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Alpha-particle scattering and Rutherford model: NEET MCQs with answers

50 questions on this topic of Atoms and Nuclei (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In the Geiger-Marsden experiment, approximately what percentage of the incident α-particles are scattered by more than 90°?

  1. About 1%
  2. About 0.0125%
  3. About 0.14%
  4. About 4%
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Answer: (B) About 0.0125%

According to NCERT, only about 0.14% of the incident α-particles scatter by more than 1°, and about 1 in 8000 (which is 0.0125%) deflect by more than 90°.

medium

2. In an α-particle scattering experiment, the number of α-particles scattered at an angle θ = 60° is 112 per minute. The number of α-particles scattered per minute at an angle θ = 90° for the same setup is:

  1. 28
  2. 14
  3. 56
  4. 224
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Answer: (A) 28

The number of scattered particles N(θ) is proportional to 1/sin⁴(θ/2). At θ = 60°, sin(30°) = 1/2, so 1/sin⁴(30°) = 16. At θ = 90°, sin(45°) = 1/√2, so 1/sin⁴(45°) = 4. Therefore, N(90°) = N(60°) × (4/16) = 112 / 4 = 28 per minute.

hard

3. In a Rutherford scattering experiment, an α-particle is scattered by an angle of 90°. The impact parameter of the α-particle is b. What will be the new scattering angle if the impact parameter is reduced to b / √3?

  1. 60°
  2. 120°
  3. 135°
  4. 150°
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Answer: (B) 120°

The relationship between impact parameter and scattering angle is b = C cot(θ/2). For θ = 90°, cot(45°) = 1, so b = C. When the new impact parameter is b' = b / √3, we have cot(θ'/2) = 1 / √3, which gives θ'/2 = 60°, so θ' = 120°.

medium

4. An α-particle is scattered through an angle θ = 90° by a heavy nucleus. If the initial kinetic energy of the α-particle is doubled while keeping the impact parameter unchanged, the new scattering angle θ' satisfies:

  1. θ' = 45°
  2. θ' = 120°
  3. cot(θ'/2) = 2
  4. tan(θ'/2) = 2
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Answer: (C) cot(θ'/2) = 2

The impact parameter is related to scattering angle by b = [1/(4πε₀)] · [Z e² cot(θ/2) / K]. Since b is unchanged and K is doubled, cot(θ'/2) / (2K) = cot(θ/2) / K. Since θ = 90°, cot(45°) = 1, so cot(θ'/2) = 2 cot(45°) = 2.

easy

5. How does the impact parameter (b) vary with the scattering angle (θ) in Rutherford scattering?

  1. b ∝ cot(θ/2)
  2. b ∝ tan(θ/2)
  3. b ∝ sin(θ/2)
  4. b ∝ cos(θ/2)
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Answer: (A) b ∝ cot(θ/2)

According to Rutherford's scattering formula, the impact parameter b is given by b = [1 / (4πε₀)] × [Z e² cot(θ/2)] / K. Therefore, the impact parameter is directly proportional to cot(θ/2).

hard

6. According to Rutherford's planetary model, an electron of mass m moves in a stable circular orbit of radius r around a nucleus of charge Ze. What is the total mechanical energy of the electron?

  1. + Ze² / (8πε₀ r)
  2. - Ze² / (8πε₀ r)
  3. - Ze² / (4πε₀ r)
  4. - Ze² / (2πε₀ r)
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Answer: (B) - Ze² / (8πε₀ r)

Centripetal force is provided by the Coulomb attraction: m v² / r = Ze² / (4πε₀ r²), which yields kinetic energy K = (1/2) m v² = Ze² / (8πε₀ r). The potential energy is U = - Ze² / (4πε₀ r). Thus, total mechanical energy E = K + U = Ze² / (8πε₀ r) - Ze² / (4πε₀ r) = - Ze² / (8πε₀ r).

easy

7. According to classical electromagnetic theory, which of the following is an inevitable consequence for an electron revolving in a circular orbit around the nucleus in Rutherford's model?

  1. The electron continuously emits electromagnetic radiation and spirals into the nucleus.
  2. The electron maintains a constant orbital radius indefinitely.
  3. The electron absorbs energy continuously from the surrounding space.
  4. The electron experiences zero acceleration because its speed is constant.
Show answer and explanation

Answer: (A) The electron continuously emits electromagnetic radiation and spirals into the nucleus.

An electron in a circular orbit undergoes continuous centripetal acceleration. Classical electromagnetic theory predicts that an accelerated charge must radiate electromagnetic energy continuously, causing its orbit to shrink until it collapses into the nucleus.

medium

8. In an α-particle scattering experiment, the number of particles detected per unit time at a scattering angle θ = 60° is N₀. If the detector is moved to an angle θ = 120°, the number of particles detected per unit time will be:

  1. N₀ / 9
  2. 9 N₀
  3. 3 N₀
  4. N₀ / 3
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Answer: (A) N₀ / 9

The number of scattered particles per unit area is proportional to 1 / sin⁴(θ/2). At θ = 60°, sin(30°) = 1/2, so sin⁴(30°) = 1/16. At θ = 120°, sin(60°) = √3/2, so sin⁴(60°) = 9/16. The ratio N(120°)/N(60°) = (1/16) / (9/16) = 1/9. Thus, the count becomes N₀ / 9.

easy

9. In Rutherford's α-particle scattering experiment, what is the shape of the trajectory traced by an alpha particle approaching a target nucleus under the influence of the repulsive Coulomb force?

  1. Parabolic
  2. Hyperbolic
  3. Elliptical
  4. Circular
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Answer: (B) Hyperbolic

Under the action of an inverse-square repulsive central force (electrostatic repulsion between the α-particle and the positively charged nucleus), the trajectory of the scattered projectile is a hyperbola with the nucleus located at one of the outer foci.

medium

10. A beam of α-particles of kinetic energy K is incident on a gold foil (atomic number Z₁ = 79). The distance of closest approach is d₁. When the experiment is repeated using a silver foil (atomic number Z₂ = 47) with α-particles of kinetic energy 2K, the new distance of closest approach d₂ satisfies:

  1. d₂ = (47 / 79) d₁
  2. d₂ = (94 / 79) d₁
  3. d₂ = (47 / 158) d₁
  4. d₂ = (79 / 94) d₁
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Answer: (C) d₂ = (47 / 158) d₁

The distance of closest approach is proportional to Z / K. Therefore, d₂ / d₁ = (Z₂ / Z₁) × (K₁ / K₂) = (47 / 79) × (K / 2K) = 47 / 158. Hence, d₂ = (47 / 158) d₁.

medium

11. An α-particle approaches a stationary gold nucleus (Z = 79) from infinity with initial kinetic energy K. At the distance of closest approach d in a head-on collision, what is the electrostatic potential energy of the system?

  1. K
  2. K / 2
  3. 0
  4. 2K
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Answer: (A) K

By conservation of mechanical energy, the initial total energy is solely the kinetic energy K (potential energy at infinity is zero). At the point of closest approach, the α-particle momentarily stops (instantaneous velocity = 0, so kinetic energy is zero). Therefore, the entire initial kinetic energy is converted into electrostatic potential energy, U = K.

medium

12. An α-particle of kinetic energy E is projected directly towards a heavy nucleus of atomic number Z. If the distance of closest approach is d, what will be the distance of closest approach when an α-particle of kinetic energy 2E is projected towards a nucleus of atomic number 2Z?

  1. d / 4
  2. d
  3. d / 2
  4. 2d
Show answer and explanation

Answer: (B) d

The distance of closest approach for a head-on collision is given by d = (1 / 4πε₀) · (2Ze² / E). If both Z and E are doubled, the ratio (2Z) / (2E) remains unchanged. Therefore, the new distance of closest approach is equal to d.

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