Atoms – α-Particle Scattering & Rutherford Nuclear Model of Atom: NEET MCQs with Answers

9 free questions on Atoms – α-Particle Scattering & Rutherford Nuclear Model of Atom for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: α-Particle Scattering & Rutherford Nuclear Model of Atom: Alpha-particle scattering experiment; Rutherford's model of atom

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Question 1

Previous-year question
An α-particle of mass m moving with speed v strikes head-on a heavy nucleus of charge Ze. Its distance of closest approach depends on m as [NEET 2016]
  1. 1/m
  2. 1/√m
  3. 1/m²
  4. m
Show answer
Answer: (A) 1/m
Explanation: ½mv² = k(2e)(Ze)/r₀ gives r₀ = 4kZe²/(mv²). For a fixed speed, r₀ is inversely proportional to the mass m.

Question 2

Alpha particles of equal energy travel in parallel lines towards a heavy nucleus fixed at N. Which of the following reported paths is NOT physically possible?
  1. A particle with N lying to its left curves to its left, around N, and leaves on the same side as N
  2. A particle passing very far from N continues almost undeviated
  3. A particle aimed almost exactly at N comes close to it, stops momentarily and returns almost along its own path
  4. A particle with N lying to its left is bent to its right, away from N, along a curved path
Show answer
Answer: (A) A particle with N lying to its left curves to its left, around N, and leaves on the same side as N
Explanation: The alpha particle and the nucleus are both positive, so the Coulomb force on the particle is always a repulsion directed away from N. A path can therefore only bend away from the nucleus, never towards or around it. The deflection grows as the impact parameter falls, reaching nearly 180° for a head-on approach, and is negligible for a very large impact parameter.

Question 3

A gold atom has a radius of about 1.4 × 10⁻¹⁰ m and its nucleus a radius of about 7 × 10⁻¹⁵ m. Taking nearly all the mass of the atom to lie in the nucleus, the density of the nucleus is about how many times the average density of the whole atom?
  1. 1.6 × 10¹⁷
  2. 4 × 10⁸
  3. 8 × 10¹²
  4. 2 × 10⁴
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Answer: (C) 8 × 10¹²
Explanation: For the same mass, density varies inversely as volume, and so as the cube of the radius. The ratio of radii is 1.4 × 10⁻¹⁰ / 7 × 10⁻¹⁵ = 2 × 10⁴. Cubing this gives 8 × 10¹², which shows how little of the volume of an atom the nucleus occupies, and why most α-particles pass through a foil undeflected.

Question 4

An α-particle of kinetic energy 5.5 MeV heads straight for a gold nucleus (Z = 79), and the scattering observed agrees fully with Coulomb's law. Taking e²/4πε₀ = 1.44 MeV fm, what is the distance of closest approach, and what does it tell us about the radius of the gold nucleus?
  1. About 21 fm; the nuclear radius must be larger than this
  2. About 21 fm; the nuclear radius must be smaller than this
  3. About 41 fm; the nuclear radius must be smaller than this
  4. About 41 fm; the nuclear radius is equal to this
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Answer: (C) About 41 fm; the nuclear radius must be smaller than this
Explanation: At closest approach all the kinetic energy has become electrostatic potential energy, so K = 2Ze²/(4πε₀d). This gives d = 2 × 79 × 1.44 / 5.5 ≈ 41 fm. The α-particle is turned back by Coulomb repulsion before it reaches the nuclear surface, so this distance is only an upper limit on the size of the nucleus.

Question 5

In the Geiger–Marsden experiment, α-particles were directed at a gold foil only about 2.1 × 10⁻⁷ m thick. What was the main purpose of using so thin a foil?
  1. To make sure that most α-particles suffered at most one scattering, so that a large deflection could be traced to a single encounter
  2. To slow the α-particles down, so that the nuclear force could act on them for longer
  3. To stop all the α-particles inside the foil, so that none of them reached the detector screen undeflected
  4. To allow the α-particles to collide with the electrons of many gold atoms in turn, as these cause the large deflections
Show answer
Answer: (A) To make sure that most α-particles suffered at most one scattering, so that a large deflection could be traced to a single encounter
Explanation: A very thin foil ensures that an α-particle meets a gold nucleus closely no more than once on its way through. Any large deflection can then be attributed to one close encounter with a single nucleus, not to many small deflections adding up. This is what allowed the angular distribution to be compared with the calculated scattering from a single point nucleus. Electrons are far too light to deflect an α-particle appreciably.

Question 6

In Rutherford's nuclear model, the electrons of an atom cannot be at rest, and yet classical electromagnetic theory predicts that orbiting electrons also make the atom unstable. Which option gives the correct reason in each case?
  1. At rest, an electron would be pulled into the nucleus by electrostatic attraction; in orbit, it is accelerated, radiates energy and spirals into the nucleus
  2. At rest, an electron would be repelled out of the atom by the nucleus; in orbit, it gains energy continuously and escapes from the atom
  3. At rest, an electron would radiate energy continuously; in orbit, its speed is constant, so it radiates nothing and stays in place
  4. At rest, an electron would feel no force from the nucleus; in orbit, the nuclear force pulls it slowly into the nucleus
Show answer
Answer: (A) At rest, an electron would be pulled into the nucleus by electrostatic attraction; in orbit, it is accelerated, radiates energy and spirals into the nucleus
Explanation: A stationary electron would simply fall into the positive nucleus under electrostatic attraction, so the electrons must revolve. But circular motion is accelerated motion, and classical theory says an accelerated charge radiates energy. The electron would lose energy, spiral inward and collapse into the nucleus, emitting a continuous spectrum, which is not what real atoms do.

Question 7

A narrow beam of fast α-particles strikes a thin gold foil. Equal-sized detectors are placed at scattering angles of 2° (D₁), 60° (D₂) and 150° (D₃), and each counts for the same time. Which conclusion is correct?
  1. D₁ records the most counts, and particles reaching D₃ had the smallest impact parameters
  2. D₃ records the most counts, because the heavy nucleus sends most particles straight back
  3. D₁ records the most counts, and particles reaching D₃ had the largest impact parameters
  4. All three record nearly equal counts, because the positive charge is spread uniformly through the atom
Show answer
Answer: (A) D₁ records the most counts, and particles reaching D₃ had the smallest impact parameters
Explanation: The atom is mostly empty space, so most α-particles pass far from any nucleus and are deflected very little. Large deflections need a close approach to the small, heavy, positive nucleus, which means a small impact parameter. Such close approaches are rare, so the count falls sharply as the scattering angle increases.
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Question 8

Alpha particles of a fixed kinetic energy approach a stationary gold nucleus. As the impact parameter is reduced from a large value towards zero, how does the scattering angle change?
  1. It rises steadily towards 180°
  2. It stays close to 90° throughout
  3. It falls steadily towards 0°
  4. It rises to 90° and then falls back to 0°
Show answer
Answer: (A) It rises steadily towards 180°
Explanation: A smaller impact parameter means the alpha particle passes closer to the nucleus and feels a stronger Coulomb repulsion for longer. The deflection therefore grows as the impact parameter shrinks. In the limit of a head-on approach the particle stops momentarily and rebounds along its path, which is a scattering angle of 180°. A large impact parameter leaves the particle nearly undeviated.

Question 9

Suppose an electron in a Rutherford-type atom obeyed classical electromagnetism, radiating as it orbits and spiralling inward. What emission spectrum does this picture predict, and how does it compare with observation for hydrogen?
  1. A line spectrum at fixed frequencies, because the orbit radius stays constant; this agrees with hydrogen
  2. A continuous spectrum, because the electron's energy increases as it falls inward; this agrees with hydrogen
  3. A continuous spectrum, because the revolution frequency keeps rising as the orbit shrinks; hydrogen actually shows discrete lines
  4. No radiation at all, because uniform circular motion involves no acceleration; hydrogen actually shows discrete lines
Show answer
Answer: (C) A continuous spectrum, because the revolution frequency keeps rising as the orbit shrinks; hydrogen actually shows discrete lines
Explanation: An accelerating charge radiates at its frequency of revolution. As the electron loses energy and spirals inward, this frequency increases steadily, so the classical model predicts a continuous spectrum. Atomic hydrogen actually emits sharp lines at particular wavelengths, which is one failure of the classical picture.

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