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AP EAPCET (Agriculture & Pharmacy) · Chemistry (JEE & NEET)

Atomic Structure

Bohr model, dual nature of matter, quantum numbers, orbitals, electronic configuration and Heisenberg uncertainty principle.

Eight concepts from the Bohr atom to the orbital labels the rest of chemistry uses. Every spectral-line wavelength, every quantum-number set and every Cr/Cu exception in this chapter is a consequence of one of these eight ideas — not a separate fact to memorise.

  • AP EAPCET (Agriculture & Pharmacy)
  • Medium level
  • 8 concepts
  • 5 practice questions

1Bohr radius and energy

For a hydrogen-like ion of nuclear charge Z, Bohr's postulate mvr = nh/2\pi fixes a discrete set of circular orbits. The radius and energy that follow are r_n = 0.529\,n^2/Z Å and E_n = -13.6\,Z^2/n^2 eV — both measured from the nucleus and from the ionisation continuum respectively.

Energy becomes less negative as n rises, so the levels pack toward E = 0. The ionisation energy from the ground state is exactly the magnitude of E_1: +13.6\,Z^2 eV. Velocity falls as v_n = 2.18\times10^6\,Z/n m/s.

Figure. Levels drawn for hydrogen (Z=1). Energy is measured down from the continuum at E=0; the n=3\to2 arrow is the first Balmer photon at 13.6(1/4-1/9)=1.89 eV. The levels are to scale in energy — n=1 sits much deeper than the eye expects from a textbook sketch that spreads them evenly.

How it works

  1. Quantise LSet mvr = nh/2\pi. Together with the centripetal Coulomb force this kills every radius except one per n.
  2. Read r_n and E_nSubstitute. Radius grows as n^2/Z; energy deepens as Z^2/n^2. Both formulas share the same n.
  3. Ionise from n=1The energy to reach E=0 from the ground state is +13.6\,Z^2 eV — the same number with the sign flipped.

He⁺ ground-state radius and IE

For He⁺ (Z=2), find the ground-state radius and the ionisation energy from n=1.

  • r₁ = 0.529 × 1² / 20.2645 Å
  • E₁ = −13.6 × 2² / 1²−54.4 eV
  • IE = −E₁54.4 eV
  • Check: IE = 13.6 Z² = 13.6 × 454.4 eV ✓

Pro tip. Every H-like ionisation energy from the ground state is 13.6\,Z^2 eV. He⁺ is four times H; Li²⁺ is nine times H. The radius shrinks by the same Z.

The energy of the n=2 level of Li²⁺ (Z=3) is
  1. −13.6 eV
  2. −30.6 eV
  3. −122.4 eV

E_n = -13.6 Z^2/n^2 = -13.6 × 9 / 4 = -30.6 eV. Option 1 is hydrogen's n=1; option 3 is Li²⁺'s ground state.

2Rydberg formula and spectral series

A photon is emitted when an electron falls from n_2 to n_1. Its wavenumber is fixed by the Rydberg formula \frac{1}{\lambda}=R_H Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) with R_H = 1.097\times10^7 m^{-1} (or 109677 cm^{-1}).

The series is named by the lower level alone: Lyman (n_1=1, UV), Balmer (n_1=2, visible), Paschen (n_1=3, IR), Brackett (n_1=4), Pfund (n_1=5). The same \frac{Z^2}{n^2} scaling makes the He⁺ 4\to2 line coincide with hydrogen's Lyman \alpha.

Figure. Balmer series of hydrogen: wavenumber against \frac{1}{n_1^2}-\frac{1}{n_2^2} with n_1=2. The line is forced through the origin because \frac{1}{\lambda}=R_H\times(\ldots) — every series lies on that same ray (slope R_H); only the span of x changes with n_1. Points are to scale in the Rydberg term; the Balmer series limit is at 1/n_1^2=0.25.

How it works

  1. Name n_1The lower level decides the series and the spectral region.
  2. Insert n_2 and ZApply the Rydberg formula. Larger Z compresses every wavelength by Z^2.
  3. Invert for \lambdaTake the reciprocal of the wavenumber. Check units: R_H in m^{-1} gives \lambda in metres.

Wavelength of He⁺ (4 → 2)

Find the wavelength of the photon emitted when an electron in He⁺ (Z=2) drops from n_2=4 to n_1=2. Take R_H=1.097\times10^7 m^{-1}.

  • 1/λ = R_H Z² (1/n₁² − 1/n₂²)R_H × 4 × (1/4 − 1/16)
  • 1/4 − 1/16 = 3/160.1875
  • 1/λ = 1.097×10⁷ × 4 × 0.18758.2275×10⁶ m⁻¹
  • λ = 1 / (8.2275×10⁶)1.2154×10⁻⁷ m = 1215.4 Å

Pro tip. He⁺ (4\to2) matches H Lyman \alpha (2\to1) because both evaluate Z^2(1/n_1^2-1/n_2^2) to 0.75. Same wavenumber, same wavelength — different atoms.

The Balmer series of hydrogen has lower level
  1. n₁ = 1 (UV)
  2. n₁ = 2 (visible)
  3. n₁ = 3 (IR)

Balmer is n_1=2 and lies in the visible. Lyman is n_1=1 (UV); Paschen is n_1=3 (IR).

3Counting spectral lines

When an electron falls from n_2 to n_1, every intermediate level is available, so every pair (n_i, n_j) with n_2 \ge n_i > n_j \ge n_1 produces a line. The count is \frac{(n_2-n_1)(n_2-n_1+1)}{2}.

The special case of a fall all the way to the ground state is \frac{n(n-1)}{2} lines from level n. The formula counts distinct transitions, not photons from one atom — one atom emits one photon per jump.

Figure. All six transitions from n=4 down to n=1. Three leave n=4, two leave n=3, one leaves n=2 — that is the triangular number 6, not a cascade of three steps.

How it works

  1. Fix the windowRead n_2 (start) and n_1 (floor). Every level between them is fair game.
  2. Count the pairsThere are \Delta n = n_2-n_1 steps down; the triangular number \Delta n(\Delta n+1)/2 is the line count.
  3. Name the seriesLines that share the same n_1 belong to one series; the total mixes every series the window touches.

Lines from n = 4 to ground

An electron in hydrogen falls from n=4 to n=1. How many spectral lines are possible, and which transitions are they?

  • Δn = n₂ − n₁ = 4 − 13
  • lines = Δn(Δn+1)/2 = 3×4/26
  • Check: n(n−1)/2 from n=4 to ground4×3/2 = 6 ✓
  • Pairs4→3, 4→2, 4→1, 3→2, 3→1, 2→1

Pro tip. From n=5 to ground the count is 10, not 5. Students who report n-1 are counting only the direct cascade 5\to4\to3\to2\to1 and missing every skip.

Number of spectral lines when an electron falls from n=5 to n=2 is
  1. 3
  2. 6
  3. 10

\Delta n=3, so 3\times4/2=6. Option 1 is \Delta n itself; option 3 is the count from n=5 all the way to ground.

4de Broglie wavelength

Matter carries a wavelength \lambda = h/p = h/(mv). For the same speed a proton's wavelength is about 1836 times shorter than an electron's — which is why electron diffraction is a laboratory tool and proton diffraction is not a classroom demonstration.

For an electron accelerated through V volts the kinetic energy is eV, so \lambda = h/\sqrt{2m_e eV}. In ångströms the working form is \lambda\text{(Å)} = 12.27/\sqrt{V}.

Figure. de Broglie wavelength of an electron against speed. The curve is \lambda\propto 1/v projected into the plot frame — not a sketch. A proton at the same speed would sit 1836\times lower, below the resolution of this axis; that comparison is left to the prose.

How it works

  1. Find the momentumFrom speed (p=mv) or from kinetic energy (p=\sqrt{2mE}).
  2. Divide h by pThat is the whole de Broglie relation. Units: joule-seconds over kg·m/s gives metres.
  3. Use the volt shortcutFor electrons alone, \lambda\text{(Å)} = 12.27/\sqrt{V} skips the constants.

Electron accelerated through 150 V

Find the de Broglie wavelength of an electron accelerated from rest through 150 V.

  • λ(Å) = 12.27 / √V12.27 / √150
  • √15012.247
  • λ = 12.27 / 12.2471.002 Å ≈ 1.00 Å
  • Check: h/√(2m_e eV) with eV = 2.403×10⁻¹⁷ J1.00 Å ✓

Pro tip. 150 V is the classic drill because the answer is almost exactly 1 Å — a crystal-lattice scale, which is why electron diffraction works on solids.

If an electron's speed doubles, its de Broglie wavelength
  1. doubles
  2. halves
  3. falls by a factor of four

\lambda = h/(mv), so doubling v halves \lambda. Energy would bring a \sqrt{} if the question had doubled KE instead.

5Heisenberg uncertainty

Position and momentum of the same particle cannot both be sharp: \Delta x\cdot\Delta p \ge h/(4\pi). With \Delta p = m\Delta v the bound on position is \Delta x \ge h/(4\pi m \Delta v).

For a macroscopic mass the right-hand side is absurdly small and classical trajectories survive. For an electron the same \Delta v leaves \Delta x on the ångström scale — which is why Bohr's sharp orbits are an approximation the uncertainty principle forbids as exact.

Figure. Δx·Δp ≥ h/(4π): sharpening position inflates the allowed momentum spread, and sharpening momentum inflates position. That product floor is why a Bohr orbit cannot be an exact classical trajectory.

How it works

  1. Read \Delta vThe question usually gives a velocity uncertainty, not a momentum one.
  2. Form \Delta p = m\Delta vUse the electron mass 9.1\times10^{-31} kg unless a different particle is named.
  3. Apply the bound\Delta x \ge h/(4\pi\Delta p). The answer is a minimum; the actual spread can only be larger.

Minimum Δx for an electron

An electron has speed uncertainty \Delta v = 5.0\times10^5 m/s. Find the minimum uncertainty in its position. Take h=6.626\times10^{-34} J s and m_e=9.1\times10^{-31} kg.

  • Δp = m_e Δv = 9.1×10⁻³¹ × 5.0×10⁵4.55×10⁻²⁵ kg m/s
  • 4π Δp = 4 × 3.1416 × 4.55×10⁻²⁵5.717×10⁻²⁴
  • Δx ≥ h / (4π Δp) = 6.626×10⁻³⁴ / 5.717×10⁻²⁴1.16×10⁻¹⁰ m
  • In ångströms1.16 Å

Pro tip. 1 Å is a bond-length scale. An electron whose speed is uncertain by 5\times10^5 m/s cannot be localised inside one atomic orbital — that is the quantitative content of "orbits are not trajectories".

If Δv of an electron is halved, the minimum Δx
  1. halves
  2. doubles
  3. stays the same

\Delta x \ge h/(4\pi m\Delta v), so halving \Delta v doubles the floor on \Delta x.

6Quantum numbers and Pauli exclusion

Four quantum numbers label every electron in an atom: the principal number n (shell), the azimuthal number l=0,1,\ldots,n-1 (subshell shape: s,p,d,f), the magnetic number m_l=-l,\ldots,+l (orbital orientation), and the spin m_s=\pm\frac{1}{2}.

Pauli exclusion says no two electrons in an atom share the same set of four. One orbital — fixed (n,l,m_l) — therefore holds at most two electrons, of opposite spin. A shell of principal number n holds n^2 orbitals and at most 2n^2 electrons.

Figure. Each subshell holds 2(2l+1) electrons: one orbital for s, three for p, five for d, seven for f, two spins each. Pauli forbids a second electron sharing all four quantum numbers.

How it works

  1. Pick nShell. n=1,2,3,\ldots with no upper bound in the isolated atom.
  2. Restrict l and m_ll runs from 0 to n-1; for each l, m_l runs through 2l+1 values.
  3. Apply PauliEach (n,l,m_l) takes two spins. Count: n^2 orbitals, 2n^2 electrons in the shell.

Orbitals and electrons in n = 3

For the shell n=3, how many orbitals are there and what is the maximum number of electrons?

  • l can be 0, 1, 2s, p, d subshells
  • orbitals: 1 + 3 + 59
  • Check: orbitals = n² = 3²9 ✓
  • max electrons = 2n² = 2×918

Pro tip. The n^2 rule is faster than listing subshells, and it is what the question usually wants. Listing is the check, not the method.

Which set of quantum numbers is not allowed?
  1. n=3, l=2, m_l=−2, m_s=+1/2
  2. n=2, l=2, m_l=0, m_s=−1/2
  3. n=4, l=0, m_l=0, m_s=+1/2

For n=2, l may be only 0 or 1. Option 2 has l=2, which needs n\ge3. The other two sets sit inside their ranges.

7Radial and angular nodes

A node is a surface where the wave function is identically zero. An orbital with quantum numbers n and l has n-l-1 radial nodes (spherical surfaces of zero amplitude) and l angular nodes (planes or cones through the nucleus). The total is n-1 — independent of which subshell you are in.

Any orbital with n-l-1=0 — 1s, 2p, 3d, 4f — has no radial node. That is why the 3d orbital, despite sitting in the third shell, has a nodeless radial factor.

Figure. Radial nodes n-l-1 for the first six orbitals. Length follows the count by construction: 2s and 3p share height 1; 3s alone reaches 2; 1s, 2p and 3d sit on the baseline at zero. Angular nodes are l and are not on this chart.

How it works

  1. Read n and lFrom the orbital label: 3d means n=3, l=2; 2p means n=2, l=1.
  2. Radial nodesn-l-1. Zero is allowed and common for the first orbital of each type.
  3. Angular and totalAngular nodes =l; total nodes =n-1. The two pieces always add to n-1.

Nodes of a 3d orbital

How many radial and angular nodes does a 3d orbital have?

  • 3d means n=3, l=2d ↔ l=2
  • radial nodes = n − l − 1 = 3 − 2 − 10
  • angular nodes = l2
  • total nodes = n − 1 = 3 − 12 (= 0 + 2) ✓

Pro tip. Any nl orbital with n-l-1=0 (1s, 2p, 3d, 4f) has zero radial nodes. Counting total nodes as n-1 is the fastest check that your split into radial and angular adds up.

A 4p orbital has how many radial nodes?
  1. 1
  2. 2
  3. 3

n=4, l=1, so n-l-1=2. Option 1 is the angular-node count; option 3 is n-1 (total nodes).

8Aufbau, Hund and the Cr/Cu exceptions

Electrons fill subshells in order of increasing n+l (Aufbau), and within a subshell they spread across orbitals with parallel spins before pairing (Hund). Both rules are energetic: unpaired parallel spins lower repulsion, and a lower n+l subshell sits lower in energy.

Half-filled and fully filled d subshells are extra-stable, so chromium and copper break the naive Aufbau order: \mathrm{Cr}=[\mathrm{Ar}]\,3d^5\,4s^1 rather than 3d^4\,4s^2, and \mathrm{Cu}=[\mathrm{Ar}]\,3d^{10}\,4s^1 rather than 3d^9\,4s^2. The exception is the configuration that wins, not a third rule to memorise in isolation.

Figure. Aufbau order is increasing n+l (bar length), with lower n first on a tie — so 4s (n+l=4) fills before 3d (n+l=5), and 3p before 4s even though both score 4. Cr and Cu then reshuffle inside the 3d/4s pair; the bar chart is the rule they break.

How it works

  1. Order by n+lFill 1s, 2s, 2p, 3s, 3p, 4s, then 3d. Equal n+l prefers the lower n.
  2. Spread spinsHund: put one electron in each orbital of a subshell, same spin, before any pairing.
  3. Watch Cr and CuAt d^4 s^2 and d^9 s^2 the atom borrows one s electron to make d^5 s^1 or d^{10} s^1.

Ground-state configuration of Cr

Write the ground-state electron configuration of Cr (Z=24).

  • Z = 24 → fill past [Ar] (18 e⁻)6 electrons left
  • Naive Aufbau: 4s² 3d⁴[Ar] 3d⁴ 4s²
  • Half-filled 3d⁵ is lower; move one 4s electron3d⁵ 4s¹
  • Ground state[Ar] 3d⁵ 4s¹

Pro tip. Cu (Z=29) makes the same move to finish 3d^{10}. Mo and Ag are the 4d analogues. Nothing else in the 3d row does this in the ground state.

The ground-state configuration of Cu is
  1. [Ar] 3d⁹ 4s²
  2. [Ar] 3d¹⁰ 4s¹
  3. [Ar] 3d¹⁰ 4s²

Copper takes the fully filled 3d^{10} and leaves 4s^1. Option 1 is the naive Aufbau order; option 3 has 30 electrons.

Notes

  • Bohr model (H-like species, charge Z): angular momentum is quantized mvr=\frac{nh}{2\pi}, giving r_n=0.529\frac{n^2}{Z} Å and E_n=-13.6\frac{Z^2}{n^2} eV.
  • Dual nature: de Broglie wavelength \lambda=\frac{h}{mv}; Heisenberg's uncertainty principle \Delta x\cdot\Delta p\geq\frac{h}{4\pi} forbids exact simultaneous position and momentum.
  • Quantum numbers: n (shell), l=0 to n-1 (subshell shape), m_l=-l to +l (orientation), and m_s=\pm\frac{1}{2} (spin); by Pauli exclusion each orbital holds at most 2 electrons.
  • Filling rules: Aufbau (increasing n+l), Hund's rule (maximum unpaired spins), and extra stability of half/fully filled subshells, e.g. Cr=[Ar]3d^5 4s^1, Cu=[Ar]3d^{10}4s^1.
  • Nodes: radial nodes =n-l-1, angular nodes =l, total nodes =n-1.

Formulas

  • r_n=0.529\frac{n^2}{Z} Å,\quad E_n=-13.6\frac{Z^2}{n^2} eV,\quad v_n=2.18\times10^6\frac{Z}{n} m/s
  • \frac{1}{\lambda}=R_H Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),\quad R_H=109677\,\text{cm}^{-1}
  • \lambda=\frac{h}{mv},\quad \Delta x\,\Delta p\geq\frac{h}{4\pi}
  • Spectral lines from level n to ground state =\frac{n(n-1)}{2}
  • Orbitals in shell =n^2; maximum electrons =2n^2

Exam traps & shortcuts

  • Spectral series by lower level n_1: Lyman (1, UV), Balmer (2, visible), Paschen (3, IR), Brackett (4), Pfund (5).
  • Number of spectral lines when an electron falls from n_2 to n_1 is \frac{(n_2-n_1)(n_2-n_1+1)}{2}.
  • Ionisation energy from ground state of an H-like ion =+13.6\,Z^2 eV.

Reference tables

Named by the lower level n_1. The region is for hydrogen; every H-like ion of charge Z compresses the same series by Z^2.

Spectral series
Seriesn₁Region (H)
Lyman1Ultraviolet
Balmer2Visible
Paschen3Infrared
Brackett4Infrared
Pfund5Infrared

The four numbers that label one electron. Pauli exclusion is the statement that the four-tuple is unique inside an atom.

Quantum number ranges
SymbolNameAllowed valuesWhat it fixes
nPrincipal1, 2, 3, …Shell; energy in H-like ions
lAzimuthal0 … n−1Subshell shape (s,p,d,f)
m_lMagnetic−l … +lOrbital orientation
m_sSpin+1/2 or −1/2Electron spin

The relations this topic actually computes with. R_H=1.097\times10^7 m^{-1} =109677 cm^{-1}.

Formula sheet
QuantityFormula
Bohr radiusr_n=0.529\,n^2/Z Å
Bohr energyE_n=-13.6\,Z^2/n^2 eV
Bohr speedv_n=2.18\times10^6\,Z/n m/s
Rydberg1/\lambda=R_H Z^2(1/n_1^2-1/n_2^2)
de Broglie\lambda=h/mv; electrons: \lambda(\mathrm{Å})=12.27/\sqrt{V}
Heisenberg\Delta x\,\Delta p\ge h/(4\pi)
Line count(n_2-n_1)(n_2-n_1+1)/2
Shell capacityorbitals n^2; electrons 2n^2
Nodesradial n-l-1; angular l; total n-1

Recap

Read only this the night before.

Bohr pair
r_n=0.529\,n^2/Z Å and E_n=-13.6\,Z^2/n^2 eV. Ground-state IE is +13.6\,Z^2 eV.
Rydberg
1/\lambda=R_H Z^2(1/n_1^2-1/n_2^2). Series named by n_1: Lyman 1, Balmer 2, Paschen 3. He⁺ 4\to2 matches H Lyman α.
Line count
From n_2 to n_1: \Delta n(\Delta n+1)/2 lines. From n to ground: n(n-1)/2.
Wave and bound
\lambda=h/mv; electrons \lambda(\mathrm{Å})=12.27/\sqrt{V}. \Delta x\,\Delta p\ge h/4\pi — Bohr orbits are not exact trajectories.
Labels and fill
Pauli: one orbital, two electrons. Nodes: radial n-l-1, angular l. Cr and Cu steal an s electron to make d^5 or d^{10}.

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