CUET UG Physics Booster Test 2-Mass-Energy and Binding -

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CUET UG Physics Booster Test 2-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 2-Mass-Energy and Binding

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QUESTION 1 OF 20

Statements based on Einstein's special relativity applied to nuclear reactions:
1. Initial energy and final energy are equal provided the energy associated with mass is included.
2. The classical separate conservation law of energy fails if mass conversion is ignored.
3. Mass is treated as merely another form of energy.
4. Kinetic energy cannot ever be converted back into mass.

Answer Review

1 Statements based on Einstein's special relativity applied to nuclear reactions:
1. Initial energy and final energy are equal provided the energy associated with mass is included.
2. The classical separate conservation law of energy fails if mass conversion is ignored.
3. Mass is treated as merely another form of energy.
4. Kinetic energy cannot ever be converted back into mass.

A (1), (3), (4)

B (2), (3), (4)

C (1), (2), (3)

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

💡 Short Explanation

�� Mass and energy are equivalent. �� Total mass-energy is conserved. �� Kinetic energy can be converted into mass.

✅ Correct Answer Explanation

Statement 1 is correct because total energy remains conserved when mass energy is included. Statement 2 is correct because considering energy alone while ignoring mass conversion gives incorrect results in nuclear reactions. Statement 3 is correct because Einstein showed mass is a form of energy. Statement 4 is incorrect because kinetic energy can be converted into mass in particle creation processes. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Check each statement using the principle of mass-energy equivalence.

Final Logic:

Only statement 4 is incorrect.

🧠 Memory Trick

Mass ⇄ Energy

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

2 Incorrect statement about the equivalence relation

A It explains why the mass of an oxygen nucleus is less than the free masses of its constituents.

B The energy equivalent of mass m is given mathematically by E = mc².

C It demonstrates that chemical reactions involve mass defects a million times larger than nuclear reactions.

D It provides the fundamental theoretical basis for calculating binding energy.

💡 Short Explanation

�� Nuclear mass defects are much larger than chemical ones. �� E = mc² links mass and energy. �� Binding energy arises from mass defect.

✅ Correct Answer Explanation

Chemical reactions involve extremely small mass defects compared to nuclear reactions. Nuclear reactions release far greater energies because their mass defects are much larger. Therefore, statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct consequence of mass defect.
�� Option B → Correct statement of Einstein's relation.
�� Option D → Binding energy calculations depend on E = mc².

🎯 Strategy

Used

Elimination

Application:

Compare chemical and nuclear energy scales.

Final Logic:

Nuclear mass defects ≫ Chemical mass defects.

🧠 Memory Trick

Nuclear > Chemical

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

3 If a mass M gets completely converted to energy, the expression for the energy E released in terms of the velocity of light c is

A Mc²

B M²c

C M/c²

D Mc

💡 Short Explanation

�� Complete mass conversion follows Einstein's equation. �� Energy is proportional to mass. �� c² is the conversion factor.

✅ Correct Answer Explanation

Einstein's mass-energy equivalence relation is E = Mc² Thus, complete conversion of mass M produces energy Mc². Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect dimensional form.
�� Option C → Inverse relation is incorrect.
�� Option D → Missing one factor of c.

🎯 Strategy

Used

Definition Recall

Application:

Recall Einstein's mass-energy equation.

Final Logic:

E = Mc².

🧠 Memory Trick

Mass × c²

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

4 Converting 1 gram of matter completely releases an ________ amount of energy, which evaluates to ________ 9 × 10¹³ J.

A Insignificant, Equal to

B Enormous, exactly equal to

C Enormous, Proportional to

D Insignificant, Less than

💡 Short Explanation

�� 1 g = 10⁻³ kg. �� E = mc² gives huge energy. �� Value equals 9 × 10¹³ J.

✅ Correct Answer Explanation

Using E = mc² = (10⁻³)(3 × 10⁸)² = 9 × 10¹³ J This is an enormous amount of energy. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Energy is not insignificant.
�� Option C → The statement asks for the exact evaluated value.
�� Option D → Both terms are incorrect.

🎯 Strategy

Used

Substitution

Application:

Apply E = mc² for 1 g mass.

Final Logic:

1 g → 9 × 10¹³ J.

🧠 Memory Trick

1 g = 9 × 10¹³ J

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

5 Correct statements about nuclear mass difference:
1. The mass of ¹⁶₈O is experimentally found to be 15.99053 u.
2. The combined mass of 8 unbound protons and 8 unbound neutrons is greater than the oxygen nucleus.
3. The difference is mathematically termed the mass defect ΔM.
4. The mass defect implies the equivalent energy of the nucleus is less than the sum of the equivalent energies of constituents.

A (1), (2), (3)

B (2), (3), (4)

C (1), (2), (4)

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

💡 Short Explanation

�� Bound nuclei have lower mass. �� Missing mass is mass defect. �� Binding energy corresponds to missing mass.

✅ Correct Answer Explanation

Statement 1 is correct as given in NCERT data. Statement 2 is correct because free nucleons have greater combined mass. Statement 3 is correct because the difference is called mass defect. Statement 4 is correct because the nucleus possesses less mass-energy than the separated nucleons by an amount equal to binding energy. Therefore, all four statements are correct.

❓ Why Other Options Are Incorrect

�� Option A → Omits statement 4.
�� Option B → Omits statement 1.
�� Option C → Omits statement 3.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using the concept of mass defect.

Final Logic:

All four statements are correct.

🧠 Memory Trick

Missing Mass = ΔM

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

6 In the theoretical formula for mass defect [Zmₚ + (A − Z)mₙ] − M, the term (A − Z) specifically represents

A the total number of protons in the initial unbound state

B the total number of nucleons inside the nucleus

C the exact number of neutrons in the unbound state

D the combined mass of the electrons in the atom

💡 Short Explanation

�� A = Z + N. �� N = A − Z. �� N is neutron number.

✅ Correct Answer Explanation

Mass number A is the total number of nucleons. Atomic number Z is the number of protons. Therefore, N = A − Z which represents the number of neutrons. Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents Z.
�� Option B → Represents A.
�� Option D → Unrelated to mass defect formula.

🎯 Strategy

Used

Substitution

Application:

Use A = Z + N.

Final Logic:

A − Z = N.

🧠 Memory Trick

A − Z = N

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

7 Match List I with List II regarding energy conservation manifestations

List I	List II
1. Chemical reaction energy magnitude	a. 10¹⁴ J
2. Nuclear reaction energy magnitude	b. 10⁷ J
3. Fission of 1 kg Uranium	c. Scale of MeV
4. Burning of 1 kg coal	d. Scale of electron volts (eV)

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

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

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

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

💡 Short Explanation

�� Chemical reactions involve eV scale energies. �� Nuclear reactions involve MeV scale energies. �� Uranium fission releases enormous energy.

✅ Correct Answer Explanation

1 → d : Chemical reactions involve energies of the eV scale. 2 → c : Nuclear reactions involve MeV-scale energies. 3 → a : Fission of 1 kg uranium releases about 10¹⁴ J. 4 → b : Burning 1 kg coal releases about 10⁷ J. Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Chemical and nuclear energy scales are interchanged.
�� Option C → Coal and uranium energies are interchanged.
�� Option D → Multiple incorrect pairings.

🎯 Strategy

Used

Option Grouping

Application:

Match energy scales and magnitudes with their processes.

Final Logic:

Chemical → eV, Nuclear → MeV.

🧠 Memory Trick

Chemical-eV, Nuclear-MeV

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

8 In a hypothetical nuclear fission reaction, the total calculated gain in binding energy is 216 MeV. If this disintegration energy entirely appears equally as the kinetic energy of two identical fragments, what is the kinetic energy of one fragment?

A 216 MeV

B 108 MeV

C 432 MeV

D 8.5 MeV

💡 Short Explanation

�� Total energy = 216 MeV. �� Two identical fragments share energy equally. �� Divide by 2.

✅ Correct Answer Explanation

Energy of each fragment = 216/2 = 108 MeV Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Total energy, not individual energy.
�� Option C → Double the total energy.
�� Option D → Unrelated value.

🎯 Strategy

Used

Substitution

Application:

Split total kinetic energy equally between fragments.

Final Logic:

216 ÷ 2 = 108 MeV.

🧠 Memory Trick

Half of 216 = 108

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Energy Release in Fission

9 The strict algebraic conversion factor relation from Joules to electron volts (eV), where e = 1.602 × 10⁻¹⁹, can be represented as:

A E(in eV) = E(in J) × e

B E(in eV) = E(in J) / e

C E(in eV) = E(in J) × 931.5

D E(in eV) = E(in J) × c²

💡 Short Explanation

�� 1 eV = 1.602 × 10⁻¹⁹ J. �� Divide joules by e. �� Converts J to eV.

✅ Correct Answer Explanation

Since 1 eV = 1.602 × 10⁻¹⁹ J Energy in eV is E(eV) = E(J)/1.602 × 10⁻¹⁹ Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Gives incorrect conversion.
�� Option C → Related to atomic mass unit conversion.
�� Option D → Unrelated conversion.

🎯 Strategy

Used

Substitution

Application:

Use the definition of electron volt.

Final Logic:

J ÷ e = eV.

🧠 Memory Trick

J to eV → Divide by e

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

10 Statements regarding mass and energy units in nuclear physics:
1. 1 u = 1.6605 × 10⁻²⁷ kg.
2. 1 u converts to 931.5 MeV/c².
3. 1 u of energy is 1.4924 × 10⁻¹⁰ J.
4. The energy equivalent of 1 u is 931.5 eV.

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

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

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

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

💡 Short Explanation

�� 1 u has a standard SI value. �� 1 u corresponds to 931.5 MeV/c². �� Energy equivalent is 931.5 MeV, not 931.5 eV.

✅ Correct Answer Explanation

Statement 1 is correct because 1 u = 1.6605 × 10⁻²⁷ kg. Statement 2 is correct because 1 u corresponds to 931.5 MeV/c². Statement 3 is correct because the energy equivalent of 1 u is 1.4924 × 10⁻¹⁰ J. Statement 4 is incorrect because the energy equivalent is 931.5 MeV, not 931.5 eV. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 4.
�� Option C → Includes statement 4.
�� Option D → Includes statement 4.

🎯 Strategy

Used

Elimination

Application:

Verify each standard nuclear conversion constant.

Final Logic:

Statement 4 is incorrect by a factor of 10⁶.

🧠 Memory Trick

1 u = 931.5 MeV

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

11 Incorrect statement about structurally separating a nucleus

A If we separate a nucleus into nucleons, we must supply external energy equal to Eᵦ.

B A bound nucleus has greater total mass than the sum of the separated free nucleons.

C The nuclear binding energy remains a convenient measure of how well a nucleus is held together.

D This separation process is theoretically related to overcoming the mass defect.

💡 Short Explanation

�� Bound nuclei have lower mass. �� Mass defect produces binding energy. �� Separation requires energy input.

✅ Correct Answer Explanation

A bound nucleus always has a smaller mass than the sum of its free nucleons because part of the mass has been converted into binding energy. Therefore, statement B is incorrect. Statements A, C and D are correct descriptions of nuclear binding energy and mass defect.

❓ Why Other Options Are Incorrect

�� Option A → Correct; separation requires binding energy input.
�� Option C → Correct; binding energy measures nuclear stability.
�� Option D → Correct; mass defect and binding energy are directly related.

🎯 Strategy

Used

Elimination

Application:

Compare bound nuclear mass with the mass of free nucleons.

Final Logic:

Bound nucleus mass < Sum of free nucleon masses.

🧠 Memory Trick

Bound = Less Mass

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

12 When free nucleons assemble to form a bound nucleus, the total mass ________, and an amount of energy equivalent to the mass defect is ________.

A Increases, Absorbed

B Increases, Released

C Decreases, Released

D Decreases, Absorbed

💡 Short Explanation

�� Nucleons bind together. �� Mass defect appears. �� Energy is released.

✅ Correct Answer Explanation

During nucleus formation, some mass is converted into binding energy. Hence the total mass decreases and energy equal to ΔMc² is released. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Mass does not increase.
�� Option B → Increased mass is incorrect.
�� Option D → Energy is released, not absorbed.

🎯 Strategy

Used

Definition Recall

Application:

Apply the concept of mass defect.

Final Logic:

Mass decreases → Energy released.

🧠 Memory Trick

Less Mass, More Energy

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

13 Statements regarding the ratio Eᵦ/A:
1. It is defined as the binding energy per nucleon.
2. It effectively acts as an average energy per nucleon needed for total separation.
3. It reaches a distinct maximum value of about 8.75 MeV in the curve.
4. It linearly increases indefinitely with mass number A for all heavy nuclei.

A (1), (2), (3)

B (2), (3), (4)

C (1), (3), (4)

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

💡 Short Explanation

�� Eᵦ/A is binding energy per nucleon. �� Maximum occurs near iron. �� Heavy nuclei do not show continuous increase.

✅ Correct Answer Explanation

Statement 1 is correct because Eᵦ/A is the definition of binding energy per nucleon. Statement 2 is correct because it represents the average energy required per nucleon to separate the nucleus. Statement 3 is correct because the curve reaches a maximum value of about 8.75 MeV near ⁵⁶Fe. Statement 4 is incorrect because Eᵦ/A decreases slightly for heavy nuclei. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 4.
�� Option C → Includes statement 4.
�� Option D → Includes statement 4.

🎯 Strategy

Used

Elimination

Application:

Analyze the binding-energy-per-nucleon curve.

Final Logic:

Iron peak disproves indefinite increase.

🧠 Memory Trick

Eᵦ/A → Peak at Fe

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

14 Statements about Eᵦₙ as a measure of stability:
1. Eᵦₙ is practically constant for nuclei spanning 30 < A < 170.
2. The constancy is a consequence of the short-range nature of the nuclear force.
3. The nuclear force influences only nucleons relatively close to it (saturation property).
4. Eᵦₙ drops significantly for A > 170.

A (1), (2), (3)

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

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

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

💡 Short Explanation

�� Middle nuclei show a plateau. �� Nuclear force is short-ranged. �� Heavy nuclei show decreasing Eᵦₙ.

✅ Correct Answer Explanation

Statement 1 is correct because Eᵦₙ remains nearly constant for intermediate mass nuclei. Statement 2 is correct because saturation arises from the short-range nature of nuclear forces. Statement 3 is correct because each nucleon interacts mainly with nearby nucleons. Statement 4 is correct because Eᵦₙ decreases beyond A ≈ 170 due to increasing Coulomb repulsion. Therefore, all four statements are correct.

❓ Why Other Options Are Incorrect

�� Option A → Omits statement 4.
�� Option B → Omits statement 2.
�� Option C → Omits statement 1.

🎯 Strategy

Used

Elimination

Application:

Check each statement using the binding-energy curve and saturation property.

Final Logic:

All four statements are valid.

🧠 Memory Trick

Middle Flat, Heavy Fall

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

15 The constancy of the binding energy per nucleon in the mass range 30 < A < 170 is primarily a direct physical consequence of

A the saturation property and short-range nature of the strong nuclear force

B The Coulomb repulsion between protons overwhelmingly dominating the nuclear force

C the total absence of neutrons in this specific middle mass region

D the gravitational forces becoming structurally significant for these nuclei

💡 Short Explanation

�� Nuclear force is short-ranged. �� Each nucleon interacts with nearby nucleons. �� Binding per nucleon becomes nearly constant.

✅ Correct Answer Explanation

The saturation property means a nucleon interacts strongly only with its nearest neighbors. Because of this, the average binding energy per nucleon remains approximately constant over a large range of A. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Coulomb force reduces stability.
�� Option C → Neutrons are present.
�� Option D → Gravity is negligible at nuclear scales.

🎯 Strategy

Used

Definition Recall

Application:

Recall the origin of saturation property.

Final Logic:

Short-range force → Constant Eᵦₙ.

🧠 Memory Trick

Short Range → Saturation

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

16 The maximum binding energy per nucleon occurs at Iron (⁵⁶Fe). If its Eᵦₙ is exactly 8.75 MeV, what is its total approximate binding energy?

A 8.75 MeV

B 490 MeV

C 8.0 MeV

D 238 MeV

💡 Short Explanation

�� Total binding energy = Eᵦₙ × A. �� A = 56. �� Eᵦₙ = 8.75 MeV.

✅ Correct Answer Explanation

Total binding energy Eᵦ = 8.75 × 56 = 490 MeV Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Per nucleon value only.
�� Option C → Plateau value, not total energy.
�� Option D → Incorrect calculation.

🎯 Strategy

Used

Substitution

Application:

Use Eᵦ = A × Eᵦₙ.

Final Logic:

56 × 8.75 = 490.

🧠 Memory Trick

Fe-56 × 8.75

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

17 Statements regarding light nuclei:
1. For A ≤ 10, joining lighter nuclei forms a more tightly bound heavier nucleus.
2. The binding energy per nucleon of the fused heavier nucleus is strictly more.
3. The final system is much less tightly bound than the initial isolated systems.
4. Massive energy is released in such an advantageous fusion process.

A (1), (2), (3)

B (1), (2), (4)

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

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

💡 Short Explanation

�� Fusion increases Eᵦₙ. �� Final nucleus is more stable. �� Energy is released.

✅ Correct Answer Explanation

Statement 1 is correct because fusion produces a more stable nucleus. Statement 2 is correct because Eᵦₙ increases. Statement 3 is incorrect because the final nucleus is more tightly bound. Statement 4 is correct because increased binding energy results in energy release. Therefore, statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 3.
�� Option C → Includes statement 3.
�� Option D → Includes statement 3.

🎯 Strategy

Used

Elimination

Application:

Compare binding energies before and after fusion.

Final Logic:

Fusion → Higher Eᵦₙ → Energy Release.

🧠 Memory Trick

Light Join, Energy Out

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

18 Match List I with List II regarding nuclear phenomena based on the binding energy curve

List I	List II
1. Fission of A = 240 into two A = 120 fragments	a. Energy is released due to tighter binding of the final heavier nucleus
2. Fusion of two light nuclei (A ≤ 10)	b. Explained primarily by the saturation property of the nuclear force
3. Constancy of Eᵦₙ in the middle region	c. Nucleons get more tightly bound, energy released in fission
4. The main reason fusion occurs in stars	d. Relies on overcoming the Coulomb barrier at high temperatures

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

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

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

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

💡 Short Explanation

�� Fission and fusion release energy through increased binding. �� Saturation explains the plateau. �� Stellar fusion requires overcoming Coulomb repulsion.

✅ Correct Answer Explanation

1 → c : Heavy nucleus fission increases binding per nucleon and releases energy. 2 → a : Fusion produces a more tightly bound heavier nucleus. 3 → b : Middle-region constancy arises from saturation of nuclear forces. 4 → d : High stellar temperatures help overcome the Coulomb barrier. Thus, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Fission and stellar fusion are mismatched.
�� Option C → Multiple incorrect pairings.
�� Option D → Incorrect assignment of physical processes.

🎯 Strategy

Used

Option Grouping

Application:

Match each phenomenon with its physical explanation.

Final Logic:

Fission → Tighter Binding, Fusion → Heavier Stable Nucleus, Plateau → Saturation.

🧠 Memory Trick

Fission-Bind, Fusion-Grow, Plateau-Saturate

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Binding Energy Curve, Fusion and Fission

19 Let Eᵦ(He) be the total binding energy of ⁴He and Eᵦ(Li) be the total binding energy of ⁶Li. The binding curve shows a sharp dip for Li compared to He. The strict relationship of their binding energies per nucleon is:

A Eᵦ(He)/4 > Eᵦ(Li)/6

B Eᵦ(He)/4 < Eᵦ(Li)/6

C Eᵦ(He)/4 = Eᵦ(Li)/6

D Eᵦ(He) = Eᵦ(Li)

💡 Short Explanation

�� ⁴He is exceptionally stable. �� ⁶Li lies in a dip region. �� Binding energy per nucleon is higher for ⁴He.

✅ Correct Answer Explanation

The binding-energy curve shows a local peak at ⁴He. Therefore, the binding energy per nucleon of helium exceeds that of lithium. Hence, Eᵦ(He)/4 > Eᵦ(Li)/6

❓ Why Other Options Are Incorrect

�� Option B → Opposite relation.
�� Option C → Values are not equal.
�� Option D → Total binding energies are not equal.

🎯 Strategy

Used

Curve Interpretation

Application:

Use the local peak at ⁴He.

Final Logic:

Peak nucleus → Greater Eᵦ/A.

🧠 Memory Trick

He Peak > Li

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

20 The noticeable presence of peaks at ⁴He and ¹⁶O suggests that they are exceptionally ________ against separation, providing key evidence for an atom-like ________ structure in nuclei.

A Unstable, Uniform

B Stable, Shell

C Unstable, Shell

D Stable, Uniform

💡 Short Explanation

�� Peaks indicate extra stability. �� Shell effects produce special stability. �� Similar to atomic shell structure.

✅ Correct Answer Explanation

The peaks at ⁴He and ¹⁶O indicate unusually large binding energy per nucleon. This means they are especially stable against separation. Such peaks provide evidence for shell structure within nuclei. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Peaks indicate stability, not instability.
�� Option C → First term is incorrect.
�� Option D → Uniform structure does not explain the peaks.

🎯 Strategy

Used

Definition Recall

Application:

Associate local maxima with enhanced nuclear stability.

Final Logic:

Peak → Stable → Shell Structure.

🧠 Memory Trick

Peaks Mean Shells

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

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