Quick Recap — Atomic Models & Quantum Numbers

  • Bohr model: electrons occupy fixed energy levels; En=−13.6 Z2n2E_n=-\dfrac{13.6\,Z^2}{n^2} eV; rn=0.529n2Zr_n=0.529\dfrac{n^2}{Z} Å.
  • Quantum numbers: nn (shell/size), ll (subshell/shape, 0..n−10..n-1), mm (orientation, −l..+l-l..+l), ss (spin ±12\pm\tfrac12).
  • Capacities: subshell s/p/d/f=2/6/10/14s/p/d/f=2/6/10/14; shell holds 2n22n^2 electrons and n2n^2 orbitals.
  • Filling rules: Aufbau (lowest energy first), Pauli (no two electrons with all four quantum numbers equal), Hund (singly fill degenerate orbitals first).

Beyond-NCERT JEE Formulae

This is the working sheet for the fast one-electron (hydrogen-like) numericals in Section B and the tricky single-line MCQs. Unless stated otherwise every relation below assumes a single electron and nuclear charge Z (H: Z=1, He+: Z=2, Li2+: Z=3).

1. Bohr Model (hydrogen-like)

  • Radius: rn=0.529n2Zr_n=0.529\dfrac{n^2}{Z} angstrom, i.e. 0.529×10−10n2Z0.529\times10^{-10}\dfrac{n^2}{Z} m.
  • Energy: En=−13.6Z2n2E_n=-13.6\dfrac{Z^2}{n^2} eV.
  • Velocity: vn=2.18×106Znv_n=2.18\times10^6\dfrac{Z}{n} m/s.
  • Energy split: KE=−En=+13.6Z2n2KE=-E_n=+13.6\dfrac{Z^2}{n^2} eV, while PE=2EnPE=2E_n, so the total energy En=−KEE_n=-KE.
  • Quantised angular momentum mvr=nh2πmvr=\dfrac{nh}{2\pi}; orbital time period Tn∝n3Z2T_n\propto\dfrac{n^3}{Z^2}.

When to use: any "radius / energy / speed of the nth orbit" or "ionize from this orbit" question on a one-electron species (H, He+, Li2+, Be3+).

[JEE Tip] Scale, do not re-plug: r∝n2Zr\propto\dfrac{n^2}{Z}, E∝Z2n2E\propto\dfrac{Z^2}{n^2}, v∝Znv\propto\dfrac{Z}{n} and T∝n3Z2T\propto\dfrac{n^3}{Z^2}. Ground-state energies are simply −13.6Z2-13.6Z^2 eV, so for H that is −13.6-13.6, for He+ −54.4-54.4 and for Li2+ −122.4-122.4 eV.

2. Spectra — Rydberg Equation

  • 1λ=RHZ2(1n12−1n22)\dfrac{1}{\lambda}=R_H Z^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right) with RH=1.097×107R_H=1.097\times10^7 per m and n2>n1n_2>n_1.
  • Series by lower level: Lyman n1=1n_1=1 (UV), Balmer n1=2n_1=2 (visible), Paschen n1=3n_1=3 (IR).
  • Spectral lines emitted when an electron falls from level nn to the ground state =n(n−1)2=\dfrac{n(n-1)}{2}.
  • Between any two levels: number of lines =(n2−n1)(n2−n1+1)2=\dfrac{(n_2-n_1)(n_2-n_1+1)}{2}.

When to use: wavelength or frequency of an emission/absorption line, a series limit, or "how many lines appear".

[JEE Tip] Always compute 1λ\dfrac{1}{\lambda} fully first and invert only at the very end. The longest wavelength of a series is the smallest jump (n1+1→n1n_1+1\to n_1); the series limit is n2→∞n_2\to\infty, which just deletes the second bracket term.

3. Dual Nature and Uncertainty

  • de Broglie: λ=hmv=h2mKE=h2mqV\lambda=\dfrac{h}{mv}=\dfrac{h}{\sqrt{2mKE}}=\dfrac{h}{\sqrt{2mqV}}, with h=6.626×10−34h=6.626\times10^{-34} J s.
  • Electron accelerated through V volts: λ=12.27V\lambda=\dfrac{12.27}{\sqrt{V}} angstrom.
  • Heisenberg: Δx⋅Δp≥h4π\Delta x\cdot\Delta p\ge\dfrac{h}{4\pi}, equivalently Δx⋅Δv≥h4πm\Delta x\cdot\Delta v\ge\dfrac{h}{4\pi m}.

When to use: wavelength of any moving particle (from speed, kinetic energy or accelerating voltage) and any "minimum uncertainty" estimate.

[JEE Tip] Pick the form that matches the data: if the speed is given use hmv\dfrac{h}{mv}; if the kinetic energy is given use h2mKE\dfrac{h}{\sqrt{2mKE}}; if an accelerating voltage is given use h2mqV\dfrac{h}{\sqrt{2mqV}} or, for an electron, the 12.27V\dfrac{12.27}{\sqrt{V}} angstrom shortcut. Heavier or faster always means a shorter λ\lambda.

4. Quantum Numbers, Orbitals and Nodes

  • Allowed values: l=0l=0 to n−1n-1; m=−lm=-l to +l+l giving 2l+12l+1 orbitals; spin =±12=\pm\dfrac{1}{2}.
  • Counts: per shell, orbitals =n2=n^2 and electrons =2n2=2n^2; per subshell, orbitals =2l+1=2l+1 and electrons =2(2l+1)=2(2l+1).
  • Nodes: radial (spherical) =n−l−1=n-l-1; angular (planar) =l=l; total =n−1=n-1.

When to use: "how many orbitals / electrons / nodes", checking whether a set of quantum numbers is legal, or comparing two orbitals.

[JEE Tip] Total nodes depend only on nn and equal n−1n-1, never on ll: a 3s, 3p and 3d orbital each have 2 total nodes, only split differently between radial and angular. A quantum-number set is valid only if l<nl<n and ∣m∣≤l|m|\le l.

5. Photoelectric Effect

  • Einstein equation hν=W+KEmaxh\nu=W+KE_{max}, where the work function W=hν0W=h\nu_0 and ν0\nu_0 is the threshold frequency.
  • Handy constant hc=1240hc=1240 eV nm, so photon energy E(eV)=1240λ(nm)E(\text{eV})=\dfrac{1240}{\lambda(\text{nm})}.
  • Stopping potential from eV0=KEmaxeV_0=KE_{max}: numerically KEmaxKE_{max} in eV equals V0V_0 in volts.

When to use: any question giving a light wavelength or frequency together with a metal's work function or threshold.

[JEE Tip] Turn the wavelength straight into eV with 1240λ(nm)\dfrac{1240}{\lambda(\text{nm})}, then subtract WW. If Ephoton<WE_{photon}<W no electron escapes however intense the beam, and any surplus photon energy raises KEmaxKE_{max}, never the number of electrons.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Orbit of a one-electron ion. For the He+ ion (Z=2), find the radius, energy and electron speed in its ground state (n=1).

  • Radius: r1=0.529n2Z=0.529×12=0.2645r_1=0.529\dfrac{n^2}{Z}=0.529\times\dfrac{1}{2}=0.2645 angstrom.
  • Energy: E1=−13.6Z2n2=−13.6×41=−54.4E_1=-13.6\dfrac{Z^2}{n^2}=-13.6\times\dfrac{4}{1}=-54.4 eV.
  • Speed: v1=2.18×106×Zn=2.18×106×2=4.36×106v_1=2.18\times10^6\times\dfrac{Z}{n}=2.18\times10^6\times2=4.36\times10^6 m/s.

He+ is half the size of the H atom yet four times more tightly bound. Answer: r1=0.2645r_1=0.2645 angstrom, E1=−54.4E_1=-54.4 eV, v1=4.36×106v_1=4.36\times10^6 m/s.

Example 2 — Wavelength of a spectral line. Find the wavelength emitted when the electron in a hydrogen atom falls from n=3 to n=2, with RH=1.097×107R_H=1.097\times10^7 per m.

1λ=RHZ2(1n12−1n22)=1.097×107×1×(14−19)\dfrac{1}{\lambda}=R_H Z^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)=1.097\times10^7\times1\times\left(\dfrac{1}{4}-\dfrac{1}{9}\right)

The bracket is 9−436=536=0.1389\dfrac{9-4}{36}=\dfrac{5}{36}=0.1389, so 1λ=1.524×106\dfrac{1}{\lambda}=1.524\times10^6 per m and hence λ=6.56×10−7\lambda=6.56\times10^{-7} m, i.e. about 656 nm. Answer: the red H-alpha (Balmer) line at λ≈656\lambda\approx656 nm.

Example 3 — Counting spectral lines. A sample of hydrogen atoms is excited to the n=6 level. How many distinct emission lines can appear as the atoms return to the ground state, and how many of them belong to the Balmer series?

  • Total lines from level n to the ground state =n(n−1)2=6×52=15=\dfrac{n(n-1)}{2}=\dfrac{6\times5}{2}=15.
  • Balmer lines end at n=2, so they are the drops from 6, 5, 4 and 3 down to 2 — exactly 4 lines.

Answer: 15 lines in all, of which 4 are Balmer lines.

Example 4 — de Broglie wavelength of an electron. An electron is accelerated from rest through 100 V. Find its wavelength, taking h=6.626×10−34h=6.626\times10^{-34} J s, m=9.11×10−31m=9.11\times10^{-31} kg and q=1.6×10−19q=1.6\times10^{-19} C.

The kinetic energy gained is KE=qVKE=qV, so

λ=h2mqV=6.626×10−342×9.11×10−31×1.6×10−19×100\lambda=\dfrac{h}{\sqrt{2mqV}}=\dfrac{6.626\times10^{-34}}{\sqrt{2\times9.11\times10^{-31}\times1.6\times10^{-19}\times100}}

The denominator is 2.915×10−47=5.40×10−24\sqrt{2.915\times10^{-47}}=5.40\times10^{-24}, giving λ=1.23×10−10\lambda=1.23\times10^{-10} m, i.e. 1.23 angstrom. The electron shortcut λ=12.27V=12.2710=1.23\lambda=\dfrac{12.27}{\sqrt{V}}=\dfrac{12.27}{10}=1.23 angstrom reaches the same result in one line. Answer: λ≈1.23\lambda\approx1.23 angstrom (0.123 nm).

Example 5 — A matter wave that fits the orbit. The electron in the first Bohr orbit of hydrogen moves at v=2.18×106v=2.18\times10^6 m/s. Find its de Broglie wavelength and compare it with the orbit's circumference.

λ=hmv=6.626×10−349.11×10−31×2.18×106\lambda=\dfrac{h}{mv}=\dfrac{6.626\times10^{-34}}{9.11\times10^{-31}\times2.18\times10^6}

Evaluating gives λ=3.34×10−10\lambda=3.34\times10^{-10} m, i.e. 3.34 angstrom. The first orbit has r1=0.529r_1=0.529 angstrom, so its circumference is 2πr1=2π×0.529=3.322\pi r_1=2\pi\times0.529=3.32 angstrom. The two agree: the orbit holds exactly one de Broglie wavelength (2πr=nλ2\pi r=n\lambda with n=1), the wave picture behind Bohr's quantisation. Answer: λ≈3.34\lambda\approx3.34 angstrom, equal to 2πr12\pi r_1.

Example 6 — Nodes in an orbital. State the radial, angular and total nodes for the 3d, 4p and 5f orbitals, using radial =n−l−1=n-l-1, angular =l=l and total =n−1=n-1.

  • 3d, with n=3n=3 and l=2l=2: radial =0=0, angular =2=2, total =2=2.
  • 4p, with n=4n=4 and l=1l=1: radial =2=2, angular =1=1, total =3=3.
  • 5f, with n=5n=5 and l=3l=3: radial =1=1, angular =3=3, total =4=4.

The total is n−1n-1 in every case, and radial plus angular always rebuilds it. Answer: (3d) 0, 2, 2; (4p) 2, 1, 3; (5f) 1, 3, 4.

Example 7 — Ionization energy of a hydrogen-like ion. Find the energy needed to (a) fully ionize Li2+ (Z=3) from its ground state and (b) remove the electron from its first excited state (n=2).

The ionization energy is 0−En=13.6Z2n20-E_n=13.6\dfrac{Z^2}{n^2} eV.

  • (a) Ground state, n=1: IE=13.6×91=122.4IE=13.6\times\dfrac{9}{1}=122.4 eV.
  • (b) First excited state, n=2: IE=13.6×94=30.6IE=13.6\times\dfrac{9}{4}=30.6 eV.

The ground-state value is just 13.6Z213.6Z^2, and every hydrogen-like ionization energy scales as Z2Z^2. Answer: 122.4 eV from the ground state and 30.6 eV from n=2.

Example 8 — Photoelectric effect. Ultraviolet light of wavelength 248 nm strikes a metal of work function 3.0 eV. Find the maximum kinetic energy of the photoelectrons, the stopping potential and the threshold wavelength.

The photon energy is 1240λ(nm)=1240248=5.0\dfrac{1240}{\lambda(\text{nm})}=\dfrac{1240}{248}=5.0 eV.

  • Kinetic energy: KEmax=hν−W=5.0−3.0=2.0KE_{max}=h\nu-W=5.0-3.0=2.0 eV.
  • Stopping potential: from eV0=KEmaxeV_0=KE_{max}, V0=2.0V_0=2.0 V.
  • Threshold wavelength: λ0=1240W=12403.0=413\lambda_0=\dfrac{1240}{W}=\dfrac{1240}{3.0}=413 nm.

Light longer than 413 nm carries too little energy to eject any electron. Answer: KEmax=2.0KE_{max}=2.0 eV, V0=2.0V_0=2.0 V and λ0≈413\lambda_0\approx413 nm.