Rutherford's model is often revised as though it already contains Bohr's rules. It does not. The scattering data fix where the charge and mass sit; they say nothing about which orbits an electron may occupy. Any statement about allowed radii, quantised angular momentum or discrete energy levels belongs to the next model, not this one.
What the model asserts (NCERT Class 12 Physics, Chapter 12 — Atoms, page 293): all of the atom's positive charge and nearly all of its mass are concentrated in a central nucleus of radius about 1 × 10⁻¹⁵ m, while the atom as a whole is about 1 × 10⁻¹⁰ m across. The nucleus therefore occupies roughly 10⁻¹⁵ of the atom's volume — the atom is overwhelmingly empty space. The electrons revolve around this nucleus, held by its electrostatic attraction. Gravity is ignored, and the worked example shows by how large a margin that is justified.
The dynamics follow from a single line. Coulomb attraction supplies the centripetal force:
mv²/r = (1/4πε₀)(Ze²/r²)
so K = Ze²/(8πε₀r), |U| = Ze²/(4πε₀r) = 2K, and E = K + U = −K. Three consequences worth holding: the total energy is negative, so the electron is bound; its magnitude equals the kinetic energy; and both depend on r, which the model leaves entirely free.
The model's failure is examinable too. A revolving electron is accelerating, and classical electromagnetism requires an accelerating charge to radiate. Losing energy makes E more negative, so r shrinks while K rises — the electron spirals inward and the emitted frequency climbs continuously, predicting a continuous spectrum and an atom that collapses. Bohr's postulates, covered in a separate lesson, are the response to exactly this.
Watch out: in this topic, "allowed" means not restricted. If an option quantises anything, it is describing a different model.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Rutherford's nuclear model of the atom, as inferred from the scattering data, asserts that
Answer: C. The nuclear model places the whole positive charge and almost the entire mass in a tiny central nucleus, with the electrons outside it — NCERT Class 12 Physics, Chapter 12 (Atoms), page 293.
Why A is wrong: A describes Thomson's uniform-sphere picture, which is the model the large-angle scattering displaced.
Why B is wrong: B reverses the charges: the nucleus is positive, and the negative electrons lie outside it.
Why D is wrong: D fails on size: the nucleus is smaller than the atom by a factor of about 10⁵, and the electrons are not inside it.
In Rutherford's model the electron moves in a circular orbit around the nucleus. The force that supplies the required centripetal force is
Answer: A. The model holds the electron in orbit by the Coulomb attraction of the positive nucleus, which plays the role of the centripetal force — NCERT Class 12 Physics, Chapter 12 (Atoms), page 295.
Why B is wrong: B underestimates the gap: the gravitational attraction between an electron and a proton is smaller than the electrostatic attraction by about 10³⁹ (see the worked example), so it cannot sustain the orbit.
Why C is wrong: C misapplies the strong force, which acts between nucleons and does not act on electrons at all.
Why D is wrong: D invokes a magnetic force with no external field present, and a charge cannot supply its own centripetal force this way.
Rutherford's model, taken on its own, restricts the radius of the electron's orbit to
Answer: D. The model contains no selection rule for the orbit; a continuous range of radii, and therefore of energies, is allowed — NCERT Class 12 Physics, Chapter 12 (Atoms), page 296.
Why A is wrong: A treats 5.3 × 10⁻¹¹ m as a derived result, but it is the observed size scale of a hydrogen atom, not something this model predicts.
Why B is wrong: B invents a spacing rule; nothing in the model selects integer multiples of any length.
Why C is wrong: C imports Bohr's angular-momentum postulate, which was introduced later precisely because Rutherford's model imposes no such condition.
Take the nucleus to have a radius of about 1 × 10⁻¹⁵ m and the atom about 1 × 10⁻¹⁰ m. On Rutherford's picture, the fraction of the atom's volume occupied by the nucleus is of the order of
Answer: B. Volume scales as the cube of the radius, so the fraction is (10⁻¹⁵/10⁻¹⁰)³ = (10⁻⁵)³ = 10⁻¹⁵ — the emptiness of the atom stated on NCERT Class 12 Physics, Chapter 12 (Atoms), page 294.
Why A is wrong: A stops at the ratio of the radii and never cubes it to reach a volume ratio.
Why C is wrong: C comes from combining the two length scales without cubing, and matches neither a radius nor a volume ratio.
Why D is wrong: D squares the volume ratio instead of cubing the radius ratio once.
An electron of charge magnitude e is held in a circular orbit of radius r by the Coulomb attraction of a nucleus of charge +Ze. Its kinetic energy K and the magnitude of its potential energy |U| are related by
Answer: A. Balancing mv²/r against Ze²/4πε₀r² gives mv² = Ze²/4πε₀r, so K = Ze²/8πε₀r while |U| = Ze²/4πε₀r — exactly twice K, from the electron-orbit treatment in NCERT Class 12 Physics, Chapter 12 (Atoms).
Why B is wrong: B drops the factor of 2 that appears because the kinetic energy carries the ½ in ½mv² while mv² equals the full Coulomb term.
Why C is wrong: C inverts the relation: the potential energy is the larger of the two in magnitude, not the smaller.
Why D is wrong: D introduces a factor of 4 that no step in the Coulomb-centripetal balance produces.
For the same orbiting electron, the total energy E is
Answer: D. With U = −2K, the total is E = K + U = −K: negative, so the electron is bound, and equal in magnitude to the kinetic energy — from the electron-orbit treatment in NCERT Class 12 Physics, Chapter 12 (Atoms).
Why A is wrong: A gets the magnitude right but the sign wrong; a positive total energy would describe an unbound electron that escapes the nucleus.
Why B is wrong: B assumes the two energies cancel, but K = ½|U|, so a residue of magnitude K is left over.
Why C is wrong: C adds K and |U| instead of subtracting, and loses the sign as well.
In Rutherford's model, the electron of a hydrogen atom moves in a circular orbit of radius 5.3 × 10⁻¹¹ m about a proton. Taking e = 1.6 × 10⁻¹⁹ C, mₑ = 9.1 × 10⁻³¹ kg and 1/4πε₀ = 9.0 × 10⁹ N m² C⁻², the orbital speed is closest to
Answer: C. From mv²/r = e²/4πε₀r², v² = (9.0 × 10⁹)(1.6 × 10⁻¹⁹)²/[(9.1 × 10⁻³¹)(5.3 × 10⁻¹¹)] = 4.8 × 10¹² m² s⁻², so v = 2.2 × 10⁶ m s⁻¹ — the classical orbit speed implied by the model of NCERT Class 12 Physics, Chapter 12 (Atoms).
Why A is wrong: A is the correct value divided by 10², an arithmetic slip in the powers of ten during the substitution.
Why B is wrong: B is v² in m² s⁻², reported before the square root was taken — note that it is not even a speed dimensionally.
Why D is wrong: D is the speed of light; the orbital speed is smaller than this by more than two orders of magnitude, which is what makes the non-relativistic treatment legitimate.
Classical electromagnetism requires the orbiting electron, being accelerated, to radiate energy continuously. In Rutherford's model this means the orbit radius and the electron's kinetic energy would respectively
Answer: B. Radiating makes E = −Ze²/8πε₀r more negative, which requires a smaller r; since K = Ze²/8πε₀r, the kinetic energy rises as the orbit shrinks — the spiralling collapse that the model of NCERT Class 12 Physics, Chapter 12 (Atoms) cannot avoid.
Why A is wrong: A has the radius right but not the kinetic energy: K = Ze²/8πε₀r rises as r falls, so it cannot decrease alongside the radius.
Why C is wrong: C would require the total energy to become less negative, which is the opposite of what radiating away energy does.
Why D is wrong: D reverses both changes; losing energy cannot expand the orbit.
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Given.
Electron–proton separation r = 5.3 × 10⁻¹¹ m; e = 1.6 × 10⁻¹⁹ C; mₑ = 9.1 × 10⁻³¹ kg; m_p = 1.67 × 10⁻²⁷ kg; 1/4πε₀ = 9.0 × 10⁹ N m² C⁻²; G = 6.67 × 10⁻¹¹ N m² kg⁻².
Required.
The ratio of the electrostatic force to the gravitational force between the electron and the nucleus.
Concept.
Rutherford's model attributes the electron's orbit entirely to the nucleus's electrostatic attraction and ignores gravity. The ratio quantifies whether that omission is defensible.
Formula.
F_e = (1/4πε₀)·e²/r² and F_g = G·mₑm_p/r². Both fall as 1/r², so in the ratio r² cancels:
F_e/F_g = e²/(4πε₀·G·mₑ·m_p).
Substitution.
F_e = (9.0 × 10⁹)(1.6 × 10⁻¹⁹)²/(5.3 × 10⁻¹¹)²
F_g = (6.67 × 10⁻¹¹)(9.1 × 10⁻³¹)(1.67 × 10⁻²⁷)/(5.3 × 10⁻¹¹)²
Calculation.
r² = 2.8 × 10⁻²¹ m².
Numerator of F_e: (9.0 × 10⁹)(2.56 × 10⁻³⁸) = 2.3 × 10⁻²⁸ N m², so F_e = 8.2 × 10⁻⁸ N.
Numerator of F_g: (6.67 × 10⁻¹¹)(1.52 × 10⁻⁵⁷) = 1.0 × 10⁻⁶⁷ N m², so F_g = 3.6 × 10⁻⁴⁷ N.
Ratio = 8.2 × 10⁻⁸ / 3.6 × 10⁻⁴⁷ = 2.3 × 10³⁹.
Final answer.
F_e/F_g ≈ 2.3 × 10³⁹ — the electrostatic attraction exceeds the gravitational attraction by about thirty-nine orders of magnitude, so neglecting gravity in the model costs nothing. The exponents 2 on r and on e are counting exponents, not measured quantities, and contribute nothing to the significant-figure count; the supplied data carry two significant figures, so the answer is quoted to two.
Common trap.
Because r² cancels, the ratio is the same at every separation — it is not true that gravity "starts to matter" at larger radii. Students who instead evaluate both forces separately frequently mis-handle r² = 2.8 × 10⁻²¹ m² and lose a power of ten in one force only, which shifts the ratio by an order of magnitude while leaving the answer superficially plausible. This pitfall is not a corpus trap ID; it comes from the calculation itself.
Similar NEET-style question.
Two electrons are held 1.0 × 10⁻¹⁰ m apart. The ratio of the electrostatic repulsion between them to their gravitational attraction is of the order of (A) 10³⁹ (B) 10⁴² (C) 10²⁰ (D) 10⁻⁴². Answer: B — e²/(4πε₀·G·mₑ²) ≈ 4 × 10⁴², larger than the electron–proton case because the proton mass in the denominator is replaced by the much smaller electron mass.
Rutherford's experiment (1911): most α particles passed through gold foil with little deflection; few were scattered through large angles. Conclusion: atom has a small dense nucleus.
-- NCERT Class 12 Physics, Ch. 12, p. 291More in Atoms and Nuclei: 3 exam traps and mistakes · 6 formulas · 5 question patterns from its other lessons.
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