Replace the electron’s imagined track with quantum states, probability and a disciplined filling model.
JEE Main · JEE Advanced · NEET UG48 min study guide10 indexed sections
BY THE END, YOU CAN
Calculate matter wavelengths and apply position–momentum uncertainty with correct assumptions.
Interpret wavefunctions, probability density, radial probability and nodes.
Validate quantum-number sets and count orbitals and electron capacities.
Explain hydrogen-like degeneracy and many-electron energy splitting.
Write neutral and ionic configurations using Pauli, Hund and the limits of Aufbau ordering.
BEFORE YOU START
Hydrogen-like energy levels and photon energy.
Atomic number, ionic charge and electron counting.
Basic algebra, units 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.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. Momentum sets the wavelength of matter
The de Broglie relation λ=h/p connects wavelength to momentum for matter as well as providing the familiar photon relation. For a nonrelativistic particle of mass m and speed v, p=mv, so λ=h/(mv). Use kilograms with SI Planck’s constant. Substituting a mass in grams introduces a factor of one thousand.
For a nonrelativistic particle of kinetic energy K, p=√(2mK), giving λ=h/√(2mK). At the same speed, a heavier particle has a shorter wavelength in inverse proportion to mass. At the same kinetic energy, wavelength instead scales as the inverse square root of mass. These two comparison conditions must not be mixed.
An electron accelerated from rest through a potential difference of magnitude V gains kinetic energy eV in the ideal nonrelativistic model, neglecting losses. Substitution gives λ=h/√(2m_e eV), approximately 1.227/√V nm when V is in volts. At sufficiently high accelerating voltage, relativistic momentum must replace √(2mK).
Matter waves describe quantum amplitudes and observable interference statistics, not a little material object physically wobbling along a sinusoidal path. Individual detections can be localized even when many detections build a diffraction pattern. Wave-like propagation and particle-like detection are complementary features of the same quantum description.
λ=h/p≈h/(mv)=h/√(2mK)
The mv and kinetic-energy forms are nonrelativistic.
λ_e≈1.227/√V nm
Electron accelerated from rest through V volts; nonrelativistic, negligible energy loss.
ORIGINAL WORKED EXAMPLE
An accelerated electron wavelength
Estimate the de Broglie wavelength of an electron accelerated from rest through 100 V in the nonrelativistic model.
The electron kinetic energy is 100 eV, small compared with its rest-energy scale.
Use λ≈1.227/√100 nm.
√100=10, so λ≈0.1227 nm.
If the accelerating voltage became four times larger within the same model, wavelength would halve.
Answer Approximately 0.123 nm.
02 / UNDERSTAND & APPLY
2. A quantum state does not specify an exact classical trajectory
The position–momentum uncertainty relation for one Cartesian direction is ΔxΔp_x≥ħ/2. Here the uncertainties are statistical spreads in measurements for identically prepared states, not simply an instrument’s avoidable error bars. A state with a narrow position distribution necessarily has a broad spread of the conjugate momentum component.
For a nonrelativistic particle of fixed mass, Δp_x=mΔv_x, so Δv_x≥ħ/(2mΔx). This is a lower bound, not a prediction that every state exactly saturates it. The same position localization gives a much larger velocity spread for an electron than for a macroscopic mass.
Uncertainty does not say every physical property is unknowable, nor does it forbid a precisely defined energy in a stationary state. It constrains particular pairs of noncommuting observables. A hydrogen stationary state can have a sharply defined energy while its position and momentum distributions are spread out.
The old idea of a simultaneous exact orbit radius and exact tangential velocity is therefore not the modern atomic picture. Quantum calculations instead predict allowed states, energies and probabilities for measurement outcomes. The mathematical precision of a wavefunction does not turn its probability distribution into a hidden classical path.
ΔxΔp_x≥ħ/2; Δv_x≥ħ/(2mΔx)
The velocity form assumes nonrelativistic momentum and fixed mass.
ORIGINAL WORKED EXAMPLE
A lower bound on electron velocity spread
An electron state has Δx=0.100 nm. Estimate the minimum Δv_x using ħ=1.055×10⁻³⁴ J s and m_e=9.109×10⁻³¹ kg.
Convert Δx to 1.00×10⁻¹⁰ m.
Δp_x≥1.055×10⁻³⁴/(2×10⁻¹⁰)=5.275×10⁻²⁵ kg m s⁻¹.
Divide by electron mass: Δv_x≥5.275×10⁻²⁵/(9.109×10⁻³¹).
The lower bound is about 5.79×10⁵ m s⁻¹, not a definite electron speed.
Answer Δv_x≥5.79×10⁵ m s⁻¹.
03 / UNDERSTAND & APPLY
3. An orbital is a spatial wavefunction, not a container wall
In an atomic one-electron description, an orbital is a spatial wavefunction ψ that satisfies the appropriate quantum equation and boundary conditions. The probability density is |ψ|². Integrating that density over a region gives the probability of finding the electron there, and normalization makes the integral over all space equal to one.
The wavefunction itself can be positive, negative or complex; probability density cannot be negative. Different colors on orbital lobes conventionally show opposite signs or phases of ψ, not positive and negative electric charges. This phase distinction later controls constructive and destructive overlap in molecular bonding.
An orbital boundary surface usually encloses a chosen probability, such as a large specified fraction of the total. It is not a hard surface beyond which the electron cannot be found. Bound-state wavefunctions generally extend with decaying tails, so an atom has no single perfectly sharp quantum boundary.
For a hydrogenic 1s orbital, the probability density is largest at the nucleus, but the radial probability includes the increasing volume of spherical shells. The shell probability is proportional to 4πr²|ψ|² for a spherical orbital, and it is zero at r=0 because the shell volume vanishes. Its maximum occurs at a nonzero radius. These statements describe different probability questions and are consistent.
In many-electron atoms, a configuration built from orbitals is an approximate organized description of a correlated many-electron state. Orbital energies help rationalize filling and bonding, but they should not be mistaken for a collection of completely independent exact electron energies whose sum always equals the atom’s total energy.
04 / UNDERSTAND & APPLY
4. Quantum numbers define allowed state labels
The principal quantum number n is a positive integer. For a given n, the angular-momentum quantum number l can be 0 through n−1. The subshell labels s, p, d and f correspond to l=0, 1, 2 and 3. Thus 1p and 2d are not allowed atomic subshell labels because their proposed l values exceed n−1.
For a given l, the magnetic quantum number m_l takes integer values from −l to +l, giving 2l+1 spatial orbitals in that subshell. The electron spin projection quantum number m_s is +1/2 or −1/2. Spin is an intrinsic quantum property; the label should not be interpreted literally as a tiny charged sphere spinning on its own axis.
The first three quantum numbers identify a hydrogenic spatial orbital in the usual angular-momentum basis. Adding spin projection labels a one-electron spin-orbital. Real p_x, p_y and p_z drawings are convenient linear combinations or choices within a degenerate p subspace; one should not naively assign every familiar real-lobe drawing to a unique m_l value without specifying the basis.
A validity check should test the hierarchy in order: n first, then whether l is permitted by n, then whether m_l is permitted by l, then the spin value. Checking only that each number looks familiar misses combinations that violate the dependencies.
n=1,2,...; l=0,...,n−1; m_l=−l,...,+l; m_s=±1/2
Allowed quantum-number ranges for atomic electron state labels.
ORIGINAL WORKED EXAMPLE
Audit quantum-number sets
Which sets (n,l,m_l,m_s) are allowed: (3,2,−2,+1/2), (2,2,0,−1/2), (4,1,2,+1/2)?
(3,2,−2,+1/2) is allowed: l≤2 and m_l lies between −2 and +2.
(2,2,0,−1/2) fails because n=2 allows only l=0 or 1.
(4,1,2,+1/2) fails because l=1 allows only m_l=−1,0,+1.
A permitted spin value cannot repair an invalid spatial-number combination.
Answer Only (3,2,−2,+1/2) is allowed.
05 / UNDERSTAND & APPLY
5. Count orbitals, electrons and nodes separately
A subshell contains 2l+1 orbitals, and each orbital can accommodate at most two electrons with opposite spin projections. Hence s, p, d and f capacities are 2, 6, 10 and 14 electrons. These capacities count possible occupancy, not a claim that every atom fills each subshell completely.
For shell n, summing the orbital counts over l=0 through n−1 gives Σ(2l+1)=n² spatial orbitals. Its maximum electron capacity is 2n². The shell capacity does not determine the actual ground-state filling order across subshells of different n in many-electron atoms.
In the usual real hydrogenic atomic-orbital representation used for the node-counting questions here, the number of angular nodal surfaces is l and the number of radial nodes is n−l−1, giving n−1 total nodes in that convention. Complex m_l eigenfunctions need not have the same angular nodal-plane picture. Nodes are regions where the wavefunction vanishes. A radial node is a spherical nodal surface; angular nodes arise from the angular function and can be planes or cones.
The 1s orbital has no nodes. A 2s orbital has one radial node and no angular node, while a real 2p orbital such as 2p_z has no radial node and one angular nodal plane. A real 3p orbital has one of each. A picture showing opposite signs on two p lobes should include the nodal plane between them, rather than implying two separate electrons occupy the lobes.
An s orbital is spherically symmetric. Common real p orbitals have two lobes, and many d orbitals have four-lobe angular patterns; d_z² instead has two axial lobes and an equatorial toroidal region. Shape sketches are angular/probability representations with a stated contour convention, not rigid objects or electron trajectories.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Schematic cross-sections. Chosen outer contours do not bound all probability and are not electron trajectories. Overall wavefunction sign may be reversed without changing probability.
ORIGINAL WORKED EXAMPLE
Count a shell and one orbital’s nodes
For n=4, find the shell’s orbital count and electron capacity. For a 4d orbital, find radial and angular nodes.
The shell has n²=16 spatial orbitals and capacity 2n²=32 electrons.
A d orbital has l=2, so it has two angular nodes.
Radial nodes are n−l−1=4−2−1=1.
Total nodes are 1+2=3, agreeing with n−1.
Answer 16 orbitals, 32-electron capacity; 4d has one radial and two angular nodes.
06 / UNDERSTAND & APPLY
6. Why 2s and 2p are equal in one model and split in another
For a nonrelativistic hydrogen-like atom with no external field and ignoring fine-structure effects, orbital energy depends only on n. Thus 2s and 2p are degenerate in that model, as are 3s, 3p and 3d. Their shapes and angular momentum differ even though the energy is the same.
In many-electron atoms, electron–electron repulsion, screening and penetration make subshell energy depend on more than n. At a common n, an s orbital generally penetrates closer to the nucleus than p, then d and f, experiencing a different effective attraction. This helps rationalize the usual s<p<d<f energy trend within a shell in an introductory central-field picture.
The Aufbau or n+l guideline orders subshells approximately by increasing n+l, breaking a tie by lower n. It yields the familiar sequence 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p and onward. It is a useful mnemonic for many ground-state configurations, not an immutable spectrum of fixed orbital energies for every atom and ion.
In particular, 4s and 3d relative energies change as nuclear charge and occupancy change. A neutral-atom filling mnemonic does not imply 4s electrons must be removed after 3d electrons in transition-metal cations. Ground-state configurations and ionization should be determined with the appropriate species and known energetic behavior.
ORIGINAL WORKED EXAMPLE
Use the filling guideline with its scope
Compare the approximate Aufbau order of 4s and 3d, and of 4p and 5s.
For 4s, n+l=4; for 3d, n+l=5. The guideline places 4s first.
For 4p, n+l=5; for 5s, n+l=5. The lower n breaks the tie, placing 4p first.
These comparisons describe the mnemonic order for building many neutral ground-state configurations.
They are not a universal ranking valid for all occupancies, ions or excited states.
Answer Guideline order: 4s before 3d; 4p before 5s.
07 / UNDERSTAND & APPLY
7. Pauli and Hund control how electrons share the available orbitals
The Pauli exclusion principle forbids two electrons in an atom from having the same complete set of four quantum numbers. In an orbital-box representation, this means at most two electrons in one spatial orbital and opposite spin arrows if both occupy it. An orbital with three electrons is not merely an excited state; it violates the occupancy constraint.
Hund’s rule for degenerate orbitals gives singly occupied orbitals with parallel spins before pairing in the usual lowest-energy arrangement. For a p³ subshell, place one electron in each of the three p orbitals with parallel arrows. Pairing two in one orbital while leaving another empty is not the ordinary ground-state arrangement for that isolated degenerate subshell.
Build a configuration by counting the atom’s electrons, selecting a reasonable subshell order, enforcing Pauli and Hund, and verifying the final total. Noble-gas notation replaces a filled inner configuration by the preceding noble-gas symbol. For chlorine, [Ne]3s²3p⁵ contains 10+2+5=17 electrons.
Common neutral ground-state exceptions include chromium [Ar]3d⁵4s¹ and copper [Ar]3d¹⁰4s¹ rather than the simplest uncorrected mnemonic predictions. Near-degenerate subshell energies and electron interactions determine these configurations. The phrase “half-filled or filled stability” is a useful summary, but it is not a universal algorithm that predicts every transition-element exception.
Unpaired electrons can be counted from an appropriate orbital occupancy diagram. They often indicate paramagnetism in an isolated-atom or ion model, while fully paired configurations give diamagnetic behavior in that elementary treatment. For coordination compounds, ligand-field splitting can alter the occupancy, so the free-ion count is not automatically the complex’s count.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Orbital-box model for stated free atoms or free ions. Coordination-complex ligand fields may change d-electron occupancy; arrows denote spin projections, not physical trajectories.
ORIGINAL WORKED EXAMPLE
Write configuration and count unpaired electrons
Write the neutral nitrogen ground-state configuration and count unpaired electrons in its orbital-box model.
Nitrogen has seven electrons. Fill 1s² and 2s², leaving three for 2p.
The configuration is 1s²2s²2p³.
The three degenerate p orbitals receive one electron each with parallel spins under Hund’s rule.
There are three unpaired electrons; the two filled s orbitals contain paired electrons.
Answer 1s²2s²2p³; three unpaired electrons.
08 / UNDERSTAND & APPLY
8. Electron removal is not simply reverse diagonal filling
For a main-group cation, remove electrons from the highest occupied principal shell first in the usual ground-state configuration treatment. For transition-metal cations, remove the ns electrons before the (n−1)d electrons. Thus Fe starts approximately [Ar]3d⁶4s², Fe²⁺ is [Ar]3d⁶ and Fe³⁺ is [Ar]3d⁵.
For anions, add the required electron number and construct the resulting configuration with the appropriate orbital order and occupancy constraints. O²⁻ has ten electrons and is isoelectronic with Ne, but equal electron counts do not mean equal nuclear charge or equal size. The different nucleus changes the attraction and energetics.
An excited configuration can obey Pauli while placing an electron above its ground-state arrangement. For example, a hydrogen electron in 2p is excited but allowed. Conversely, a proposed 1s³ configuration is forbidden by orbital capacity, not a legitimate excited state. Distinguish allowed state labels from whether the state is lowest in energy.
The written order of subshells in a configuration is partly convention: [Ar]3d⁶4s² and [Ar]4s²3d⁶ describe the same occupation numbers. Do not infer an ionization sequence solely from the left-to-right order of symbols. Count occupancy and identify the species before discussing removal.
ORIGINAL WORKED EXAMPLE
Form iron ions
Starting from Fe=[Ar]3d⁶4s², give Fe²⁺ and Fe³⁺ configurations and elementary free-ion unpaired counts.
Remove the two 4s electrons first to form Fe²⁺=[Ar]3d⁶.
Remove one 3d electron next for Fe³⁺=[Ar]3d⁵.
In the free-ion degenerate-d Hund picture, d⁶ has one paired orbital and four singly occupied orbitals.
d⁵ has five singly occupied orbitals. Ligand fields can change these counts in complexes.
Answer Fe²⁺: [Ar]3d⁶, four unpaired in the stated model. Fe³⁺: [Ar]3d⁵, five unpaired.
09 / UNDERSTAND & APPLY
9. Use each model for the question it can answer
Matter-wave wavelength is a momentum relation; an orbital node count is a stationary-state spatial property. They are related through quantum mechanics but are not interchangeable formulas. In particular, one cannot assign a single classical electron speed to every point in an orbital picture and treat it as a circular wave wrapped around a track.
An orbital’s sign matters for interference, while its probability density controls position statistics. A nodal surface has zero density in the ideal wavefunction, but a colored contour boundary merely marks a chosen value or enclosed probability. Be explicit about which type of surface a drawing represents.
For counting questions, distinguish shell, subshell, spatial orbital and spin-orbital. A p subshell has three spatial orbitals, six spin-orbitals and maximum six electrons. Saying simply “six orbitals” without identifying spin-orbitals confuses the conventional chemistry use of orbital.
For configuration questions, keep electron count, allowed quantum labels and energetic preference as separate checks. A configuration can have the correct total electron number but violate Pauli, or obey Pauli but describe an excited state. The Aufbau guideline alone does not establish which of these cases applies.
Do not use grams with SI electron mass calculations.
Do not interpret uncertainty as only poor equipment.
Do not call orbital phase positive or negative charge.
Do not equate radial probability with local probability density.
Do not apply the hydrogenic n-only energy rule to all many-electron atoms.
Do not remove transition-metal 3d electrons before occupied 4s electrons by blindly reversing a mnemonic.
10 / UNDERSTAND & APPLY
10. Eight solved practice checks
Check model assumptions before calculating. Where a problem asks about orbital energies or occupancy, state whether the hydrogenic, neutral-atom guideline or free-ion model is intended.
ORIGINAL PRACTICE · 1
At equal nonrelativistic kinetic energy, particle B has four times the mass of particle A. Compare wavelengths.
Reveal answer & reasoning
λ_B=λ_A/2.
λ=h/√(2mK), so at equal K wavelength scales as m⁻¹/².
ORIGINAL PRACTICE · 2
An electron accelerating voltage rises from 25 V to 100 V within the nonrelativistic regime. How does wavelength change?
Reveal answer & reasoning
It halves.
λ is proportional to V⁻¹/² and the voltage increases by a factor of four.
ORIGINAL PRACTICE · 3
Can n=3,l=3 identify an atomic orbital?
Reveal answer & reasoning
No.
For n=3, allowed l values are 0,1,2. The proposed f label would require at least n=4.
ORIGINAL PRACTICE · 4
How many spatial orbitals and maximum electrons occur in an f subshell?
Reveal answer & reasoning
Seven orbitals and fourteen electrons.
l=3 gives 2l+1=7 orbitals; each holds at most two opposite-spin electrons.
ORIGINAL PRACTICE · 5
How many radial and angular nodes does a hydrogenic 3p orbital have?
Reveal answer & reasoning
One radial and one angular node.
l=1 gives one angular node; n−l−1=3−1−1=1 radial node.
ORIGINAL PRACTICE · 6
Are 3s and 3d degenerate in the nonrelativistic hydrogenic model without external fields or fine structure?
Reveal answer & reasoning
Yes.
Energy depends only on n in that stated model. The two orbitals differ in shape and angular momentum, not the model energy.
ORIGINAL PRACTICE · 7
Write the ground-state configuration of Cu⁺ given neutral Cu=[Ar]3d¹⁰4s¹.
Reveal answer & reasoning
[Ar]3d¹⁰.
Remove the occupied 4s electron first; the ion has 28 electrons.
ORIGINAL PRACTICE · 8
Is a drawing with two same-spin electrons in one spatial orbital an allowed excited state?
Reveal answer & reasoning
No.
Both electrons would share the same spatial quantum numbers and spin projection, violating Pauli. Excitation does not remove that restriction.
KEEP THE BIG IDEAS
Quick revision
Matter wavelength is set by momentum, with nonrelativistic shortcuts used only in their regime.
An orbital is a wavefunction; |ψ|² is probability density and phase is not charge.
Quantum-number ranges depend on one another.
A shell has n² spatial orbitals and capacity 2n² electrons.
Hydrogen-like energy degeneracy and many-electron filling order are different models.
Pauli, Hund, electron count and species-specific energies all matter for configurations.
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.