AtomsNEET MCQs with solutions

Not quite — the answer is C.

The 3→2 transition belongs to the Balmer series (Hα line) with wavelength ≈656 nm. Option A is Lyman-α (2→1) and Option B is Lyman-β (3→1), both UV. Option D (486 nm) is the 4→2 Hβ line.

Not quite — the answer is A.

Using rₙ=n²a₀, r₂=4a₀. Radius scales as n², not n. Choosing 2a₀ is the classic linear-scaling trap.

Not quite — the answer is C.

Most atomic volume is empty space with positive charge and mass concentrated in a tiny nucleus, so most alpha particles pass through undeflected. Uniform charge distribution (Thomson model) would cause only small, uniform deflections — not the observed pattern.

Not quite — the answer is D.

For hydrogen, r_n = n²a₀. For n=2, r₂ = 4a₀. The common trap is assuming r ∝ n, giving 2a₀ — this error arises from confusing linear scaling with quadratic scaling of orbit radius.

Not quite — the answer is A.

Bohr's quantisation condition: angular momentum L = mvr = nh/(2π), where n = 1,2,3,... Only these discrete values are permitted. Option C (KE is multiple of h) is dimensionally wrong — h has units J·s, not joules.

Not quite — the answer is A.

Bohr's first postulate: electrons occupy only certain stationary orbits and do not radiate while in those orbits. Radiation occurs only during transitions between two stationary states. Option D is wrong — "every possible orbit" implies arbitrary orbits, contradicting quantisation.

Not quite — the answer is A.

Ground state is the lowest allowed energy level, n=1 with E₁ = −13.6 eV. n=∞ has E = 0 and is the ionisation limit, not the ground state. Students who confuse lowest energy with smallest n value may still choose correctly here.

Not quite — the answer is C.

Eₙ = −13.6/n² eV. The gap between levels n and n+1 equals 13.6[1/n² − 1/(n+1)²], which shrinks toward zero as n → ∞. Options A and D are physically wrong; Option B is incorrect because nuclear charge is constant.

Not quite — the answer is B.

A stable orbit requires the circumference to contain an integer number of de Broglie wavelengths, giving 2πr = nλ. This produces a standing matter wave with no destructive interference. Option A (λ/n) and Option D (λ/n²) incorrectly invert or square the n-dependence.

Not quite — the answer is C.

For n=2: r₂ = a₀n² = 52.9 × 4 = 211.6 pm. Using 2πr = nλ: λ = 2πr/n = πr = 211.6π pm. Option B (211.6 pm) is the radius r₂ itself — students who omit the π factor choose this.

Not quite — the answer is C.

rn = n²a₀, so r₂ = 4a₀ = 4r₁. Option A (×2) applies r ∝ n instead of n². Option D (r₁/4) inverts the ratio — a trap for students who write r ∝ 1/n².

Not quite — the answer is A.

r₃ = 9×0.529 = 4.77 Å. Option D (2.38 Å) arises from r ∝ n (linear), giving 3×0.529/√3 — a ×n error. Option C (9.54 Å) doubles the correct answer — a factor-of-2 slip.

Not quite — the answer is B.

The Lyman series consists of all transitions from higher levels to n=1, lying in the ultraviolet region. Option A (n=2) is the Balmer series terminal level — the most common confusion in series identification.

Not quite — the answer is A.

Shortest wavelength = maximum photon energy = series limit. At n=∞→2, ΔE is maximum for the Balmer series. Option B (3→2) is the longest Balmer wavelength — students frequently invert shortest/longest.

Not quite — the answer is A.

A very thin target was needed so alpha particles interact with individual atoms. Gold's exceptional malleability allows it to be beaten into sheets thin enough without tearing. Options B, C, and D are factually incorrect.

Not quite — the answer is C.

The overwhelming majority of alpha particles passed through with negligible deviation, showing the atom is mostly empty space. Only a tiny fraction passed close enough to the nucleus to suffer significant deflection.

Not quite — the answer is A.

Rutherford concluded that nearly all positive charge and most atomic mass are in a tiny central nucleus. Most of the atom is empty space with electrons orbiting outside. Option B describes the Thomson model, which scattering disproved.

Not quite — the answer is B.

Large-angle deflections require a strong localised electric field. Thomson's diffuse positive charge cannot generate such a field, making large deflections impossible in his model. Option A (most passing through) is actually consistent with Thomson's model.

Not quite — the answer is B.

Excited hydrogen emits photons only at wavelengths corresponding to transitions between discrete energy levels. Since only certain photon energies are possible, sharp well-separated spectral lines appear on a dark background — a line emission spectrum, not a continuous spread.

Not quite — the answer is C.

Atomic energy levels are quantised, so only transitions between specific pairs of levels are possible. Each transition releases a photon whose energy equals the level difference. Fixed level differences produce fixed photon energies and therefore fixed wavelengths — a discrete line spectrum.

Not quite — the answer is A.

Ionisation = taking electron from E₁ = −13.6 eV to E = 0. Energy required = 0 − (−13.6) = 13.6 eV. Option B (10.2 eV) is excitation energy n=1→2, the standard confusion.

Not quite — the answer is C.

E₁ = −13.6 eV, E₂ = −3.4 eV. Excitation energy = E₂ − E₁ = 10.2 eV. Option B (3.4 eV) is ionisation energy from n=2, a frequent confusion between excitation and ionisation.

Not quite — the answer is C.

Bohr's model applies to all one-electron systems: H, He+, Li2+, Be3+. It is not restricted to hydrogen alone. Helium and multi-electron atoms are outside its scope due to electron-electron interactions.

Not quite — the answer is D.

Bohr's model is valid only for one-electron systems. H, He+, and deuterium each have one electron. Ne+ has nine electrons; electron-electron interactions invalidate the one-body Bohr picture entirely.