Use scattering to locate the nucleus and energy differences to decode atomic spectra.
JEE Main · JEE Advanced · NEET UG65 min study guide9 indexed sections
BY THE END, YOU CAN
Interpret Rutherford scattering with explicit Coulomb-model assumptions.
Derive hydrogen-like radius, energy and kinetic/potential relations.
Calculate emission, absorption, excitation and ionisation energies.
Distinguish population line counts from individual cascades and Bohr models from orbitals.
BEFORE YOU START
Coulomb force, circular motion and energy conservation.
Photon energy hν=hc/λ and scientific notation.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Advanced 2020. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Matched topics: Atomic Structure and Spectra.
JEE Advanced · 2020 · Paper 1 · Physics · Q7
Paraphrased exam question
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
For a particle of mass m in V(r)=Fr with F>0, apply Bohr quantization to circular orbits. Determine how radius R and speed v depend on orbit number n, and find total energy E. Here h denotes Planck’s constant.
Reveal answer & worked solution
Answer: R∝n^(2/3), v∝n^(1/3); E=(3/2)[n²h²F²/(4π²m)]^(1/3), matching options B and C (independently derived).
The inward force has magnitude F, so mv²/R=F. Bohr quantization gives mvR=nh/(2π).
Combining these yields R³=n²h²/(4π²mF), hence R∝n^(2/3) and v=√(FR/m)∝n^(1/3).
Kinetic energy is FR/2 and potential energy is FR, so E=3FR/2. Substitute R to obtain the stated expression.
Source: JEE (Advanced) 2020, Paper 1, Physics Q7; independently derived solution; key finality unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. Scattering reveals where an atom’s mass and charge are concentrated
An atom is electrically neutral when its electron count equals its nuclear charge number Z. Rutherford-type alpha-particle scattering showed that most incident particles passed through thin foil with small deflection while a small fraction scattered through large angles. This pattern supported a small, concentrated positive nucleus rather than positive charge spread uniformly throughout the whole atom.
The interpretation depends on a model: a fast charged projectile interacts mainly through Coulomb repulsion with a much heavier target nucleus, and multiple scattering is limited by a thin target. Large-angle events probe close approaches. The fact that most particles pass through does not mean atoms contain no fields; it means close encounters with the compact nucleus are comparatively rare.
For a head-on nonrelativistic projectile of charge ze and kinetic energy K approaching a fixed nucleus of charge Ze, conservation of energy gives K = kZze²/rmin at its turning point. This closest-approach estimate neglects nuclear recoil, electron screening at the relevant short distance and the strong interaction. It is a probe scale, not automatically the exact nuclear radius.
At nonzero impact parameter, angular momentum prevents a head-on stop and the projectile follows a deflected orbit. In the ideal Rutherford model, scattering intensity varies strongly as 1/sin⁴(θ/2) and decreases with the square of projectile energy for fixed geometry and charges. These trends explain why backscattering is rare and higher-energy projectiles are harder to deflect.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Schematic, not a nuclear size scale. The horizontal closest-approach bracket measures to the nucleus centre; it is not a nuclear radius. Nuclear recoil and non-Coulomb interactions are neglected.
ORIGINAL WORKED EXAMPLE
Head-on closest approach
An alpha particle of charge +2e and kinetic energy 5.0 MeV approaches a fixed nucleus with Z = 50 head-on. Use ke² = 1.44 MeV fm. Estimate closest approach.
The Coulomb potential energy is k(2e)(50e)/r = 100ke²/r.
At the turning point, 5.0 MeV = (100 × 1.44 MeV fm)/r.
Thus r = 28.8 fm. This is an electrostatic turning distance under the fixed-nucleus approximation.
Answer Approximately 29 fm.
02 / UNDERSTAND & APPLY
2. A classical orbit cannot explain stable line spectra
A naive planetary atom combines a positive nucleus with orbiting electrons held by Coulomb attraction. Classical circular dynamics can balance centripetal acceleration, but classical electrodynamics predicts that an accelerating charge radiates. An orbiting electron would lose energy and spiral inward, so this picture alone does not explain stable atoms.
Experiments also show discrete emission and absorption lines rather than an arbitrary continuum from a slowly shrinking orbit. Different elements have characteristic spectra, and hydrogen’s line frequencies follow a simple pattern. A successful model must explain both stability and discrete energy differences, not merely draw electrons around a nucleus.
Bohr’s model introduced allowed stationary states in which the electron does not radiate continuously, plus transitions that emit or absorb photons with energy equal to a state-energy difference. This was a historically important limited model, not a derivation from classical mechanics. Modern quantum mechanics replaces literal fixed circular electron paths with quantum states and probability distributions.
03 / UNDERSTAND & APPLY
3. Quantised angular momentum selects hydrogen-like states
For one electron bound to a nucleus of charge +Ze, Bohr combined Coulomb circular motion with angular-momentum quantisation mvr = nħ, where n = 1,2,3… and ħ = h/(2π). Using mv²/r = kZe²/r² gives rₙ proportional to n²/Z and vₙ proportional to Z/n.
With the usual fixed-nucleus approximation, rₙ = a₀n²/Z, where a₀ ≈ 0.0529 nm, and Eₙ = −13.6Z²/n² eV. The zero of energy is an electron and nucleus at infinite separation with zero relative kinetic energy. Negative energy means the electron is bound; a higher n is a less negative, less tightly bound state.
Coulomb potential energy is U = −kZe²/r and the circular kinetic energy is K = kZe²/(2r), so U = −2K and total E = −K. Increasing orbit radius therefore decreases kinetic energy and raises total energy toward zero. Confusing the signs of total and kinetic energy is a frequent error.
The model applies directly to hydrogen-like one-electron ions such as He⁺ and Li²⁺, not to neutral helium or arbitrary many-electron atoms with electron–electron interactions. Finite nuclear mass can be included through reduced mass for more precise predictions. Relativistic, spin and radiative effects produce refinements beyond these elementary formulas.
Bohr circular-state relations; the modern quantum model does not assign the same literal orbit picture to all states.
ORIGINAL WORKED EXAMPLE
Compare a hydrogen-like ion with hydrogen
Find radius and energy of the n = 2 state of He⁺, using a₀ = 0.0529 nm.
He⁺ has one electron and nuclear charge Z = 2, so the hydrogen-like formulas apply.
Radius is a₀n²/Z = 0.0529 × 4/2 = 0.1058 nm.
Energy is −13.6 × 4/4 = −13.6 eV. This equals hydrogen’s ground-state energy, but its radius differs from hydrogen’s ground-state radius.
Answer Radius 0.1058 nm; energy −13.6 eV.
04 / UNDERSTAND & APPLY
4. A photon measures a difference between levels
For a downward transition from ni to nf with ni > nf, emitted photon energy is Eni − Enf = 13.6Z²(1/nf² − 1/ni²) eV. The electron ends at the lower, more negative energy. For absorption, the photon must supply the positive gap from a lower occupied level to an allowed higher state, subject to the detailed selection rules of quantum theory.
Photon frequency is ΔE/h and vacuum wavelength hc/ΔE. Do not use the magnitude of either individual level energy unless the transition is ionisation to the zero-energy continuum. The wavelength decreases as the energy gap increases. An atom’s emitted spectrum therefore records its energy-level differences rather than the frequency of a classical electron’s revolution.
The Rydberg form is 1/λ = RZ²(1/nf² − 1/ni²) in the same approximation. For hydrogen, transitions ending at n = 1 form the Lyman series, n = 2 the Balmer series and n = 3 the Paschen series. The first line of a series comes from ni = nf + 1, while its short-wavelength limit comes from ni → ∞.
Within one series, successive lines crowd together toward the limit because high-n levels become closely spaced. The longest wavelength corresponds to the smallest energy gap in that series. The phrase “series limit” is therefore a specific transition limit, not a largest wavelength or a largest finite principal quantum number.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Level labels and photon energies are quantitative; vertical spacing is schematic and not a linear energy scale. The zero-energy line is the continuum threshold.
ORIGINAL WORKED EXAMPLE
A Balmer-line calculation
A hydrogen atom falls from n = 3 to n = 2. Find photon energy and approximate vacuum wavelength using hc = 1240 eV nm.
Energy gap is 13.6(1/4 − 1/9) = 13.6 × 5/36 ≈ 1.889 eV.
Wavelength is 1240/1.889 ≈ 656.5 nm.
The final level n=2 identifies a Balmer-series transition.
Answer Approximately 1.89 eV and 656 nm.
05 / UNDERSTAND & APPLY
5. Excitation and ionisation ask for different final states
Excitation moves the electron to another bound state; ionisation removes it to the continuum. From hydrogen level n, the minimum ionisation energy in the fixed-nucleus model is 13.6/n² eV. From ground state, the first excitation to n=2 requires 10.2 eV, while ionisation requires 13.6 eV.
A photon below the ionisation threshold is not necessarily absorbed merely because its energy is positive. It must match an allowed bound-state gap within the relevant line width. A photon above the threshold can ionise, leaving excess energy as kinetic energy of the outgoing particles. In the simple heavy-nucleus approximation, the electron receives approximately hν − Eion.
An incident electron or other massive projectile can transfer only part of its kinetic energy in a collision, so excitation by impact has a threshold rather than the same exact photon-energy matching rule. Energy and momentum conservation still apply. Do not assert that a 12 eV electron cannot excite a 10.2 eV transition simply because the two energies differ.
Ionisation energy is positive even though the bound-state energy is negative. It is the work required to raise the system from En to zero. Removing an electron from an excited state costs less than removing it from the ground state; spontaneous downward emission does the opposite, releasing energy while making the remaining atom more tightly bound.
ORIGINAL WORKED EXAMPLE
An atom already excited
Hydrogen starts in n = 3. Find its ionisation energy and the kinetic energy of an emitted electron if a 4.0 eV photon ionises it, neglecting recoil.
Initial binding magnitude is 13.6/9 ≈ 1.511 eV.
The photon exceeds that threshold, so the simple outgoing-electron kinetic energy is 4.0 − 1.511 ≈ 2.489 eV.
Using 13.6 eV would incorrectly assume a ground-state atom.
Answer Ionisation energy 1.51 eV; electron kinetic energy about 2.49 eV.
06 / UNDERSTAND & APPLY
6. Count possible lines without inventing a cascade
If a population of hydrogen-like atoms occupies level n and can cascade through all lower levels under an elementary all-pairs counting model, the maximum number of distinct downward energy differences among levels 1 through n is n(n−1)/2. This counts unordered level pairs, not the number of photons emitted by one atom in a single path.
A single atom descending from n to 1 can emit at most n−1 photons if it steps through every intermediate level, or fewer if it makes larger jumps. A large population can follow different paths and produce more distinct lines collectively. Detailed quantum selection rules and state populations can restrict which transitions actually appear, so the elementary maximum is not a guarantee of observed intensity for every pair.
For absorption by atoms all initially in one lower state, only transitions originating from that state are immediately available. The emission spectrum after excitation can be richer because cascades populate intermediate states. State preparation is therefore part of a line-counting problem, not background wording to ignore.
ORIGINAL WORKED EXAMPLE
Population maximum versus one cascade
In an elementary level-counting model, many hydrogen atoms are prepared at n = 4 and may decay through all lower levels. Find the maximum number of distinct lines and maximum photons from one atom on a fully stepwise cascade.
There are four levels in the set, so distinct downward pairs number 4×3/2 = 6.
A single stepwise cascade can take 4→3→2→1, emitting three photons.
The six-line population maximum does not mean each atom emits six photons.
Answer Six possible distinct lines collectively; at most three photons in one fully stepwise cascade.
07 / UNDERSTAND & APPLY
7. What the modern description keeps and changes
Modern quantum mechanics retains discrete stationary-state energies and photon energy differences but describes electrons using wavefunctions rather than definite Bohr paths. Quantum numbers specify properties of states; probability density describes where an electron may be detected. An orbital is not a small classical orbit drawn around a nucleus.
For hydrogen-like systems, the simple energy dependence on n is a strong leading approximation. Additional quantum numbers, angular momentum, spin and interactions are needed to explain selection rules, fine structure, magnetic splitting and many-electron spectra. The Bohr angular momentum nħ should not be substituted for the modern orbital magnitude √[l(l+1)]ħ without identifying which model is intended.
Spectral measurements can identify elements because their energy structures differ, but an observed line can also be shifted by motion, fields or environment. In elementary problems these effects are often neglected. A useful answer states the atomic species, charge state, initial/final levels and approximation before giving a number.
08 / UNDERSTAND & APPLY
8. Original practice with level and sign checks
Use the hydrogen-like fixed-nucleus formulas and hc=1240 eV nm unless stated otherwise. The questions are original teaching practice.
ORIGINAL PRACTICE · 1
For hydrogen, compare r₃ with r₁.
Reveal answer & reasoning
r₃ = 9r₁.
Radius scales as n² at fixed Z.
ORIGINAL PRACTICE · 2
For hydrogen, find total, kinetic and potential energy at n=2.
Reveal answer & reasoning
E=−3.4 eV; K=+3.4 eV; U=−6.8 eV.
E=−13.6/4 and the Coulomb circular-state relations give K=−E,U=2E.
ORIGINAL PRACTICE · 3
Can the hydrogen-like formula be directly applied to neutral helium?
Reveal answer & reasoning
No.
Neutral helium has two interacting electrons; He⁺ is the one-electron ion.
ORIGINAL PRACTICE · 4
What energy is needed to excite ground-state hydrogen to n=2?
Reveal answer & reasoning
10.2 eV.
Subtract −13.6 eV from −3.4 eV.
ORIGINAL PRACTICE · 5
What is the ionisation energy of He⁺ in its ground state?
Reveal answer & reasoning
54.4 eV.
Binding energy scales as Z², giving13.6×4.
ORIGINAL PRACTICE · 6
Which hydrogen transition has greater photon energy:3→2 or2→1?
Reveal answer & reasoning
2→1.
Its gap is10.2 eV, compared with about1.89 eV for3→2.
ORIGINAL PRACTICE · 7
Find the hydrogen Lyman-series limiting wavelength.
Reveal answer & reasoning
Approximately91.2 nm.
The limiting gap to n=1 is13.6 eV, so λ=1240/13.6.
ORIGINAL PRACTICE · 8
A population can use levels1 through5 in the elementary all-pairs model. What is maximum distinct emission-line count?
Reveal answer & reasoning
10.
n(n−1)/2=5×4/2.
ORIGINAL PRACTICE · 9
An11 eV photon meets ground-state hydrogen. Does the elementary discrete-level model guarantee absorption?
Reveal answer & reasoning
No.
It does not match the10.2 eV first excitation or another allowed ground-state bound gap, and it is below13.6 eV ionisation.
ORIGINAL PRACTICE · 10
Can an11 eV incident electron excite ground-state hydrogen to n=2 in principle?
Reveal answer & reasoning
Yes, if collision energy/momentum conditions permit.
It can transfer10.2 eV and retain excess kinetic energy; impact excitation is not restricted to exact photon-like energy matching.
ORIGINAL PRACTICE · 11
For the same head-on nuclear target, projectile kinetic energy doubles. What happens to Coulomb closest approach?
Reveal answer & reasoning
It halves.
rmin=kZze²/K in the fixed-target model.
ORIGINAL PRACTICE · 12
Why does a negative bound-state energy not mean negative kinetic energy?
Reveal answer & reasoning
The energy reference includes negative potential energy.
For a Coulomb circular state,K is positive,U is twice as negative as totalE, and E=K+U<0.
09 / UNDERSTAND & APPLY
9. Read every spectrum as a map of energy differences
Identify the atom or ion before choosing Z, then decide whether the event is excitation, emission or ionisation. Use signed level energies to obtain a positive transferred photon energy. Check wavelength units and compare with the expected series. For a population count, separate possible lines from photons along one cascade.
Finally name the model boundary. Rutherford scattering is a Coulomb probe approximation; Bohr states describe one-electron energy scaling; the modern orbital picture is not a literal planetary path. Recognising what each model explains is as important as remembering its compact formula.
Rutherford scattering supports a concentrated nucleus but does not by itself determine every nuclear property.
Hydrogen-like means one electron, not merely an element with a small atomic number.
Negative total energy signals binding, not negative kinetic energy.
A spectral photon carries an energy difference.
Excitation, ionisation and line counting have different final-state conditions.
Bohr orbits and quantum orbitals are different descriptions.
KEEP THE BIG IDEAS
Quick revision
Atomic stability and discrete spectra require more than a classical orbit model.
One-electron energies scale asZ²/n² under the stated approximation.
Transitions measure level differences; ionisation reaches the continuum.
Counting possible lines does not count photons emitted by one atom.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.