUET UG Physics Booster Test 3-Mass-Energy and Binding -

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CUET UG Physics Booster Test 3-Mass-Energy and Binding

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 3-Mass-Energy and Binding

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QUESTION 1 OF 20

In a high-energy nuclear reaction, the strict classical conservation of total rest mass is ________, entirely because mass and energy are ________ according to special relativity.

Answer Review

1 In a high-energy nuclear reaction, the strict classical conservation of total rest mass is ________, entirely because mass and energy are ________ according to special relativity.

A Maintained, Independent

B Violated, Interconvertible

C Maintained, Interconvertible

D Violated, Independent

💡 Short Explanation

�� Mass and energy are equivalent. �� Nuclear reactions involve mass-energy conversion. �� Total mass alone is not conserved.

✅ Correct Answer Explanation

According to Einstein's special relativity, mass and energy are interconvertible and related by E = mc². In nuclear reactions, a small amount of mass can be converted into energy or vice versa. Therefore, the classical law of strict rest-mass conservation is violated. However, the total mass-energy of the system remains conserved. Option B correctly states that strict mass conservation is violated because mass and energy are interconvertible.

❓ Why Other Options Are Incorrect

�� Option A → Mass and energy are not independent according to relativity.
�� Option C → Strict rest mass is not conserved in nuclear reactions.
�� Option D → Mass and energy are not independent quantities.

🎯 Strategy

Used

Elimination

Application:

Check whether mass-energy equivalence is accepted.

Final Logic:

Mass and energy are interconvertible, so strict mass conservation is violated.

🧠 Memory Trick

E = mc² → Mass ↔ Energy

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Mass-Energy Equivalence

2 Match List I with List II regarding the practical implications of E = mc²

List I	List II
1. Chemical reaction mass defect	a. Extra energy ΔMc² has to be supplied
2. Nuclear reaction mass defect	b. Almost a million times smaller than in nuclear reactions
3. Exothermic fusion reaction	c. Final binding energy > Initial binding energy
4. Endothermic separation process	d. Directly dictates energy scale of MeV per reaction

A 1-a, 2-b, 3-c, 4-d

B 1-b, 2-a, 3-c, 4-d

C 1-a, 2-c, 3-b, 4-d

D 1-c, 2-d, 3-a, 4-b

💡 Short Explanation

�� Chemical mass defects are tiny. �� Nuclear reactions release MeV-scale energy. �� Fusion increases binding energy.

✅ Correct Answer Explanation

1 → a : Chemical reaction mass defects are almost a million times smaller than nuclear mass defects. 2 → b : Nuclear mass defects determine the MeV energy scale through E = mc². 3 → c : In exothermic fusion, final nuclei have greater binding energy than initial nuclei. 4 → d : Endothermic separation requires external energy equal to ΔMc². Thus, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Statements 1 and 2 are interchanged.
�� Option C → Nuclear mass defect does not correspond directly to binding-energy comparison.
�� Option D → Multiple incorrect pairings occur.

🎯 Strategy

Used

Option Grouping

Application:

Match each physical process with its energy implication.

Final Logic:

Chemical → tiny defect, Nuclear → MeV scale, Fusion → higher binding energy, Separation → energy input.

🧠 Memory Trick

Chemical = Small, Nuclear = MeV

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Mass-Energy Equivalence and Binding Energy

3 If an isolated nuclear fragment of mass m travels with classical velocity v, its kinetic energy is (1/2)mv². If an identical mass m is completely converted to rest energy E, the algebraic ratio of the converted energy to the kinetic energy is

A 2c² / v²

B c² / 2v²

C 2v² / c²

D v² / 2c²

💡 Short Explanation

�� Rest energy = mc². �� Kinetic energy = ½mv². �� Mass cancels in ratio.

✅ Correct Answer Explanation

Rest energy: E = mc² Kinetic energy: K = ½mv² Ratio: E/K = mc²/(½mv²) = 2c²/v² Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Inverse factor is incorrect.
�� Option C → Reciprocal of correct expression.
�� Option D → Incorrect reciprocal and factor.

🎯 Strategy

Used

Substitution

Application:

Substitute standard formulas directly.

Final Logic:

mc² ÷ (½mv²) = 2c²/v².

🧠 Memory Trick

Rest/Kinetic → Double c²/v²

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Einstein Mass-Energy Relation

4 Burning 1 kg of coal gives 10⁷ J. Fission of 1 kg of uranium generates 10¹⁴ J. How many kilograms of coal must be completely burned to exactly match the energy released by the total conversion of merely 1 gram of matter into pure energy? (Note: complete mass conversion of 1g = 9 × 10¹³ J).

A 9 × 10³ kg

B 9 × 10⁶ kg

C 10⁷ kg

D 9 × 10⁹ kg

💡 Short Explanation

�� 1 g mass → 9 × 10¹³ J. �� Coal gives 10⁷ J/kg. �� Divide total energy by energy per kg.

✅ Correct Answer Explanation

Energy from complete conversion of 1 g: = 9 × 10¹³ J Coal energy output: = 10⁷ J/kg Required coal: = (9 × 10¹³)/(10⁷) = 9 × 10⁶ kg Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Underestimates by factor 1000.
�� Option C → Does not match exact calculation.
�� Option D → Overestimates by factor 1000.

🎯 Strategy

Used

Substitution

Application:

Apply direct ratio of energies.

Final Logic:

9 × 10¹³ ÷ 10⁷ = 9 × 10⁶ kg.

🧠 Memory Trick

13 − 7 = 6

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Mass-Energy Conversion

5 Incorrect statement about mass defect in isotopes

A The mass of ¹⁶₈O is strictly less than the total unbound mass of 8 protons and 8 neutrons.

B The measured difference in mass between isolated constituents and the nucleus is the mass defect.

C The mass of an atom's electrons is subtracted from the atomic mass to precisely find the experimental nuclear mass.

D The mass of the bound nucleus is always strictly equal to the simple total mass of its individual protons and neutrons.

💡 Short Explanation

�� Bound nucleus has lower mass. �� Mass defect produces binding energy. �� Equality never holds for stable nuclei.

✅ Correct Answer Explanation

A bound nucleus possesses less mass than the sum of its free nucleons because part of the mass is converted into binding energy. This difference is called mass defect. Option D is incorrect because the mass of a bound nucleus is not equal to the simple sum of proton and neutron masses.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement about mass defect.
�� Option B → Correct definition of mass defect.
�� Option C → Correct method for obtaining nuclear mass from atomic mass.

🎯 Strategy

Used

Elimination

Application:

Identify statement contradicting mass defect concept.

Final Logic:

Bound nucleus mass is always less than free nucleon mass sum.

🧠 Memory Trick

Bound = Less Mass

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Mass Defect

6 Regarding the analytical mass defect formula
ΔM = [Zmₚ + (A − Z)mₙ] − M
Statements:
1. M represents the rest mass of the fully bound nucleus.
2. The term Zmₚ represents the total mass of isolated protons.
3. ΔMc² represents the binding energy holding the nucleus together.
4. The formula assumes nucleons are in an unbound free state initially.

A (1), (2), (4)

B (1), (3), (4)

C (2), (3), (4)

D (1), (2), (3), (4)

💡 Short Explanation

�� Formula compares free and bound states. �� Mass defect gives binding energy. �� All statements are correct.

✅ Correct Answer Explanation

Statement 1 is correct because M denotes the mass of the bound nucleus. Statement 2 is correct because Zmₚ represents total mass of Z free protons. Statement 3 is correct because binding energy = ΔMc². Statement 4 is correct because the formula compares free nucleons with the bound nucleus. Therefore all four statements are correct.

❓ Why Other Options Are Incorrect

�� Option A → Omits statement 3, which is correct.
�� Option B → Omits statement 2, which is correct.
�� Option C → Omits statement 1, which is correct.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using the mass defect formula.

Final Logic:

All four statements correctly describe the formula.

🧠 Memory Trick

Defect × c² = Binding Energy

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Mass Defect and Binding Energy

7 Statements regarding the unified conservation law conceptually
1. Strict mass and strict energy are always conserved entirely separately in high energy nuclear reactions.
2. The measurable difference in chemical binding energies technically implies tiny mass-energy conversion occurs in chemical reactions too.
3. What is strictly conserved in realm of very high energies is total charge and total baryon number.
4. The unified law proves that mass defect conceptually exists in molecules as well, just a million times smaller.

A (2), (3)

B (1), (2), (3)

C (2), (3), (4)

D (1), (3), (4)

💡 Short Explanation

�� Mass-energy is conserved jointly. �� Chemical reactions also involve tiny mass defects. �� Molecular binding energy implies mass defect.

✅ Correct Answer Explanation

Statement 1 is incorrect because mass and energy are not conserved separately in high-energy processes. Statement 2 is correct because chemical binding energy corresponds to extremely small mass changes. Statement 3 is correct because charge and baryon number remain conserved in nuclear reactions. Statement 4 is correct because molecular binding energy also implies a tiny mass defect. Hence statements 2, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Omits statement 4.
�� Option B → Includes statement 1, which is incorrect.
�� Option D → Includes statement 1, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Reject statements that assume separate conservation of mass and energy.

Final Logic:

Only statements 2, 3 and 4 remain valid.

🧠 Memory Trick

Energy Bound → Mass Found

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Conservation of Mass-Energy

8 In a nuclear reaction where the total binding energy of the nuclei on the right side is substantially greater than that on the left side, the physical difference in these binding energies

A appears actively as energy released in the process (positive Q-value)

B strictly means the reaction is endothermic and requires external energy

C results in an unnatural increase in the total rest mass of the final system

D completely violates the unified law of conservation of mass and energy

💡 Short Explanation

�� Greater binding energy means greater stability. �� Excess energy is released. �� Positive Q-value results.

✅ Correct Answer Explanation

If final nuclei possess greater total binding energy than initial nuclei, the difference appears as released energy. Such reactions are exothermic and have positive Q-values. Mass-energy conservation remains valid. Therefore option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Opposite of actual situation.
�� Option C → Final rest mass decreases rather than increases.
�� Option D → Conservation law remains valid.

🎯 Strategy

Used

Elimination

Application:

Higher binding energy corresponds to energy release.

Final Logic:

Increase in binding energy leads to positive Q-value.

🧠 Memory Trick

More Binding → More Release

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Reactions and Q-Value

9 Let EJ be the calculated energy in Joules and EeV be the identical energy in eV. Given the fundamental charge of an electron is e, the strict algebraic relation is

A EJ = EeV × e

B EeV = EJ × e

C EJ = EeV / e

D EJ = EeV − e

💡 Short Explanation

�� 1 eV = e joules. �� Convert eV to joules by multiplication. �� e = 1.6 × 10⁻¹⁹ C.

✅ Correct Answer Explanation

By definition, 1 eV = 1.6 × 10⁻¹⁹ J Therefore, EJ = EeV × e Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Gives incorrect dimensional conversion.
�� Option C → Uses inverse relation.
�� Option D → Physical meaning is invalid.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Use the definition of electron volt.

Final Logic:

eV × e gives energy in joules.

🧠 Memory Trick

eV → Multiply by e

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Energy Units

10 Statements regarding energy units mapped to nuclear physics scales
1. 1 u of mass defect equates to 931.5 MeV of binding energy.
2. The binding energy of ¹⁶₈O is roughly 127.5 MeV.
3. 1 MeV is defined identically as 1.6 × 10⁻¹³ J.
4. Energy released in typical nuclear fission is of the macroscopic order of purely electron volts.

A (1), (2), (4)

B (1), (2), (3)

C (2), (3), (4)

D (1), (3), (4)

💡 Short Explanation

�� 1 u corresponds to 931.5 MeV. �� Oxygen-16 binding energy is about 127 MeV. �� MeV is the standard nuclear energy unit.

✅ Correct Answer Explanation

Statement 1 is correct because 1 u corresponds to approximately 931.5 MeV. Statement 2 is correct because the total binding energy of Oxygen-16 is approximately 127.5 MeV. Statement 3 is correct because 1 MeV = 1.6 × 10⁻¹³ J. Statement 4 is incorrect because fission releases energies of the order of MeV, not eV. Therefore statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 4, which is incorrect.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4, which is incorrect and omits statement 2.

🎯 Strategy

Used

Elimination

Application:

Check each statement using standard nuclear energy values.

Final Logic:

Only statements 1, 2 and 3 are correct.

🧠 Memory Trick

1 u → 931.5 MeV

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Binding Energy and Energy Units

11 Incorrect statement about conceptual nucleus separation

A To fully separate a nucleus, one must supply external energy equal to E_b.

B Separating a nucleus drastically decreases the total mass of the resulting unbound system.

C The separated nucleons have a greater total rest mass than the original bound nucleus.

D The binding energy reflects the stability of the bound state against this separation.

💡 Short Explanation

�� Binding energy must be supplied for separation. �� Free nucleons have greater mass. �� Bound nuclei are more stable.

✅ Correct Answer Explanation

When a nucleus is separated into free nucleons, energy equal to the binding energy must be supplied. This added energy increases the total mass-energy of the system. Therefore, the total mass of separated nucleons is greater than that of the original bound nucleus. Hence, statement B is incorrect because separation does not decrease the total mass; it increases it.

❓ Why Other Options Are Incorrect

�� Option A → Correct because binding energy must be supplied to break the nucleus.
�� Option C → Correct because free nucleons possess greater total rest mass.
�� Option D → Correct because binding energy is a measure of nuclear stability.

🎯 Strategy

Used

Elimination

Application:

Compare masses before and after nuclear separation.

Final Logic:

Separated nucleons have greater mass, not lower mass.

🧠 Memory Trick

Break Nucleus → Add Energy

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Binding Energy

12 Assembling a tightly bound nucleus directly from free protons and neutrons leads to a ________ in total system mass, thereby functioning fundamentally as an ________ process.

A Reduction, Exothermic

B Increase, Endothermic

C Reduction, Endothermic

D Increase, Exothermic

💡 Short Explanation

�� Binding releases energy. �� Released energy corresponds to mass defect. �� Final nucleus has lower mass.

✅ Correct Answer Explanation

During nucleus formation, free nucleons combine and release binding energy. Since energy leaves the system, the final bound nucleus has less mass than the sum of the individual nucleons. Thus mass decreases and the process is exothermic.

❓ Why Other Options Are Incorrect

�� Option B → Mass does not increase during binding.
�� Option C → Binding is not endothermic.
�� Option D → Exothermic is correct but mass does not increase.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Nuclear binding always corresponds to energy release.

Final Logic:

Energy released ⇒ mass reduced ⇒ exothermic process.

🧠 Memory Trick

Bind → Release → Defect

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Mass Defect and Binding Energy

13 A heavy nucleus X of mass number A = 240 with E_bn = 7.6 MeV breaks into two identical fragments Y of A = 120 with E_bn = 8.5 MeV. What is the absolute total energy released in the process?

A 216 MeV

B 0.9 MeV

C 108 MeV

D 240 MeV

💡 Short Explanation

�� Binding energy per nucleon increases. �� Final nuclei are more stable. �� Energy released equals increase in total binding energy.

✅ Correct Answer Explanation

Initial total binding energy: = 240 × 7.6 = 1824 MeV Final total binding energy: = 240 × 8.5 = 2040 MeV Energy released: = 2040 − 1824 = 216 MeV Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Only difference per nucleon is considered.
�� Option C → Half of the actual energy released.
�� Option D → Incorrect calculation.

🎯 Strategy

Used

Substitution

Application:

Calculate total binding energies before and after fission.

Final Logic:

240 × (8.5 − 7.6) = 216 MeV.

🧠 Memory Trick

ΔE = A × Δ(E/A)

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Binding Energy per Nucleon

14 Statements about E_bn as an analytical measure of stability
1. The saturation property of nuclear forces limits E_bn in large nuclei.
2. Continually adding more nucleons to a heavy nucleus linearly increases its E_bn.
3. The value of E_bn directly determines which exothermic reactions are energetically possible.
4. Fission is possible because heavy nuclei have higher E_bn than middle-mass nuclei.

A (1), (3)

B (1), (2), (3)

C (2), (4)

D (3), (4)

💡 Short Explanation

�� Saturation limits binding energy. �� Binding energy per nucleon determines stability. �� Heavy nuclei have lower E_bn than medium nuclei.

✅ Correct Answer Explanation

Statement 1 is correct because nuclear force saturation prevents unlimited growth of E_bn. Statement 2 is incorrect because E_bn does not increase linearly for heavy nuclei. Statement 3 is correct because energy release depends on changes in binding energy. Statement 4 is incorrect because heavy nuclei have lower E_bn than nuclei near iron. Therefore, statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 2, which is incorrect.
�� Option C → Includes statements 2 and 4, both incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Check each statement using the binding-energy-per-nucleon curve.

Final Logic:

Only statements 1 and 3 agree with the stability curve.

🧠 Memory Trick

Iron Peak = Maximum Stability

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Binding Energy per Nucleon Curve

15 If a nucleon can have a maximum of p neighbours within the limited range of the nuclear force, and the interaction energy with one such neighbour is k, the binding energy of a single nucleon well inside the plateau region of the nucleus is approximately

A pk

B p/k

C k/p

D p²k

💡 Short Explanation

�� Nuclear force is short ranged. �� Each nucleon interacts with limited neighbours. �� Total interaction energy adds approximately.

✅ Correct Answer Explanation

Inside a large nucleus, a nucleon interacts with approximately p neighbouring nucleons. If interaction energy with each neighbour is k, total binding contribution becomes: E ≈ p × k Therefore option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect dimensional relationship.
�� Option C → Interaction energies do not divide.
�� Option D → Double counting interaction contribution.

🎯 Strategy

Used

Substitution

Application:

Multiply number of interactions by energy per interaction.

Final Logic:

Total energy = p interactions × k energy.

🧠 Memory Trick

Neighbours × Energy

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Saturation Property of Nuclear Forces

16 Characteristics of the prominent maximum at Iron (A = 56)
Statements:
1. It represents the single most tightly bound nucleus per nucleon.
2. Its experimentally derived E_bn value is approximately 8.75 MeV.
3. Fission of iron into lighter nuclei would heavily release spontaneous energy.
4. It lies physically within the flat middle region of the overall curve.

A (1), (2), (4)

B (1), (3), (4)

C (2), (3), (4)

D (1), (2), (3)

💡 Short Explanation

�� Iron lies near the peak. �� Binding energy per nucleon is maximum. �� Extremely stable nucleus.

✅ Correct Answer Explanation

Statement 1 is correct because nuclei around iron are the most tightly bound. Statement 2 is correct because E_bn is approximately 8.75 MeV. Statement 3 is incorrect because fission of iron would require energy rather than release it. Statement 4 is correct because iron lies within the broad middle plateau region. Therefore statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 3, which is incorrect.
�� Option C → Includes statement 3, which is incorrect.
�� Option D → Includes statement 3, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Recall the binding-energy-per-nucleon curve maximum.

Final Logic:

Iron is already near maximum stability.

🧠 Memory Trick

Iron = Peak Stability

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Binding Energy Curve

17 The markedly low binding energy per nucleon for A < 30 implies that when two such light nuclei successfully fuse to form a heavier nucleus,

A the final system achieves a higher binding energy per nucleon, leading to a massive energy release

B the final system achieves a lower binding energy per nucleon, continually absorbing energy

C the strict mass defect of the final system is smaller than the initial isolated systems

D the prevailing Coulomb barrier entirely prevents the reaction at all theoretical temperatures

💡 Short Explanation

�� Light nuclei have lower E_bn. �� Fusion increases stability. �� Energy is released.

✅ Correct Answer Explanation

For light nuclei, fusion generally moves the products toward the peak of the binding-energy curve. As binding energy per nucleon increases, the excess energy is released. Thus option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Opposite of fusion energetics.
�� Option C → Final mass defect is larger, not smaller.
�� Option D → High temperatures can overcome the Coulomb barrier.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Move toward higher E_bn on the stability curve.

Final Logic:

Fusion of light nuclei increases binding energy per nucleon.

🧠 Memory Trick

Light Fuse → Energy Loose

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Nuclear Fusion

18 For analytically mapping heavy nuclei (A > 170)
Statements:
1. They possess significantly lower E_bn compared to nuclei with A = 120.
2. Their breaking into two lighter fragments naturally increases the total binding energy of the system.
3. Fission functions as an entirely endothermic process for them.
4. They are far more tightly bound per nucleon than iron-56.

A (1), (2)

B (1), (3)

C (2), (4)

D (1), (2), (4)

💡 Short Explanation

�� Heavy nuclei lie on descending side of curve. �� Fission increases total binding energy. �� Energy is released.

✅ Correct Answer Explanation

Statement 1 is correct because E_bn decreases for very heavy nuclei. Statement 2 is correct because fission products are more tightly bound. Statement 3 is incorrect because fission is generally exothermic. Statement 4 is incorrect because iron-56 is more tightly bound. Hence statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 3, which is incorrect.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Use the heavy-nuclei region of the binding-energy curve.

Final Logic:

Heavy nuclei release energy through fission.

🧠 Memory Trick

Heavy Split → Energy Emit

📚 Correct. For Match the Following questions, I should only convert List II from i, ii, iii, iv to a, b, c, d while preserving the exact mappings given in the original options. I should not reshuffle or reorder the matching pairs.

19 Match List I with List II regarding nuclear binding curves and conceptual features

List I	List II
1. Smooth E_bn curve feature	a. Decreased binding due to Coulomb repulsion dominating strong force
2. Local peak at ⁴He	b. Considered evidence of atom-like shell structure
3. Gradual drop at A > 170	c. Due strictly to saturation property of nuclear force
4. Constancy for 30 < A < 170	d. Shows nuclear force does not depend on electric charge directly

A (1)-d, (2)-b, (3)-a, (4)-c

B (1)-b, (2)-d, (3)-c, (4)-a

C (1)-d, (2)-a, (3)-b, (4)-c

D (1)-c, (2)-b, (3)-d, (4)-a

💡 Short Explanation

�� Binding-energy curve reflects nuclear stability. �� Helium-4 shows exceptional stability. �� Saturation produces the plateau region.

✅ Correct Answer Explanation

1 → d : The smooth nature of the binding-energy curve indicates that nuclear force is largely independent of electric charge. 2 → b : The local peak at ⁴He is considered evidence of shell-like nuclear structure and exceptional stability. 3 → a : For very heavy nuclei, Coulomb repulsion becomes increasingly important, reducing binding energy per nucleon. 4 → c : The near constancy of E_bn for 30 < A < 170 arises from the saturation property of nuclear forces. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrectly matches smooth curve feature and heavy-nucleus behavior.
�� Option C → Incorrectly associates the helium peak with Coulomb-repulsion effects.
�� Option D → Incorrectly matches the plateau and heavy-nucleus regions.

🎯 Strategy

Used

Option Grouping

Application:

Match each curve feature with the physical principle responsible for it.

Final Logic:

Smooth curve → charge independence, He-4 peak → shell effect, heavy drop → Coulomb repulsion, plateau → saturation.

🧠 Memory Trick

Peak → Shell, Plateau → Saturation

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Binding Energy per Nucleon Curve and Nuclear Stability

20 Incorrect statement about shell evidence extrapolated from the binding energy curve

A The primary binding curve is overall smooth and slowly varying.

B Specific sharp peaks at ⁴He and ¹⁶O locally contradict the smooth trend.

C These local peaks imply that ¹⁶O and ⁴He are relatively more tightly bound than their immediate neighbours.

D The peaks overwhelmingly indicate that nuclear mass density is much higher in oxygen than in iron.

💡 Short Explanation

�� Local peaks indicate enhanced stability. �� Peaks support shell structure. �� Density information is not inferred.

✅ Correct Answer Explanation

The binding-energy curve primarily provides information about nuclear stability and binding energy per nucleon. Peaks at ⁴He and ¹⁶O indicate unusually stable nuclei, often associated with shell effects. These peaks do not imply that oxygen has a much higher nuclear mass density than iron. Therefore option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of the overall curve.
�� Option B → Correctly describes local deviations from the smooth trend.
�� Option C → Correctly interprets enhanced local stability.

🎯 Strategy

Used

Elimination

Application:

Separate stability information from density-related properties.

Final Logic:

Binding-energy peaks indicate stability, not unusually high density.

🧠 Memory Trick

Peak = Stable, Not Dense

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Shell Effects and Binding Energy Curve

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