Thirty multiple choice questions. Three of them come from the 2023 papers, recovered by running optical character recognition over the scanned answer sheets and then re-solving each one independently.
How to use this worksheet. Learn which level each series ends on — Lyman on n = 1 in the ultraviolet, Balmer on n = 2 in the visible, Paschen on n = 3 in the infrared. Two of the three previous year questions are answered by that fact alone. For the shortest wavelength of any series, take n = infinity.
Section A · Level 1
Rutherford, Bohr and the hydrogen spectrum
Ten questions. None should take more than twenty seconds.
Q1
The alpha particle scattering experiment was performed under the direction of:
Geiger and Marsden carried out the experiment under Rutherford, whose interpretation of the rare large-angle deflections established the nuclear model of the atom.
Q2
The alpha particle scattering experiment established that an atom has:
Most alpha particles passed straight through, but a few bounced back sharply. That could only happen if the positive charge and nearly all the mass were concentrated in a tiny nucleus.
Q3
According to Bohr, the angular momentum of an electron in a stationary orbit is an integral multiple of:
Bohr's quantisation condition is mvr = nh ∕ 2π. This single postulate is what selects the allowed orbits.
Q4
While an electron remains in a stationary orbit, it:
Bohr postulated that stationary orbits are non-radiating, which is exactly what classical electromagnetism could not explain and why the model needed a new assumption.
Q5
The energy of the electron in the ground state of the hydrogen atom is:
En = −13.6 ∕ n² eV, so for n = 1 the energy is −13.6 eV. The negative sign means the electron is bound.
Q6
The ionisation energy of a hydrogen atom in its ground state is:
Ionisation means lifting the electron from −13.6 eV to zero, so 13.6 eV of energy is required.
Q7
The radius of the n-th Bohr orbit is proportional to:
rn = 0.529 n² Å, so the orbits spread out rapidly as n rises. The energy, by contrast, goes as 1 ∕ n².
Q8
Which series of the hydrogen spectrum lies in the ultraviolet region?
The Lyman series ends on n = 1 and carries the largest energy jumps, so it falls in the ultraviolet.
Q9
An emission spectral line is produced when an electron:
Falling to a lower orbit releases the energy difference as a photon. Absorbing a photon to jump upward produces an absorption line instead.
Q10
The size of a nucleus compared with the size of an atom is roughly:
An atom is about 10−10 m across while a nucleus is about 10−15 m, a factor of roughly 100000. Almost all of an atom is empty space.
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Section B · Level 2
Energy levels, radii and spectral series
Ten questions. Know which level each series ends on.
Q11
The energy of the electron in the n = 2 state of hydrogen is:
E2 = −13.6 ∕ 4 = −3.4 eV.
Q12
The energy required to excite a hydrogen atom from n = 1 to n = 2 is:
The gap is −3.4 − (−13.6) = 10.2 eV. This is the first line of the Lyman series.
Q13
The radius of the second Bohr orbit of hydrogen is closest to:
r2 = 0.529 × 2² = 2.12 Å. Option 2 comes from multiplying by n instead of n².
Q14
For the Balmer series, the electron falls to the level:
Balmer lines all end on n = 2, which places them in the visible region.
Q15
For the Paschen series, the electron falls to the level:
Paschen lines end on n = 3 and lie in the infrared.
Q16
The shortest wavelength line of a spectral series corresponds to a transition from:
The largest energy jump, and therefore the shortest wavelength, comes from the series limit where the electron falls from n = infinity.
Q17
The velocity of an electron in the n-th Bohr orbit is proportional to:
Combining the quantisation condition with the force balance gives v proportional to 1 ∕ n, so outer electrons move more slowly.
Q18
A photon of energy 10.2 eV is absorbed by a hydrogen atom in its ground state. The electron moves to:
10.2 eV is exactly the gap from n = 1 to n = 2, so the atom is excited to the first excited state rather than ionised.
Q19
According to the Rydberg formula, the wavelength of a spectral line depends on:
1 ∕ λ = R(1 ∕ n1² − 1 ∕ n2²) involves both levels, so a line is identified by the pair of states, not by one alone.
Q20
The Bohr model successfully explains the spectrum of:
The model works for one-electron systems such as hydrogen, singly ionised helium and doubly ionised lithium. It fails once electron-electron repulsion enters.
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Section C · Level 3
Transitions, limits and the model's reach
Seven questions. Each needs a formula plus a piece of reasoning.
Q21
In the Bohr model, the total energy of the electron is related to its kinetic energy by:
The potential energy is twice the kinetic energy in magnitude and negative, so the total energy equals minus the kinetic energy. This is a general result for inverse square attraction.
Q22
The number of spectral lines produced when hydrogen atoms de-excite from the n = 4 level is:
The number of possible downward transitions is n(n − 1) ∕ 2 = 4 × 3 ∕ 2 = 6.
Q23
The wavelength of the series limit of the Lyman series in terms of the Rydberg constant R is:
At the limit n2 = infinity and n1 = 1, so 1 ∕ λ = R and λ = 1 ∕ R.
Q24
A limitation of the Bohr model is that it:
The model gives the line positions for hydrogen correctly but says nothing about their relative brightness, and it fails entirely for multi-electron atoms.
Q25
As the principal quantum number n increases, the spacing between successive energy levels:
Since E goes as −1 ∕ n², the levels crowd together as n rises and converge on zero at the ionisation limit.
Q26
For a hydrogen-like ion of atomic number Z, the energy of the n-th level is:
The energy scales with the square of the nuclear charge, giving En = −13.6 Z² ∕ n² eV. For He⁺ with Z = 2 the ground state is therefore −54.4 eV.
Q27
The impact parameter for an alpha particle that is scattered straight back through 180° is:
A head-on approach means zero impact parameter, and the particle retraces its path after reaching the distance of closest approach.
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Section D · Previous year questions
Straight from the AAI ATC papers
Three questions from the 2023 papers, recovered by optical character recognition of the scanned answer sheets and independently re-solved.
Q28PYQ 2023 · Morning
Which of the following spectral series lies in the visible region of the hydrogen spectrum?
Only the Balmer series, which ends on n = 2, falls in the visible range. Lyman lies in the ultraviolet, while Paschen and Brackett lie in the infrared. Learning which level each series ends on settles every question of this type.
Q29PYQ 2023 · Morning
An electron in a hydrogen atom (Bohr model) makes a transition from the n = 2 state to the n = 1 state. The frequency of the emitted photon is (take R = 1.0 × 107 m−1):
Using the Rydberg formula, 1 ∕ λ = R(1 ∕ 1 − 1 ∕ 4) = 1.0 × 107 × 0.75 = 7.5 × 106 m−1, so λ = 1.33 × 10−7 m. Then f = c ∕ λ = (3 × 108) ∕ (1.33 × 10−7) = 2.25 × 1015 Hz. Work out the wavelength first and convert at the end — going straight for frequency is where the powers of ten go wrong.
Q30PYQ 2023 · Evening
The shortest wavelength in the Paschen series, in terms of the Rydberg constant R, is:
The shortest wavelength is the series limit, where the electron falls from n = infinity. For Paschen, n1 = 3, so 1 ∕ λ = R(1 ∕ 9 − 0) = R ∕ 9, giving λ = 9 ∕ R. Options 1 and 2 are the Lyman and Balmer limits respectively.
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↓ Free download — Atoms PDF
All thirty questions with the formula sheet printed on the front page and ruled working space for your circuit reductions. Print it, solve it, keep it.
Seven questions from one chapter in a single year.
Live classes cover the full Class 11 and 12 physics and maths syllabus for AAI ATC, taught from the real question papers with the method for each question type drilled until it is automatic.