CUET UG Physics Booster Test 3- Nuclear Forces and Stability -

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CUET UG Physics Booster Test 3- Nuclear Forces and Stability

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3- Nuclear Forces and Stability

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QUESTION 1 OF 20

Match List I with List II regarding internal nuclear force dynamics

List I	List II
1. Short-range Strong Nuclear Force	a. Approximately 8 MeV per nucleon for middle mass range
2. Coulomb Force Repulsion	b. Inherently opposes the tight binding of the nucleus
3. Nuclear Binding Energy	c. Essential to bind both protons and neutrons tightly
4. Force Magnitude Ratio	d. Nuclear force dominates over electrostatic completely

Answer Review

1 Match List I with List II regarding internal nuclear force dynamics

List I	List II
1. Short-range Strong Nuclear Force	a. Approximately 8 MeV per nucleon for middle mass range
2. Coulomb Force Repulsion	b. Inherently opposes the tight binding of the nucleus
3. Nuclear Binding Energy	c. Essential to bind both protons and neutrons tightly
4. Force Magnitude Ratio	d. Nuclear force dominates over electrostatic completely

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

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

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

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

💡 Short Explanation

�� Strong force binds nucleons. �� Coulomb force opposes binding. �� Binding energy is a few MeV per nucleon.

✅ Correct Answer Explanation

1 → c : Strong nuclear force binds protons and neutrons tightly. 2 → b : Coulomb repulsion opposes nuclear binding. 3 → a : Binding energy is about 8 MeV per nucleon for medium-mass nuclei. 4 → d : Strong nuclear force dominates electrostatic repulsion inside the nucleus. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Strong force and binding energy are mismatched.
�� Option C → Coulomb force is incorrectly matched.
�� Option D → Multiple pairings are incorrect.

🎯 Strategy

Used

Option Grouping

Application:

Match each nuclear concept with its physical role.

Final Logic:

Strong → Binding, Coulomb → Opposition, Binding Energy → ~8 MeV, Strong > Coulomb.

🧠 Memory Trick

Strong Binds, Coulomb Pushes

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

2 Correct statements interpreting the interplay of constituent forces in heavy nuclei
1. As overall mass increases, relatively more neutrons are needed to help overcome accumulating proton repulsion.
2. The Coulomb force plays virtually no role in defining the stability boundary of heavy nuclei.
3. Coulomb repulsion acts across the whole nucleus, unlike the short-ranged strong nuclear force.
4. The strong binding force must consistently counteract the additive Coulomb repulsion.

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

B (1), (2)

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

D (1), (4)

💡 Short Explanation

�� Heavy nuclei require extra neutrons. �� Coulomb force acts throughout the nucleus. �� Strong force is short-ranged.

✅ Correct Answer Explanation

Statement 1 is correct because heavy nuclei require more neutrons for stability. Statement 2 is incorrect because Coulomb repulsion strongly influences stability. Statement 3 is correct because Coulomb force is long-ranged. Statement 4 is correct because strong force must overcome proton repulsion. Therefore, statements 1, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 2, which is incorrect.
�� Option C → Includes statement 2, which is incorrect.
�� Option D → Omits statement 3, which is correct.

🎯 Strategy

Used

Elimination

Application:

Compare long-range Coulomb force with short-range strong force.

Final Logic:

Only statements 1, 3 and 4 correctly describe heavy nuclei.

🧠 Memory Trick

Heavy Nuclei Need More Neutrons

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Stability of Heavy Nuclei

3 Analytical statements on the potential energy minimum (r₀) between two nucleons
1. It strictly occurs at approximately 0.8 fm.
2. At exactly r₀, the attractive and repulsive force gradients balance to zero.
3. Forcing the nucleons closer than r₀ requires massive external energy.
4. At r₀, the nuclear force becomes infinitely strong.

A (2), (3), (4)

B (1), (4)

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

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

💡 Short Explanation

�� r₀ corresponds to minimum potential energy. �� Net force is zero at equilibrium. �� Compression below r₀ faces strong repulsion.

✅ Correct Answer Explanation

Statement 1 is correct because r₀ is approximately 0.8 fm. Statement 2 is correct because net force is zero at the minimum-energy point. Statement 3 is correct because repulsive forces increase rapidly below r₀. Statement 4 is incorrect because the force does not become infinite. 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:

Interpret the equilibrium point on the potential-energy curve.

Final Logic:

At r₀, energy is minimum and force is balanced.

🧠 Memory Trick

r₀ = Balance Point

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

4 Incorrect statement regarding nucleon interactions when forced below the 0.8 fm distance,

A The nuclear force transitions abruptly to become strongly repulsive.

B The potential energy rises sharply into positive values as separation decreases.

C This inherent repulsion prevents the atomic nucleus from collapsing in on itself.

D The potential energy drops smoothly to its absolute lowest possible negative value.

💡 Short Explanation

�� Repulsive core exists below 0.8 fm. �� Potential energy rises sharply. �� Repulsion prevents collapse.

✅ Correct Answer Explanation

When nucleons are pushed below approximately 0.8 fm, a strong repulsive force develops. Consequently, the potential energy increases sharply rather than decreasing further. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct description of the repulsive core.
�� Option B → Correct behaviour of the potential-energy curve.
�� Option C → Correct role of short-range repulsion.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement contradicting the nucleon potential-energy graph.

Final Logic:

Below 0.8 fm, potential energy rises, not falls.

🧠 Memory Trick

Too Close → Repel

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Repulsive Core of Nuclear Force

5 The exceptionally short-range nature of the nuclear force directly implies that

A Adding a nucleon to a large nucleus proportionally increases the binding energy of all other distant nucleons

B An individual nucleon essentially interacts only with its nearest neighbours, leading to energy saturation

C The force follows a strict inverse-square dependence globally

D The force works effectively over inter-atomic distances

💡 Short Explanation

�� Nuclear force acts over a few femtometres. �� Only nearby nucleons interact strongly. �� Saturation results naturally.

✅ Correct Answer Explanation

Because the nuclear force is highly short-ranged, a nucleon interacts significantly only with nearby neighbours. This limited interaction causes the binding energy per nucleon to become nearly constant in medium and heavy nuclei. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Distant nucleons are not significantly affected.
�� Option C → Nuclear force does not obey a simple inverse-square law.
�� Option D → Nuclear force is ineffective at interatomic distances.

🎯 Strategy

Used

Elimination

Application:

Use the concept of short-range interaction.

Final Logic:

Short range produces neighbour interaction and saturation.

🧠 Memory Trick

Short Range → Saturation

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

6 Given a nucleus with a radius expressed as R = 1.2A^(1/3) fm, the sphere of influence of the nuclear force for a single inner nucleon mathematically extends to a distance of:

A A few femtometres, covering only a tiny local fraction of a heavy nucleus

B Hundreds of femtometres, covering the entirety of the macroscopic atom

C Exactly 10⁻¹⁰ m irrespective of the value of A

D A uniform 1.2 fm spanning the entire atomic volume

💡 Short Explanation

�� Nuclear force is short-ranged. �� Influence extends only a few fm. �� Large nuclei contain many unaffected distant nucleons.

✅ Correct Answer Explanation

The strong nuclear force acts effectively only over a few femtometres. Even inside a large nucleus, a nucleon interacts significantly only with nearby neighbours. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Nuclear force is not long-ranged.
�� Option C → 10⁻¹⁰ m corresponds to atomic scales.
�� Option D → 1.2 fm is not the range throughout the atom.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Compare nuclear-force range with atomic dimensions.

Final Logic:

Strong force influences only a few femtometres.

🧠 Memory Trick

Strong Force = Few fm

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

7 If an interior nucleon can conceptually have a maximum of exactly p neighbours actively within the functional range of the nuclear force, its total binding energy contribution is analytically proportional to:

A p²

B p

C 1/p

D p/2

💡 Short Explanation

�� Each neighbour contributes attraction. �� Total contribution is additive. �� Saturation limits neighbour count.

✅ Correct Answer Explanation

If each neighbour contributes approximately the same interaction energy, the total binding contribution of a nucleon is proportional to the number of neighbouring nucleons within range. Hence, total binding energy contribution ∝ p. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Interaction energy is not proportional to p².
�� Option C → Inverse dependence is incorrect.
�� Option D → No factor of 1/2 arises here.

🎯 Strategy

Used

Substitution

Application:

Sum equal interaction contributions.

Final Logic:

More neighbours → proportionally greater binding contribution.

🧠 Memory Trick

More Neighbours → More Binding

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

8 Because the vast majority of nucleons in a sufficiently large nucleus are deeply embedded inside rather than exposed on the surface,

A the total binding energy per nucleon fluctuates drastically across the middle mass isotopes

B the binding energy per nucleon remains practically constant as A increases

C the surface nucleons unilaterally dictate the entire structural volume of the atom

D the nucleus rapidly and spontaneously undergoes beta decay to shed surface mass

💡 Short Explanation

�� Most nucleons are interior nucleons. �� Interior nucleons have similar environments. �� Binding energy per nucleon becomes nearly constant.

✅ Correct Answer Explanation

In large nuclei, most nucleons are surrounded by approximately the same number of neighbours. Due to saturation of nuclear forces, the binding energy per nucleon becomes nearly constant over a wide mass range. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Binding energy per nucleon changes gradually.
�� Option C → Surface nucleons do not dominate structure.
�� Option D → Beta decay is unrelated to surface nucleons.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Apply saturation-property concepts.

Final Logic:

Similar interior environments produce nearly constant binding energy per nucleon.

🧠 Memory Trick

Large Nucleus → Constant E/A

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

9 The remarkable equality of strong forces between varied nucleon pairs rigorously demonstrates its __________, a trait that sets it fundamentally apart from the purely __________ force.

A Charge independence, Coulomb

B Infinite range, Gravitational

C Mathematical simplicity, Electrostatic

D Induced instability, Electromagnetic

💡 Short Explanation

�� Strong force is nearly charge independent. �� p-p, n-n and p-n interactions are similar. �� Coulomb force depends on charge.

✅ Correct Answer Explanation

The near equality of strong-force interactions among different nucleon pairs demonstrates charge independence. This contrasts with Coulomb force, which depends directly on electric charge. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Charge independence is the relevant property, not infinite range.
�� Option C → Nuclear force does not possess a simple mathematical form.
�� Option D → Induced instability is unrelated.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate nucleon-force equality to charge independence.

Final Logic:

Equal nucleon interactions imply charge independence.

🧠 Memory Trick

Equal Forces → Ignore Charge

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Charge Independence of Nuclear Force

10 Which exact pairing of equivalent interaction strengths conclusively confirms that the strong nuclear force entirely disregards electric charge?

A Neutron-electron force ≈ Proton-electron force

B Proton-proton force ≈ Neutron-neutron force

C Alpha-particle force ≈ Photon force

D Positron force ≈ Electron force

💡 Short Explanation

�� Charge independence means force strength is nearly unchanged. �� p-p and n-n interactions are nearly equal. �� Electric charge does not significantly affect strong force.

✅ Correct Answer Explanation

Experimental observations show that the strong nuclear force between proton-proton pairs is approximately equal to that between neutron-neutron pairs. This equality demonstrates charge independence. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Electron interactions do not establish nuclear-force charge independence.
�� Option C → Unrelated physical entities.
�� Option D → Does not test nucleon interactions.

🎯 Strategy

Used

Elimination

Application:

Identify the comparison directly involving nucleon pairs.

Final Logic:

p-p ≈ n-n confirms charge independence.

🧠 Memory Trick

p-p ≈ n-n

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Charge Independence of Nuclear Force

11 Match List I with List II regarding the analytical hierarchy of fundamental forces in the nucleus

List I	List II
1. Strong Nuclear Force	a. Becomes a strongly repulsive core
2. Electromagnetic Force	b. Overcomes proton-proton electrostatic repulsion
3. Gravitational Force	c. The absolute weakest interaction, essentially negligible
4. Force behavior at r < 0.8 fm	d. Causes long-range repulsion between all protons

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

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

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

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

💡 Short Explanation

�� Strong force binds nucleons. �� Electromagnetic force causes proton repulsion. �� Gravity is negligible at nuclear scales.

✅ Correct Answer Explanation

1 → b : Strong nuclear force overcomes proton-proton repulsion. 2 → d : Electromagnetic force produces long-range repulsion between protons. 3 → c : Gravity is the weakest interaction inside the nucleus. 4 → a : For r < 0.8 fm, the nuclear force becomes strongly repulsive. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Gravity and strong force are incorrectly matched.
�� Option B → Strong force and repulsive-core behavior are interchanged.
�� Option D → Multiple pairings are physically incorrect.

🎯 Strategy

Used

Option Grouping

Application:

Associate each force with its characteristic role.

Final Logic:

Strong → binding, EM → repulsion, Gravity → weakest, r < 0.8 fm → repulsive core.

🧠 Memory Trick

Strong Binds, EM Pushes

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Fundamental Forces in the Nucleus

12 Correct analytical statements explaining why progressively heavier nuclei strictly require proportionally more neutrons
1. Added neutrons increase the total strong nuclear force without contributing any extra Coulomb repulsion.
2. Cumulative Coulomb repulsion between protons acts long-range across the entire nuclear volume.
3. The strong force quickly saturates, preventing a proton from simply binding all other distant protons.
4. Neutrons provide a necessary negative structural charge to perfectly cancel the protons.

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

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

C (1), (4)

D (3), (4)

💡 Short Explanation

�� Extra neutrons increase nuclear attraction. �� Coulomb repulsion grows with proton number. �� Strong force is short-ranged and saturates.

✅ Correct Answer Explanation

Statement 1 is correct because neutrons contribute to strong-force attraction without increasing electrical repulsion. Statement 2 is correct because Coulomb force acts throughout the nucleus. Statement 3 is correct because saturation limits the number of nucleons that one proton can strongly interact with. Statement 4 is incorrect because neutrons are electrically neutral. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using nuclear-stability concepts.

Final Logic:

Heavy nuclei require extra neutrons because Coulomb repulsion grows while strong force saturates.

🧠 Memory Trick

More Protons → More Neutrons

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Stability of Heavy Nuclei

13 Incorrect statement regarding the advanced mathematical modeling of nuclear forces

A They cannot be adequately described by an equation as purely simple as F = Gm₁m₂/r².

B The mathematical complexity arises partly because it shifts sharply from attractive to repulsive at varying short ranges.

C A universally precise and mathematically simple formula identically mirrors Coulomb's law for strong forces.

D It does not mathematically depend on the specific electric charge variable of the interacting nucleons.

💡 Short Explanation

�� Strong force is mathematically complex. �� It is not described by a simple inverse-square law. �� It changes from attraction to repulsion.

✅ Correct Answer Explanation

Unlike gravitational and electrostatic forces, the strong nuclear force does not possess a simple universally applicable mathematical formula identical to Coulomb's law. Its behavior depends strongly on distance and includes both attractive and repulsive regions. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement regarding complexity.
�� Option B → Correct explanation of the mathematical difficulty.
�� Option D → Correct because strong force is largely charge independent.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement falsely attributing Coulomb-like simplicity to the strong force.

Final Logic:

Strong force has no simple Coulomb-type formula.

🧠 Memory Trick

Strong Force ≠ Coulomb Law

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

14 When critically analyzing the precise potential energy plot of two interacting nucleons

A The energy value becomes strictly zero for all possible distances greater than 0.8 fm.

B The sharp, steep upward curve for r < 0.8 fm directly indicates massive strong repulsion.

C The potential energy hits its maximum positive peak exactly at r = 0.8 fm.

D The geometrical curve is perfectly symmetric around the r = 0.8 fm axis.

💡 Short Explanation

�� Potential energy rises sharply below 0.8 fm. �� This indicates a repulsive core. �� Nucleons cannot collapse into one another.

✅ Correct Answer Explanation

For separations below approximately 0.8 fm, the nucleon potential-energy curve rises very steeply. This behavior corresponds to the strong repulsive core of the nuclear force. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Potential energy is not exactly zero for all r > 0.8 fm.
�� Option C → r = 0.8 fm corresponds approximately to minimum potential energy.
�� Option D → The curve is not symmetric.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Interpret the shape of the nucleon potential-energy curve.

Final Logic:

Steep upward rise below 0.8 fm signifies strong repulsion.

🧠 Memory Trick

Below 0.8 → Sharp Rise

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

15 For lighter nuclei, prolonged stability is generally maintained when the operational ratio of neutrons to protons is meticulously around 1:1, largely because

A the total Coulomb repulsion is relatively minimal, allowing the strong force to dominate easily

B light nuclei intrinsically possess extremely high Coulomb barriers stopping all interaction

C they naturally contain too many excess neutrons causing constant fission

D they undergo alpha decay too rapidly to form heavier elements

💡 Short Explanation

�� Light nuclei contain fewer protons. �� Coulomb repulsion is comparatively small. �� N ≈ Z provides stability.

✅ Correct Answer Explanation

In light nuclei, the number of protons is small, so Coulomb repulsion remains relatively weak. Consequently, a neutron-to-proton ratio close to 1:1 is sufficient for stable nuclear binding. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Light nuclei do not possess extremely high Coulomb barriers.
�� Option C → Light nuclei are not characterized by excess neutrons causing fission.
�� Option D → Most light stable nuclei do not undergo rapid alpha decay.

🎯 Strategy

Used

Elimination

Application:

Relate stability to Coulomb repulsion and strong-force dominance.

Final Logic:

Low proton count allows N/Z ≈ 1.

🧠 Memory Trick

Light Nuclei → N = Z

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Stability of Light Nuclei

16 Analytical statements on the stability threshold for heavy nuclei
1. The fundamentally required N/Z ratio increases steadily to about 3:2.
2. Extra incorporated neutrons actively help dilute the accumulating Coulomb repulsion.
3. The stable ratio remains strictly fixed at 1:1 all the way up to A = 240.
4. Extremely heavy nuclei may spontaneously undergo decay to restore optimal stability.

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

B (2), (3)

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

D (2), (4)

💡 Short Explanation

�� Heavy nuclei require extra neutrons. �� N/Z ratio increases with mass number. �� Very heavy nuclei often decay spontaneously.

✅ Correct Answer Explanation

Statement 1 is correct because the stability ratio increases toward about 3:2. Statement 2 is correct because neutrons contribute strong-force attraction without electrical repulsion. Statement 3 is incorrect because heavy nuclei are not stable at N/Z = 1. Statement 4 is correct because many very heavy nuclei are radioactive. 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 → Omits statement 1, which is correct.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using the stability-band concept.

Final Logic:

Heavy nuclei require larger N/Z ratios and often decay.

🧠 Memory Trick

Heavy → 3:2 → Decay

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Stability of Heavy Nuclei

17 Nuclei populated with an __________ of either specific protons or neutrons deviate sharply from the defined stability band, rendering them highly susceptible to spontaneous __________.

A Equilibrium, Fission

B Equality, Stability

C Absence, Fusion

D Excess, Radioactivity

💡 Short Explanation

�� Stable nuclei follow the stability band. �� Excess nucleons create imbalance. �� Imbalance causes radioactive decay.

✅ Correct Answer Explanation

Nuclei containing an excess of neutrons or protons relative to the stable ratio become unstable. Such instability generally leads to radioactive decay processes. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Equilibrium does not cause fission.
�� Option B → Equality promotes stability, not radioactivity.
�� Option C → Absence of nucleons does not imply fusion.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Associate stability-band deviation with radioactivity.

Final Logic:

Excess nucleons lead to radioactive instability.

🧠 Memory Trick

Excess → Decay

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Radioactivity and Stability

18 Match List I with List II regarding the precise nature of radioactive decay triggers

List I	List II
1. Alpha decay mechanism	a. Deviating from stable nucleon ratio (excess nucleons)
2. Beta decay mechanism	b. Direct emission of a heavily bound helium nucleus
3. Gamma decay mechanism	c. Secondary emission of high energy electromagnetic photons
4. Primary instability cause	d. Emission of elemental electrons or positrons

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

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

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

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

💡 Short Explanation

�� Alpha decay emits helium nuclei. �� Beta decay emits electrons or positrons. �� Gamma decay emits photons.

✅ Correct Answer Explanation

1 → b : Alpha decay emits a helium nucleus. 2 → d : Beta decay emits electrons or positrons. 3 → c : Gamma decay emits high-energy photons. 4 → a : Instability often arises from deviation from the stable neutron-proton ratio. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Alpha and instability matches are incorrect.
�� Option C → Multiple decay modes are mismatched.
�� Option D → All principal decay processes are incorrectly matched.

🎯 Strategy

Used

Option Grouping

Application:

Match each decay type with its emitted particle.

Final Logic:

Alpha → He, Beta → e±, Gamma → Photon.

🧠 Memory Trick

α-He, β-e, γ-Photon

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Radioactive Decay

19 Statistically, out of the entire catalog of known isotopes spanning the universe and laboratory creations, what is the approximate fractional representation of perfectly stable ones?

A ~ 1/10

B ~ 1/2

C ~ 9/10

D ~ 1/100

💡 Short Explanation

�� Most isotopes are radioactive. �� Stable isotopes are comparatively few. �� Approximately 10% are stable.

✅ Correct Answer Explanation

Among all known isotopes, only about one-tenth are stable. The remaining majority are radioactive and undergo decay. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Greatly overestimates stable isotopes.
�� Option C → Represents approximately the unstable fraction.
�� Option D → Underestimates the stable fraction.

🎯 Strategy

Used

Memory-Based Recall

Application:

Recall the approximate distribution of stable isotopes.

Final Logic:

Stable isotopes comprise about 10% of known isotopes.

🧠 Memory Trick

Stable ≈ 10%

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Stable Isotopes

20 When physicists artificially produce unstable isotopes by bombarding stable targets, they add to the pool of known elements. If it is established that 90% of known isotopes are currently unstable, and theoretically there are roughly 3000 known isotopes documented, approximately how many of these are stable species?

A 30

B 1500

C 2700

D 300

💡 Short Explanation

�� Stable isotopes ≈ 10%. �� Total isotopes = 3000. �� Calculate 10% of 3000.

✅ Correct Answer Explanation

If 90% are unstable, then 10% are stable. Stable isotopes: = 10% × 3000 = 0.10 × 3000 = 300 Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to only 1%.
�� Option B → Corresponds to 50%.
�� Option C → Corresponds approximately to 90%.

🎯 Strategy

Used

Substitution

Application:

Use percentage calculation directly.

Final Logic:

10% of 3000 = 300.

🧠 Memory Trick

10% of 3000 → Move Decimal → 300

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Stable and Unstable Isotopes

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