CUET UG Physics Booster Test 3- Rutherford Nuclear Model -

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CUET UG Physics Booster Test 3- Rutherford Nuclear Model

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 3- Rutherford Nuclear Model

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QUESTION 1 OF 20

If the dimensions of the solar system had the same proportions as those of the atom (radius of orbit to nucleus is 10⁵), and the radius of the sun is 7 × 10⁸ m, what would be the scaled radius of the earth's orbit?

Answer Review

1 If the dimensions of the solar system had the same proportions as those of the atom (radius of orbit to nucleus is 10⁵), and the radius of the sun is 7 × 10⁸ m, what would be the scaled radius of the earth's orbit?

A 7 × 10¹³ m

B 1.5 × 10¹¹ m

C 7 × 10¹¹ m

D 1.5 × 10¹³ m

💡 Short Explanation

�� Atomic ratio ≈ 10⁵. �� Sun radius = 7 × 10⁸ m. �� Multiply to obtain scaled orbit radius.

✅ Correct Answer Explanation

Using the atomic proportion: Orbit radius = 10⁵ × (Sun radius) = 10⁵ × 7 × 10⁸ = 7 × 10¹³ m Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Actual Earth orbit radius, not scaled value.
�� Option C → Uses incorrect scaling factor.
�� Option D → Incorrect numerical result.

🎯 Strategy

Used

Substitution

Application:

Apply the given proportional relationship directly.

Final Logic:

10⁵ × 7 × 10⁸ = 7 × 10¹³ m.

🧠 Memory Trick

Atomic Ratio = 10⁵

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Planetary Analogy of Atom

2 Match List I (Atomic vs Solar System) with List II (Forces & Ratios)

List I	List II
1. Planet-sun primary force	a. Coulomb electrostatic force
2. Electron-nucleus primary force	b. Gravitational force
3. Atom orbit-to-nucleus ratio	c. ~10⁵ times the core radius
4. Earth actual orbit-to-sun ratio	d. ~200 times the core radius

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

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

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

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

💡 Short Explanation

�� Gravity governs planets. �� Coulomb force governs atoms. �� Atomic size ratio is much larger.

✅ Correct Answer Explanation

1 → b : Planetary motion is governed by gravity. 2 → a : Electron motion is governed by electrostatic attraction. 3 → c : Atomic orbit radius is about 10⁵ times nuclear size. 4 → d : Earth's orbit is about 200 times the Sun's radius. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reverses gravitational and electrostatic forces.
�� Option C → Incorrect ratio assignments.
�� Option D → Reverses force assignments.

🎯 Strategy

Used

Option Grouping

Application:

Match each system with its governing force and scale.

Final Logic:

Only option A correctly matches all pairs.

🧠 Memory Trick

Planet → Gravity, Atom → Coulomb

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Planetary Analogy

3 Rutherford argued that to physically deflect an incoming high-speed alpha particle by a large angle

A the atom must possess a uniformly distributed positive charge and light electrons

B the greater part of the mass and positive charge must be concentrated tightly at the centre

C the atomic electrons must have sufficient mass to directly cause a massive rebound

D the impact parameter must be vastly larger than the atomic radius itself

💡 Short Explanation

�� Large deflections require strong forces. �� Strong forces require concentrated charge. �� Rutherford proposed the nucleus.

✅ Correct Answer Explanation

A large-angle deflection can occur only if the positive charge and most of the atomic mass are concentrated in a tiny central region. This produces the intense electric field needed to repel alpha particles strongly. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Thomson's model failed to explain large deflections.
�� Option C → Electrons are too light.
�� Option D → Large impact parameters produce small deflections.

🎯 Strategy

Used

Elimination

Application:

Identify the condition required for strong scattering.

Final Logic:

Large-angle scattering requires concentrated positive charge.

🧠 Memory Trick

Big Scatter → Tiny Nucleus

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

4 Positive core interaction statements
1. An incoming alpha particle can get very close to the positive charge without penetrating it.
2. The alpha particle penetration directly results in a large angle deflection.
3. The intense electric field of the concentrated core scatters the alpha particle.
4. The atomic electrons do not appreciably affect the alpha particle's trajectory.

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

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

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

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

💡 Short Explanation

�� Alpha particles do not enter the nucleus. �� Strong nuclear field causes scattering. �� Electron effect is negligible.

✅ Correct Answer Explanation

Statements 1, 3 and 4 are correct. An alpha particle can approach very close to the nucleus and be strongly repelled without penetrating it. The intense electric field causes scattering, while electrons contribute negligibly. Statement 2 is incorrect because penetration is not required for large-angle deflection.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 2.
�� Option B → Includes statement 2.
�� Option C → Includes statement 2.

🎯 Strategy

Used

Elimination

Application:

Identify the incorrect statement regarding scattering.

Final Logic:

Large deflection occurs without nuclear penetration.

🧠 Memory Trick

Close, Not Inside

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

5 The magnitude of the electrostatic force F between an alpha particle and a gold nucleus is mathematically given by

A F = 2Ze² / (4πε₀r²)

B F = Ze² / (4πε₀r²)

C F = 2Ze / (4πε₀r²)

D F = Ze² / (8πε₀r²)

💡 Short Explanation

�� Alpha particle charge = +2e. �� Gold nucleus charge = +Ze. �� Apply Coulomb's law.

✅ Correct Answer Explanation

Using Coulomb's law: F = (1 / 4πε₀) × [(2e)(Ze)] / r² F = 2Ze² / (4πε₀r²) Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Missing factor 2.
�� Option C → Missing one factor of e.
�� Option D → Incorrect coefficient.

🎯 Strategy

Used

Substitution

Application:

Substitute charges into Coulomb's law.

Final Logic:

(+2e)(+Ze) gives 2Ze².

🧠 Memory Trick

2e × Ze = 2Ze²

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Coulomb Force in Rutherford Scattering

6 Incorrect statement about the force balance in a dynamically stable hydrogen atom

A The required centripetal force is solely provided by the electrostatic force of attraction.

B The magnitude of the required centripetal force is mv²/r.

C The electrostatic force vector is naturally directed away from the nucleus.

D The dynamic equilibrium relation mv²/r = e²/(4πε₀r²) holds true.

💡 Short Explanation

�� Electron and nucleus attract each other. �� Force points toward the nucleus. �� Centripetal force is inward.

✅ Correct Answer Explanation

The electrostatic force between the positively charged nucleus and negatively charged electron is attractive and directed toward the nucleus. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Correct centripetal force expression.
�� Option D → Correct force-balance equation.

🎯 Strategy

Used

Odd One Out

Application:

Check force direction.

Final Logic:

Attractive force must point inward.

🧠 Memory Trick

Electron → Pulled Inward

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Force Balance in Hydrogen Atom

7 Correct statements concerning the impact parameter and trajectory
1. A given beam of alpha particles inherently has a distribution of impact parameters.
2. A small impact parameter directly results in a small angle of deflection.
3. A large impact parameter results in nearly undeviated alpha particle trajectories.
4. The classical trajectory is computed by employing Newton's second law and Coulomb's law.

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

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

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

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

💡 Short Explanation

�� Impact parameters vary. �� Large impact parameter gives small scattering. �� Newton and Coulomb laws are used.

✅ Correct Answer Explanation

Statements 1, 3 and 4 are correct. A beam naturally contains a range of impact parameters. Large impact parameters produce very small deflections. Rutherford trajectories are calculated using Coulomb's law and Newton's second law. Statement 2 is incorrect because small impact parameters generally produce large deflections.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 2.
�� Option C → Includes statement 2.
�� Option D → Includes statement 2.

🎯 Strategy

Used

Elimination

Application:

Relate impact parameter to scattering angle.

Final Logic:

Small b → Large θ; Large b → Small θ.

🧠 Memory Trick

Small b, Big θ

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

8 During the mathematical computation of the scattering of an alpha particle by a single gold nucleus

A Both the alpha particle and the nucleus move considerably around their centre of mass

B The trajectory depends solely on the mass of the alpha particle and its charge

C The heavy gold nucleus is practically assumed to remain stationary throughout the process

D The force of interaction dynamically changes its fundamental nature from electrostatic to gravitational

💡 Short Explanation

�� Gold nucleus is much heavier. �� Approximation simplifies calculations. �� Motion of nucleus is neglected.

✅ Correct Answer Explanation

The gold nucleus is approximately 50 times heavier than the alpha particle. Therefore, Rutherford's analysis treats it as stationary while calculating the alpha-particle trajectory. Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not assumed in Rutherford theory.
�� Option B → Trajectory depends on more than mass and charge.
�� Option D → Interaction remains electrostatic.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify Rutherford's simplifying assumption.

Final Logic:

Heavy nucleus ≈ stationary target.

🧠 Memory Trick

Heavy Gold, Fixed Gold

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

9 For a beam of alpha particles passing through a thin metal foil, the observed frequencies of zero/small deflection and large deflection (>90°) are respectively:

A High, Low

B Low, High

C High, High

D Low, Low

💡 Short Explanation

�� Most particles pass through. �� Large deflections are rare. �� Backscattering is extremely uncommon.

✅ Correct Answer Explanation

Most alpha particles travel through the foil with no or only small deflection because atoms are mostly empty space. Only about 1 in 8000 particles scatter by more than 90°. Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Opposite of observations.
�� Option C → Large-angle scattering is not frequent.
�� Option D → Small-angle scattering is very common.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use Rutherford's experimental observations.

Final Logic:

Small deflections are common; large deflections are rare.

🧠 Memory Trick

Many Through, Few Back

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

10 In a head-on collision, if the initial kinetic energy K of the alpha particle is given, the distance of closest approach d will mathematically vary as:

A K

B K²

C 1/K²

D 1/K

💡 Short Explanation

�� Closest approach comes from energy conservation. �� d appears in the denominator. �� Higher kinetic energy means smaller d.

✅ Correct Answer Explanation

For a head-on collision: K = (1 / 4πε₀) × (2Ze² / d) Rearranging: d = (1 / 4πε₀) × (2Ze² / K) Thus, d ∝ 1/K Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Opposite relationship.
�� Option B → Incorrect dependence.
�� Option C → Incorrect inverse-square relation.

🎯 Strategy

Used

Substitution

Application:

Rearrange the closest-approach equation.

Final Logic:

d is inversely proportional to K.

🧠 Memory Trick

More KE → Less d

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

11 Initial kinetic energy and approach statements
1. The maximum kinetic energy of a 7.7 MeV alpha particle is mathematically equivalent to 1.2 × 10⁻¹² J.
2. Increasing the initial kinetic energy proportionally increases the distance of closest approach.
3. At the stopping point, the initial mechanical kinetic energy is completely transformed into potential energy.
4. The initial mechanical energy is entirely kinetic before the particle and nucleus interact.

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

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

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

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

💡 Short Explanation

�� 7.7 MeV = 1.23 × 10⁻¹² J. �� At closest approach, K.E. becomes zero. �� Distance of closest approach decreases with increasing kinetic energy.

✅ Correct Answer Explanation

Statement 1 is correct because: 7.7 MeV = 7.7 × 1.6 × 10⁻¹³ J = 1.23 × 10⁻¹² J ≈ 1.2 × 10⁻¹² J Statement 3 is correct because at the stopping point all kinetic energy is converted into electrostatic potential energy. Statement 4 is correct because before significant interaction, the alpha particle's mechanical energy is essentially kinetic. Statement 2 is incorrect because: d = (2Ze²)/(4πε₀K) Therefore d ∝ 1/K, not d ∝ K.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 2 which is incorrect.
�� Option B → Includes statement 2 which is incorrect.
�� Option D → Includes statement 2 which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Use the closest approach relation d ∝ 1/K.

Final Logic:

Statements 1, 3 and 4 are correct while statement 2 is incorrect.

🧠 Memory Trick

More K → Less d

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

12 At the critical moment an alpha particle momentarily stops and reverses its direction during a head-on collision

A its electric potential energy is at a theoretical minimum

B its kinetic energy reaches a local maximum

C the total mechanical energy of the system is entirely equal to its electric potential energy

D the electrostatic repulsive force acting on it is exactly zero

💡 Short Explanation

�� Velocity becomes zero. �� Kinetic energy becomes zero. �� Total energy equals potential energy.

✅ Correct Answer Explanation

At the distance of closest approach: K.E. = 0 Therefore: Total Energy = Potential Energy Since energy is conserved, Initial K.E. = Final Potential Energy Thus the entire mechanical energy of the system is present as electrostatic potential energy. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Potential energy is maximum, not minimum.
�� Option B → Kinetic energy becomes zero.
�� Option D → Electrostatic force is very large near the nucleus.

🎯 Strategy

Used

Energy Conservation

Application:

Analyze the turning point condition.

Final Logic:

At closest approach, K = 0 and E = U.

🧠 Memory Trick

Turning Point → K = 0

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

13 If an alpha particle of initial kinetic energy K = 8.0 × 10⁻¹³ J approaches a gold nucleus (Z = 79), calculate the distance of closest approach d.
(1/4πε₀ = 9.0 × 10⁹ N m²/C², e = 1.6 × 10⁻¹⁹ C)

A 2.1 × 10⁻⁷ m

B 3.0 × 10⁻¹⁴ m

C 6.0 × 10⁻¹⁵ m

D 4.55 × 10⁻¹⁴ m

💡 Short Explanation

�� Use energy conservation. �� K = (1/4πε₀)(2Ze²/d). �� Solve for d.

✅ Correct Answer Explanation

Using: d = (1/4πε₀)(2Ze²/K) Substituting values: d = [(9 × 10⁹) × 2 × 79 × (1.6 × 10⁻¹⁹)²] / (8 × 10⁻¹³) d = 4.55 × 10⁻¹⁴ m Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Equal to foil thickness scale, not closest approach.
�� Option B → Corresponds to a different kinetic energy value.
�� Option C → Comparable to nuclear radius, not calculated d.

🎯 Strategy

Used

Substitution

Application:

Substitute numerical values into the closest approach formula.

Final Logic:

Direct calculation gives 4.55 × 10⁻¹⁴ m.

🧠 Memory Trick

Use d ∝ 1/K

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

14 The discrepancy between the theoretical distance of closest approach and the actual radius of the gold nucleus indicates that

A The alpha particle physically touches the nucleus before dramatically rebounding

B The calculated distance of closest approach is an exact measure of the actual nuclear radius

C The alpha particle safely reverses its motion without ever actually touching the gold nucleus

D The nuclear model is fundamentally flawed and classical concepts fail completely here

💡 Short Explanation

�� Closest approach exceeds nuclear radius. �� Alpha particle turns back earlier. �� Nuclear size is much smaller.

✅ Correct Answer Explanation

For gold: Distance of closest approach ≈ 30 fm Actual nuclear radius ≈ 6 fm Since the turning point occurs well before the alpha particle reaches the nuclear surface, the particle reverses direction without touching the nucleus. Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Physical contact is not required.
�� Option B → Closest approach gives only an upper limit.
�� Option D → Rutherford's model is supported by the observation.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Compare the closest approach with actual nuclear radius.

Final Logic:

d > Nuclear Radius.

🧠 Memory Trick

30 fm > 6 fm

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Nuclear Size Estimation

15 Correct statements about the total energy in a dynamically stable orbit
1. The total energy E is the algebraic sum of kinetic K and potential U energies.
2. The total energy is mathematically given by -e²/(8πε₀r).
3. The total energy is strictly positive for a closed stable orbit.
4. The negative sign physically signifies that the electron is bound to the nucleus.

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

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

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

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

💡 Short Explanation

�� E = K + U. �� Bound systems have negative total energy. �� Negative energy implies binding.

✅ Correct Answer Explanation

Statement 1 is correct: E = K + U Statement 2 is correct: E = -e²/(8πε₀r) Statement 4 is correct because negative energy means external energy must be supplied to free the electron. Statement 3 is incorrect because stable bound orbits have negative total energy.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Recall the sign of total energy for bound states.

Final Logic:

Bound orbit ⇒ E < 0.

🧠 Memory Trick

Negative E = Bound Electron

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

16 If the orbital radius of an electron in a hydrogen atom is r, the orbital velocity v can be analytically represented as:

A v = e / √(4πε₀mr)

B v = e² / √(4πε₀mr)

C v = e / (4πε₀mr)

D v = e² / (4πε₀mr²)

💡 Short Explanation

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

✅ Correct Answer Explanation

From force balance: mv²/r = e²/(4πε₀r²) Therefore: v² = e²/(4πε₀mr) Taking square root: v = e/√(4πε₀mr) Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Extra factor of e.
�� Option C → Missing square root.
�� Option D → Wrong dimensions.

🎯 Strategy

Used

Substitution

Application:

Use force balance equation.

Final Logic:

Fₑ = F꜀ gives v.

🧠 Memory Trick

Balance Forces → Velocity

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Electron Velocity in Rutherford Model

17 Incorrect statement regarding the kinetic energy of an orbital electron

A It is directly proportional to the square of its orbital velocity.

B It is mathematically equal to e²/(8πε₀r).

C It is numerically equal to the negative of the total energy.

D It is twice the magnitude of the electrostatic potential energy.

💡 Short Explanation

�� K = ½mv². �� K = e²/(8πε₀r). �� |U| = 2K.

✅ Correct Answer Explanation

For a hydrogen atom: K = e²/(8πε₀r) U = -e²/(4πε₀r) Hence: |U| = 2K or K = |U|/2 Therefore statement D is incorrect because kinetic energy is half the magnitude of potential energy, not twice.

❓ Why Other Options Are Incorrect

�� Option A → Correct kinetic energy relation.
�� Option B → Correct expression.
�� Option C → Since E = -K, it is correct.

🎯 Strategy

Used

Formula Recall

Application:

Compare K and U relations.

Final Logic:

K = |U|/2.

🧠 Memory Trick

Potential Twice Kinetic

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

18 Match List I (Energy properties) with List II (Analytical Expressions)

List I	List II
1. Total Energy (E)	a. -e²/(4πε₀r)
2. Kinetic Energy (K)	b. U = 2E
3. Potential Energy (U)	c. e²/(8πε₀r)
4. Relation between U and E	d. -e²/(8πε₀r)

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

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

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

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

💡 Short Explanation

�� E is negative. �� K is positive. �� U = 2E.

✅ Correct Answer Explanation

1 → d : E = -e²/(8πε₀r) 2 → c : K = e²/(8πε₀r) 3 → a : U = -e²/(4πε₀r) 4 → b : U = 2E Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → E and K interchanged.
�� Option C → U and K interchanged.
�� Option D → Incorrect energy assignments.

🎯 Strategy

Used

Option Grouping

Application:

Match standard hydrogen energy equations.

Final Logic:

Only option A correctly matches all expressions.

🧠 Memory Trick

E Negative, K Positive, U Double

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

19 If the total energy E of an orbiting electron were somehow made zero or positive

A the electron would immediately shift to a vastly smaller orbit

B the electron would not follow a closed orbit and the atom would become unbound

C the kinetic energy would theoretically become negative

D the potential energy would completely overpower the kinetic energy

💡 Short Explanation

�� Bound states require E < 0. �� E ≥ 0 implies escape. �� No stable orbit exists.

✅ Correct Answer Explanation

A bound electron must have negative total energy. If E becomes zero or positive, the electron is no longer confined to the nucleus and can escape to infinity. Thus, the atom becomes unbound. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Opposite behavior.
�� Option C → Kinetic energy cannot be negative.
�� Option D → Not implied by positive total energy.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Interpret the physical meaning of total energy.

Final Logic:

Positive energy means unbound motion.

🧠 Memory Trick

Positive E → Escape

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

20 For an electron carefully evaluated in a bound state versus an unbound state, the total energy signs are respectively:

A Negative, Positive

B Positive, Negative

C Negative, Negative

D Positive, Positive

💡 Short Explanation

�� Bound systems have negative energy. �� Free systems have positive energy. �� Energy sign indicates confinement.

✅ Correct Answer Explanation

For a bound electron: E < 0 For an unbound electron: E > 0 Therefore, the correct sequence is Negative, Positive. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Reversed order.
�� Option C → Unbound state cannot have negative total energy.
�� Option D → Bound state cannot have positive total energy.

🎯 Strategy

Used

Odd One Out

Application:

Recall the energy criterion for bound systems.

Final Logic:

Bound → Negative, Unbound → Positive.

🧠 Memory Trick

Bound Below Zero

📚 Class XII – NCERT Physics Part II – Chapter: Atoms – Topic: Bound State Energy Criterion

CUET UG Physics Booster Test 3- Rutherford Nuclear Model

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