CUET UG Physics Booster Test 2- Rutherford Nuclear Model -

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CUET UG Physics Booster Test 2- Rutherford Nuclear Model

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2- Rutherford Nuclear Model

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QUESTION 1 OF 20

Alpha particle scattering confirms the planetary analogy.
Statements:
1. The massive central sun correlates with the atom's entire positive charge concentrated at the centre.
2. The sun's gravity is mathematically identical to the electrostatic force exerted by the nucleus.
3. The solar system has mostly empty space; similarly, electrons are at distances 10,000 to 100,000 times the nuclear size.
4. The planetary orbits are governed by electrostatic repulsion.

Answer Review

1 Alpha particle scattering confirms the planetary analogy.
Statements:
1. The massive central sun correlates with the atom's entire positive charge concentrated at the centre.
2. The sun's gravity is mathematically identical to the electrostatic force exerted by the nucleus.
3. The solar system has mostly empty space; similarly, electrons are at distances 10,000 to 100,000 times the nuclear size.
4. The planetary orbits are governed by electrostatic repulsion.

A (1), (3)

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

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

D (1), (4)

💡 Short Explanation

�� Nucleus corresponds to the Sun. �� Most atomic volume is empty space. �� Gravity and electrostatic force are not identical.

✅ Correct Answer Explanation

Statement 1 is correct because Rutherford compared the nucleus to the central Sun containing most of the mass and positive charge. Statement 3 is correct because electrons are located at distances about 10⁴–10⁵ times the nuclear size, making the atom mostly empty space. Statements 2 and 4 are incorrect. Gravity and electrostatic force are different interactions, and planetary motion is not governed by electrostatic repulsion. Therefore, statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Identify valid features of the planetary analogy.

Final Logic:

Only statements 1 and 3 correctly describe Rutherford's analogy.

🧠 Memory Trick

Sun → Nucleus, Space → Atom

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford's Planetary Model

2 Incorrect statement about the revolving planets analogy

A Electrons revolve around the nucleus just as planets revolve around the sun.

B The intense electric field of the electrons severely deflects incoming alpha particles.

C The vast distance of electrons from the nucleus makes most of the atom an empty space.

D The atomic electrons are so light they do not appreciably affect incoming alpha particles.

💡 Short Explanation

�� Nucleus dominates scattering. �� Electrons are very light. �� Electron fields produce negligible deflection.

✅ Correct Answer Explanation

The large-angle scattering observed by Rutherford is caused by the intense electric field of the positively charged nucleus, not by electrons. Electrons are too light to significantly alter alpha-particle trajectories. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct planetary analogy.
�� Option C → Correct description of atomic emptiness.
�� Option D → Correct explanation of electron negligibility.

🎯 Strategy

Used

Odd One Out

Application:

Identify the statement inconsistent with Rutherford scattering.

Final Logic:

Large deflections originate from the nucleus, not electrons.

🧠 Memory Trick

Scatter → Nucleus, Not Electrons

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Scattering Interpretation

3 Correct statements about concentrated mass
1. The rebound of a small fraction of alpha particles indicates concentrated mass.
2. The target nucleus is assumed to remain stationary because it is heavily concentrated and about 50 times heavier than an alpha particle.
3. The nucleus occupies the entire atomic volume to provide the concentrated mass.
4. Most of the atom's mass tightly resides at its centre in a small volume.

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

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

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

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

💡 Short Explanation

�� Backscattering indicates concentrated mass. �� Gold nucleus is much heavier. �� Mass is concentrated in a tiny nucleus.

✅ Correct Answer Explanation

Statements 1, 2 and 4 are correct. Large-angle scattering and rebound require a massive concentrated centre. The gold nucleus is treated as stationary because it is much heavier than the alpha particle. Most atomic mass is concentrated in the nucleus. Statement 3 is incorrect because the nucleus occupies only a tiny fraction of atomic volume.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Separate nuclear properties from atomic volume properties.

Final Logic:

Mass is concentrated, but volume is not.

🧠 Memory Trick

Tiny Volume, Huge Mass

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Nuclear Mass Concentration

4 The large angle deflection of an alpha particle

A is due to its direct penetration and absorption into the positive core

B requires the positive charge to be tightly concentrated at the centre to provide an intense electric field

C is a result of multiple sequential scatterings by different positive cores

D proves the presence of negatively charged plum pudding distribution

💡 Short Explanation

�� Large deflection needs strong repulsion. �� Strong field exists near nucleus. �� Supports nuclear model.

✅ Correct Answer Explanation

Large-angle scattering can occur only when the alpha particle approaches a highly concentrated positive charge. This produces a strong Coulomb repulsion and significant change in trajectory. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Alpha particles do not penetrate the nucleus.
�� Option C → Rutherford assumed single scattering.
�� Option D → Rutherford's results disproved the plum pudding model.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate scattering angle to electric field strength.

Final Logic:

Strong nuclear field causes large deflections.

🧠 Memory Trick

Large Deflection → Strong Nucleus

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Nuclear Hypothesis

5 When computing the mathematical trajectory of an alpha particle scattered by a nucleus

A Only Newton's second law is sufficient without defining the force of interaction

B The electrostatic attraction between the alpha particle and the nucleus is considered

C Coulomb's law for electrostatic force of repulsion is employed along with Newton's second law

D The interaction force's magnitude is strictly independent of the changing distance between them

💡 Short Explanation

�� Force must be known. �� Coulomb force causes scattering. �� Newton's law predicts motion.

✅ Correct Answer Explanation

The trajectory is calculated using Coulomb's law to determine the electrostatic repulsive force and Newton's second law to determine the resulting motion. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Force law is essential.
�� Option B → Interaction is repulsive, not attractive.
�� Option D → Coulomb force varies as 1/r².

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify the laws used in Rutherford trajectory calculations.

Final Logic:

Coulomb force + Newton's law determine motion.

🧠 Memory Trick

Coulomb Gives Force, Newton Gives Motion

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Scattering Theory

6 Match List I (Orbital concepts) with List II (Expressions/Dependencies)

List I	List II
1. Centripetal Force F꜀	a. Inversely proportional to v² in stable orbit
2. Orbit radius r	b. Has a negative sign indicating direction towards the origin
3. Kinetic Energy K	c. Required to keep the electron in its stable orbit
4. Potential Energy U	d. Equal to half the magnitude of the potential energy

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

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

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

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

💡 Short Explanation

�� Centripetal force maintains orbit. �� Radius varies inversely with v². �� Potential energy is negative.

✅ Correct Answer Explanation

1 → c : Centripetal force keeps electron in orbit. 2 → a : r ∝ 1/v². 3 → d : K = |U|/2. 4 → b : Potential energy is negative. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect force and energy matching.
�� Option C → Radius matched incorrectly.
�� Option D → Centripetal force and kinetic energy mismatched.

🎯 Strategy

Used

Option Grouping

Application:

Match orbital concepts with their physical properties.

Final Logic:

Only option A correctly pairs all entries.

🧠 Memory Trick

Force-Orbit, Radius-v², K-Half U

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Orbit Relations

7 The value of the scattering angle θ for a nearly undeviated particle (large impact parameter) will be approximately:

A 0

B π/2

C π

D 2π

💡 Short Explanation

�� Large impact parameter means distant encounter. �� Repulsion is weak. �� Deflection is negligible.

✅ Correct Answer Explanation

When the impact parameter is very large, the alpha particle remains far from the nucleus and experiences only a weak Coulomb force. Consequently, the scattering angle approaches zero. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents significant scattering.
�� Option C → Represents backscattering.
�� Option D → Physically not applicable here.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate impact parameter to scattering angle.

Final Logic:

Large impact parameter gives θ ≈ 0.

🧠 Memory Trick

Large b → Small θ

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Impact Parameter and Scattering Angle

8 For a beam of alpha particles traversing a thin gold foil, the distribution of impact parameters and the probabilities of different scattering directions are respectively:

A Constant, Different

B Constant, Uniform

C Distributed, Uniform

D Distributed, Different

💡 Short Explanation

�� Different particles have different impact parameters. �� Different impact parameters give different angles. �� Probabilities therefore vary.

✅ Correct Answer Explanation

In a beam, particles approach nuclei with a range of impact parameters. This leads to a range of scattering angles and therefore different probabilities for different directions. Hence, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Impact parameters are not constant.
�� Option B → Impact parameters are not constant.
�� Option C → Probabilities are not uniform.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Analyze beam statistics.

Final Logic:

Different impact parameters produce different scattering probabilities.

🧠 Memory Trick

Different b → Different θ

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Statistical Nature of Scattering

9 In the Geiger-Marsden experiment, out of 8,000,000 incident alpha particles, approximately how many particles are expected to deflect by more than 90°? (Assume 1 in 8000 deflect by more than 90°)

A 1400

B 8000

C 1000

D 1

💡 Short Explanation

�� Probability = 1/8000. �� Total particles = 8,000,000. �� Multiply accordingly.

✅ Correct Answer Explanation

Number scattered by more than 90° = 8,000,000 × (1/8000) = 1000 Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect calculation.
�� Option B → Eight times larger than correct value.
�� Option D → Far too small.

🎯 Strategy

Used

Substitution

Application:

Multiply total particles by the given probability.

Final Logic:

8,000,000 ÷ 8000 = 1000.

🧠 Memory Trick

8 Million ÷ 8000 = 1000

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Scattering Statistics

10 In the case of a head-on rebound

A The impact parameter is maximum and the particle passes completely undeviated

B The particle experiences zero electrostatic force at the closest approach

C The impact parameter is minimum and the alpha particle rebounds back near π

D The alpha particle effortlessly penetrates the nucleus and permanently stops

💡 Short Explanation

�� Head-on collision means minimum impact parameter. �� Maximum repulsion occurs. �� Particle reverses direction.

✅ Correct Answer Explanation

For a head-on collision, the impact parameter is minimum (ideally zero). The alpha particle approaches the nucleus, momentarily stops, and rebounds with a scattering angle approximately equal to π radians (180°). Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Maximum impact parameter gives negligible scattering.
�� Option B → Electrostatic force is strongest near closest approach.
�� Option D → Alpha particles do not penetrate the nucleus in Rutherford's analysis.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate impact parameter to scattering outcome.

Final Logic:

Minimum impact parameter gives backscattering.

🧠 Memory Trick

Head-On → θ ≈ π

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Head-On Rutherford Scattering

11 The initial kinetic energy of an incoming alpha particle

A increases exponentially as it gets closer to the target nucleus

B completely determines the theoretical distance of closest approach

C is completely lost as thermal heat during the collision

D is provided by the electrostatic repulsion of the target nucleus

💡 Short Explanation

�� Closest approach depends on initial kinetic energy. �� Higher kinetic energy gives smaller closest approach distance. �� Energy conservation is used.

✅ Correct Answer Explanation

The distance of closest approach is obtained by equating the initial kinetic energy of the alpha particle to the electrostatic potential energy at the turning point: K = (1 / 4πε₀) × (2Ze² / d) Thus, the initial kinetic energy directly determines the value of d. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Kinetic energy decreases as the particle approaches the nucleus.
�� Option C → Energy is converted into potential energy, not thermal heat.
�� Option D → The kinetic energy originates from radioactive decay, not nuclear repulsion.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate closest approach to conservation of energy.

Final Logic:

Initial kinetic energy fixes the turning point distance.

🧠 Memory Trick

More KE → Smaller d

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Distance of Closest Approach

12 During the scattering process, the final potential energy U of the system at the stopping point (distance d) is:

A U = 2Ze² / (4πε₀d)

B U = Ze² / (4πε₀d²)

C U = -2Ze² / (4πε₀d)

D U = Ze / (4πε₀d)

💡 Short Explanation

�� Alpha particle charge = +2e. �� Gold nucleus charge = +Ze. �� Potential energy is positive because both charges are positive.

✅ Correct Answer Explanation

The electrostatic potential energy between two charges is: U = (1 / 4πε₀) × (q₁q₂ / d) Substituting: q₁ = +2e q₂ = +Ze Therefore, U = (1 / 4πε₀) × (2Ze² / d) Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect dependence on distance.
�� Option C → Wrong sign; like charges give positive potential energy.
�� Option D → Missing one factor of e.

🎯 Strategy

Used

Substitution

Application:

Apply electrostatic potential energy formula.

Final Logic:

U = (1 / 4πε₀)(2Ze² / d).

🧠 Memory Trick

+2e × +Ze → Positive U

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Potential Energy in Rutherford Scattering

13 Closest approach distance statements
1. It provides a reliable upper limit to the size of the target nucleus.
2. At this distance, kinetic energy is zero and electric potential energy is maximum.
3. It depends directly on the initial kinetic energy of the alpha particle.
4. It is exactly equal to the physical sum of the radii of the gold nucleus and the alpha particle.

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

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

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

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

💡 Short Explanation

�� Closest approach estimates nuclear size. �� Turning point occurs when velocity becomes zero. �� Initial kinetic energy affects the distance.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct. The distance of closest approach gives an upper limit to nuclear size. At this point the alpha particle momentarily stops, making kinetic energy zero and potential energy maximum. The value of d depends on the initial kinetic energy. Statement 4 is incorrect because d is generally much larger than the physical nuclear radius.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Identify the incorrect statement about closest approach.

Final Logic:

Closest approach is not equal to physical nuclear contact.

🧠 Memory Trick

d = Upper Limit, Not Radius

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Distance of Closest Approach

14 Incorrect statement about atomic dimensions

A The atomic radius is about 10,000 to 100,000 times larger than the nuclear radius.

B The actual radius of a gold nucleus is approximately 6 fm.

C The distance of closest approach is considerably smaller than the actual radius of the gold nucleus.

D The atom contains a much greater fraction of empty space than our solar system does.

💡 Short Explanation

�� Closest approach exceeds nuclear radius. �� Nuclear radius is only a few femtometres. �� Atomic size is much larger.

✅ Correct Answer Explanation

For gold, the actual nuclear radius is approximately 6 fm (6 × 10⁻¹⁵ m), while the closest approach distance is roughly 30 fm (3 × 10⁻¹⁴ m). Thus, the distance of closest approach is larger, not smaller, than the nuclear radius. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct size ratio.
�� Option B → Correct nuclear radius estimate.
�� Option D → Correct implication of Rutherford's model.

🎯 Strategy

Used

Extreme Word Filter

Application:

Compare actual nuclear radius with closest approach distance.

Final Logic:

d > nuclear radius.

🧠 Memory Trick

30 fm > 6 fm

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Atomic and Nuclear Dimensions

15 In Rutherford's model, if the centripetal force was not continuously provided by the electrostatic attraction

A The electrons would still seamlessly orbit the nucleus in stable paths

B The electrons could not maintain a dynamically stable closed orbit around the nucleus

C The entire positive charge of the nucleus would disperse instantly

D The atom would immediately exhibit a continuous emission spectrum

💡 Short Explanation

�� Circular motion requires centripetal force. �� Electrostatic attraction supplies that force. �� Without it, closed orbits are impossible.

✅ Correct Answer Explanation

For an electron to move in a circular orbit, a centripetal force must continuously act toward the centre. In Rutherford's model, this force is provided by electrostatic attraction. Without this force, the electron cannot remain in a stable closed orbit. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Violates circular motion requirements.
�� Option C → Unrelated to orbital dynamics.
�� Option D → Not the direct consequence asked.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Apply the concept of centripetal force.

Final Logic:

No centripetal force → No stable orbit.

🧠 Memory Trick

Orbit Needs Centre Pull

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Rutherford Orbital Dynamics

16 The orbital velocity v of the electron in a dynamically stable orbit of radius r in a hydrogen atom is given by

A v = √[e² / (4πε₀mr)]

B v = e / (4πε₀mr)

C v = √[e² / (8πε₀mr)]

D v = e² / (4πε₀mr²)

💡 Short Explanation

�� Equate electrostatic and centripetal forces. �� Solve for v². �� Take square root.

✅ Correct Answer Explanation

Using: e² / (4πε₀r²) = mv² / r Therefore: v² = e² / (4πε₀mr) Hence: v = √[e² / (4πε₀mr)] Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Missing square root.
�� Option C → Incorrect factor of 2.
�� Option D → Incorrect dimensions.

🎯 Strategy

Used

Substitution

Application:

Use force balance equation.

Final Logic:

Fₑ = F꜀ gives the orbital velocity.

🧠 Memory Trick

Balance Forces → Get v

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Electron Orbital Velocity

17 For an orbiting electron in a hydrogen atom, the algebraic signs of its Kinetic Energy and Potential Energy are respectively:

A Positive, Negative

B Negative, Positive

C Positive, Positive

D Negative, Negative

💡 Short Explanation

�� Kinetic energy is always positive. �� Attractive potential energy is negative. �� Bound systems have negative potential energy.

✅ Correct Answer Explanation

The kinetic energy of a moving electron is positive. The electrostatic potential energy of the electron-nucleus system is negative because the interaction is attractive. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reverses the signs.
�� Option C → Potential energy is not positive.
�� Option D → Kinetic energy cannot be negative.

🎯 Strategy

Used

Odd One Out

Application:

Recall signs of kinetic and potential energy.

Final Logic:

K > 0 and U < 0.

🧠 Memory Trick

Move → Positive K, Bind → Negative U

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Energy of Hydrogen Atom

18 In terms of kinetic energy K, the potential energy U of an electron in a hydrogen atom is given by:

A U = K/2

B U = 2K

C U = -K/2

D U = -2K

💡 Short Explanation

�� Coulomb orbit relation. �� Potential energy is twice kinetic energy in magnitude. �� Sign is negative.

✅ Correct Answer Explanation

For a hydrogen atom: K = e² / (8πε₀r) U = -e² / (4πε₀r) Therefore: U = -2K Hence, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect factor and sign.
�� Option B → Wrong sign.
�� Option C → Incorrect factor.

🎯 Strategy

Used

Substitution

Application:

Compare standard energy expressions.

Final Logic:

Potential energy equals minus twice kinetic energy.

🧠 Memory Trick

U = -2K

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Energy Relations in Hydrogen Atom

19 If the total energy of an electron in a hydrogen atom is -2.2 × 10⁻¹⁸ J, what is the orbital radius?
(1 / 4πε₀ = 9.0 × 10⁹ N m²/C², e = 1.6 × 10⁻¹⁹ C)

A 5.3 × 10⁻¹¹ m

B 2.2 × 10⁻⁶ m

C 3.0 × 10⁻¹⁴ m

D 1.5 × 10⁻¹¹ m

💡 Short Explanation

�� Use total energy formula. �� E = -e² / (8πε₀r). �� Solve for r.

✅ Correct Answer Explanation

For hydrogen: E = -e² / (8πε₀r) Therefore: r = e² / (8πε₀|E|) Substituting values: r = (9 × 10⁹)(1.6 × 10⁻¹⁹)² / [2(2.2 × 10⁻¹⁸)] r ≈ 5.3 × 10⁻¹¹ m Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Far too large.
�� Option C → Nuclear scale, not orbital scale.
�� Option D → Incorrect numerical result.

🎯 Strategy

Used

Substitution

Application:

Use the total energy equation directly.

Final Logic:

The calculation gives the Bohr radius value.

🧠 Memory Trick

Ground State → 5.3 × 10⁻¹¹ m

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Total Energy and Orbital Radius

20 Correct statements about the bound state of an electron
1. The total energy of the electron is natively negative.
2. The electrostatic potential energy is strictly negative.
3. A positive total energy would mean the electron is unbound and does not follow a closed orbit.
4. The magnitude of the potential energy is exactly half the kinetic energy.

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

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

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

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

💡 Short Explanation

�� Bound electrons have negative total energy. �� Potential energy is negative. �� Positive total energy means escape.

✅ Correct Answer Explanation

Statements 1, 2 and 3 are correct. For a bound electron: E < 0 U < 0 Positive total energy implies an unbound state. Statement 4 is incorrect because: |U| = 2K not |U| = K/2 Therefore, option A is 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:

Use standard hydrogen atom energy relations.

Final Logic:

Bound state requires E < 0 and U = -2K.

🧠 Memory Trick

Bound → E Negative, U = -2K

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Bound States and Energy Relations

CUET UG Physics Booster Test 2- Rutherford Nuclear Model

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