1. Chapter at a Glance
This chapter traces how our picture of the atom developed: from the discovery of electrons, protons and neutrons, through Rutherford's nuclear model and Bohr's model of hydrogen, to the modern quantum-mechanical model with orbitals and quantum numbers. It ends with the rules for writing electronic configurations. NEET asks one or two questions every year, mostly on quantum numbers, configurations, the hydrogen spectrum and de Broglie or uncertainty calculations.
- Energy and radius of Bohr orbits; wavelength of hydrogen spectral lines.
- Photon energy, photoelectric effect and de Broglie wavelength.
- Heisenberg's uncertainty principle.
- Quantum numbers, shapes of orbitals and number of nodes.
- Aufbau, Pauli and Hund's rules; exceptional configurations (Cr, Cu).
2. Sub-atomic Particles and Early Models
| Particle | Charge | Mass | NEET Note |
|---|---|---|---|
| Electron | −1.602 × 10⁻¹⁹ C | 9.11 × 10⁻³¹ kg | Discovered in cathode-ray tubes (J.J. Thomson); e/m found |
| Proton | +1.602 × 10⁻¹⁹ C | 1.673 × 10⁻²⁷ kg | From anode (canal) rays |
| Neutron | 0 | 1.675 × 10⁻²⁷ kg | Discovered by Chadwick (1932) |
- Thomson's model ("plum pudding"): positive charge spread uniformly, with electrons embedded in it.
- Rutherford's α-scattering: most α-particles passed through gold foil, but a few bounced back. So the atom has a tiny, dense, positive nucleus and is mostly empty space.
- Rutherford's model failed to explain why electrons do not spiral into the nucleus, or why atoms give line spectra.
- Atomic number Z = number of protons; mass number A = protons + neutrons. Isotopes: same Z, different A. Isobars: same A, different Z. Isotones: same number of neutrons.
3. Electromagnetic Radiation and Quantum Theory
- Planck proposed that energy is emitted or absorbed in small packets called quanta (for light, photons).
- Photoelectric effect: electrons are ejected from a metal only if ν ≥ ν₀. Then hν = hν₀ + ½mv². The number of electrons depends on intensity; their kinetic energy depends on frequency.
- Line spectra: each element emits light only at certain wavelengths, like a fingerprint. This shows that electron energies are quantised.
4. Bohr Model and the Hydrogen Spectrum
- Electrons move in fixed circular orbits (stationary states) without losing energy. Angular momentum is quantised: mvr = nh/2π.
- Energy is emitted or absorbed only when an electron jumps between orbits: ΔE = hν.
- Series: Lyman (to n = 1, UV), Balmer (to n = 2, visible), Paschen (to n = 3), Brackett (to n = 4), Pfund (to n = 5) — the last three are infrared.
- Maximum number of lines when an electron falls from level n to the ground state = n(n − 1)/2.
- Limitations: the model works only for one-electron species (H, He⁺, Li²⁺). It cannot explain the spectra of multi-electron atoms, the splitting of lines in magnetic and electric fields (Zeeman and Stark effects), or molecular shapes. It also contradicts the uncertainty principle.
5. Dual Nature and the Uncertainty Principle
- de Broglie proposed that moving particles have wave properties. This is noticeable only for tiny particles such as electrons.
- Heisenberg: we cannot know both the exact position and the exact momentum of an electron at the same time. So the idea of a definite orbit is replaced by the probability of finding the electron — an orbital.
6. Quantum Numbers and Orbitals
| Quantum number | Symbol and values | Tells us | NEET Note |
|---|---|---|---|
| Principal | n = 1, 2, 3… | Shell, size and energy | Max electrons in shell = 2n² |
| Azimuthal | l = 0 to n − 1 (s, p, d, f) | Subshell and shape | Orbital angular momentum √(l(l + 1)) h/2π |
| Magnetic | ml = −l … +l | Orientation | 2l + 1 orbitals in a subshell |
| Spin | ms = +½ or −½ | Direction of spin | Two electrons per orbital |
- Shapes: s is spherical; p is dumbbell-shaped (px, py, pz); d orbitals are mostly double-dumbbell, and dz² has a ring (doughnut) around its axis.
- Nodes: radial nodes = n − l − 1; angular nodes = l; total nodes = n − 1.
- Number of orbitals in shell n = n².
7. Filling of Orbitals and Electronic Configuration
- Aufbau principle: orbitals fill in order of increasing energy, decided by the (n + l) rule. Lower n + l fills first; if equal, lower n fills first.
- Order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p…
- Pauli exclusion principle: no two electrons in an atom can have all four quantum numbers the same.
- Hund's rule: in orbitals of equal energy, electrons first occupy separate orbitals with parallel spins before pairing.
- Exceptions: Cr = [Ar]3d⁵4s¹ and Cu = [Ar]3d¹⁰4s¹, because half-filled and completely filled subshells are extra stable (symmetry and exchange energy).
- Magnetic moment (spin-only) μ = √(n(n + 2)) BM, where n = number of unpaired electrons.
8. Formula Master Table
| Quantity | Formula |
|---|---|
| Wave relation | c = νλ ; ν̄ = 1/λ |
| Photon energy | E = hν = hc/λ |
| Photoelectric equation | hν = hν₀ + ½mv² |
| Bohr energy | En = −2.18 × 10⁻¹⁸ Z²/n² J |
| Bohr radius | rn = 52.9 n²/Z pm |
| Rydberg equation | ν̄ = RHZ²(1/n₁² − 1/n₂²) |
| Number of spectral lines | n(n − 1)/2 |
| de Broglie | λ = h/mv |
| Uncertainty | Δx · mΔv ≥ h/4π |
| Nodes | radial n − l − 1 ; angular l |
| Orbital angular momentum | √(l(l + 1)) · h/2π |
| Spin-only magnetic moment | μ = √(n(n + 2)) BM |
9. Master Summary Table
| Topic | Key idea | NEET Note |
|---|---|---|
| Rutherford | Nuclear atom | Could not explain stability or spectra |
| Bohr | Quantised orbits | Only for one-electron species |
| de Broglie | Matter waves | λ ∝ 1/mv |
| Heisenberg | Δx·Δp ≥ h/4π | Orbit replaced by orbital |
| Quantum numbers | n, l, ml, ms | l ranges from 0 to n − 1 |
| (n + l) rule | Order of filling | 4s fills before 3d |
| Hund's rule | Maximum multiplicity | N has 3 unpaired electrons |
| Exceptions | Cr, Cu | Half/fully filled d subshells |
10. Mind Map
- e, p, n
- Thomson
- Rutherford
- En, rn
- Rydberg
- Series
- Photons
- de Broglie
- Uncertainty
- Quantum numbers
- Shapes, nodes
- Aufbau, Pauli, Hund
11. Tips and Tricks
- E (eV) = 1240/λ (nm) for quick photon energies.
- For hydrogen: E₁ = −13.6 eV, E₂ = −3.4 eV, E₃ = −1.51 eV, E₄ = −0.85 eV.
- To find a set of quantum numbers for the last electron, write the configuration and apply Hund's rule to the last subshell.
- Number of electrons with a given ml or l in an atom: count from the full configuration.
- Allowing l = n (l can only go up to n − 1).
- Writing Cr as 3d⁴4s² or Cu as 3d⁹4s².
- Removing 3d electrons before 4s when forming ions (4s electrons leave first: Fe²⁺ is [Ar]3d⁶).
- Confusing radial nodes with angular nodes.
12. Important One-liners
- The neutron was discovered by Chadwick.
- The charge-to-mass ratio of the electron was measured by J.J. Thomson.
- An orbital can hold at most two electrons with opposite spins.
- The 2p orbitals have no radial node and one angular node.
- The energy of an orbital in hydrogen depends only on n.
- In multi-electron atoms, the energy depends on both n and l.
- Isoelectronic species have the same number of electrons (e.g. N³⁻, O²⁻, F⁻, Ne, Na⁺).
- The Balmer series of hydrogen lies in the visible region.
13. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, me = 9.11 × 10⁻³¹ kg and RH = 1.097 × 10⁷ m⁻¹. Click to reveal each answer and working.
Average level
Find the frequency and energy of a photon of ultraviolet light with wavelength 300 nm.
Show answer & solution
Concept: ν = c/λ; E = hν.
- ν = 3 × 10⁸/300 × 10⁻⁹ = 1 × 10¹⁵ Hz.
- E = 6.626 × 10⁻³⁴ × 10¹⁵ = 6.626 × 10⁻¹⁹ J.
Answer: 1 × 10¹⁵ Hz; 6.626 × 10⁻¹⁹ J
Find the energy of an electron in the second Bohr orbit of a hydrogen atom, in joules.
Show answer & solution
Concept: En = −2.18 × 10⁻¹⁸/n² J for hydrogen.
- E₂ = −2.18 × 10⁻¹⁸/4.
- E₂ = −5.45 × 10⁻¹⁹ J (= −3.4 eV).
Answer: −5.45 × 10⁻¹⁹ J
For n = 3, list the possible values of l, the total number of orbitals, and the maximum number of electrons.
Show answer & solution
Concept: l = 0 to n − 1; orbitals = n²; electrons = 2n².
- l = 0, 1, 2 (3s, 3p, 3d).
- Orbitals = 1 + 3 + 5 = 9 = n².
- Electrons = 2 × 9 = 18.
Answer: l = 0, 1, 2; 9 orbitals; 18 electrons
Compare the de Broglie wavelengths of (a) a 0.1 kg ball moving at 10 m s⁻¹ and (b) an electron moving at 10⁶ m s⁻¹.
Show answer & solution
Concept: λ = h/mv.
- (a) λ = 6.626 × 10⁻³⁴/(0.1 × 10) = 6.626 × 10⁻³⁴ m — far too small to detect.
- (b) λ = 6.626 × 10⁻³⁴/(9.11 × 10⁻³¹ × 10⁶) ≈ 7.27 × 10⁻¹⁰ m, comparable to atomic sizes.
Answer: 6.6 × 10⁻³⁴ m; about 7.3 × 10⁻¹⁰ m
Find the wavenumber and wavelength of the first line of the Paschen series of hydrogen (n = 4 → n = 3).
Show answer & solution
Concept: ν̄ = RH(1/n₁² − 1/n₂²).
- ν̄ = 1.097 × 10⁷(1/9 − 1/16) = 1.097 × 10⁷ × 7/144.
- ν̄ ≈ 5.33 × 10⁵ m⁻¹.
- λ = 1/ν̄ ≈ 1.875 × 10⁻⁶ m = 1875 nm (infrared).
Answer: about 5.33 × 10⁵ m⁻¹; about 1875 nm
Advanced level
The position of an electron is known to within 0.1 nm. Find the minimum uncertainty in its velocity.
Show answer & solution
Concept: Δx · mΔv ≥ h/4π.
- Δv ≥ h/(4π m Δx).
- = 6.626 × 10⁻³⁴/(4 × 3.1416 × 9.11 × 10⁻³¹ × 10⁻¹⁰).
- Δv ≥ 5.79 × 10⁵ m s⁻¹ — a huge uncertainty, so a fixed orbit is meaningless.
Answer: about 5.8 × 10⁵ m s⁻¹
Find the ionisation energy of hydrogen in kJ mol⁻¹, and the ionisation energy of Li²⁺ from its ground state in eV (1 eV per atom = 96.49 kJ mol⁻¹).
Show answer & solution
Concept: IE = 13.6 Z² eV for a one-electron species in its ground state.
- Hydrogen: 13.6 × 96.49 ≈ 1312 kJ mol⁻¹.
- Li²⁺ (Z = 3): 13.6 × 9 = 122.4 eV.
Answer: about 1312 kJ mol⁻¹; 122.4 eV
A metal has a threshold frequency of 7.0 × 10¹⁴ Hz. Light of frequency 1.0 × 10¹⁵ Hz falls on it. Find the kinetic energy and speed of the fastest photoelectrons.
Show answer & solution
Concept: ½mv² = h(ν − ν₀).
- KE = 6.626 × 10⁻³⁴ × 3 × 10¹⁴ ≈ 1.99 × 10⁻¹⁹ J.
- v = √(2KE/m) = √(3.98 × 10⁻¹⁹/9.11 × 10⁻³¹).
- v ≈ 6.6 × 10⁵ m s⁻¹.
Answer: about 1.99 × 10⁻¹⁹ J; about 6.6 × 10⁵ m s⁻¹
A 25 W lamp emits yellow light of wavelength 0.57 μm. Assuming all its energy is emitted as this light, how many photons does it give out per second?
Show answer & solution
Concept: photons per second = power/energy per photon.
- E = hc/λ = 6.626 × 10⁻³⁴ × 3 × 10⁸/0.57 × 10⁻⁶ ≈ 3.49 × 10⁻¹⁹ J.
- n = 25/3.49 × 10⁻¹⁹ ≈ 7.2 × 10¹⁹ per second.
Answer: about 7.2 × 10¹⁹ photons per second
(a) Find the number of radial and angular nodes in a 4d orbital. (b) Write the configuration of Fe³⁺ (Z = 26) and find its spin-only magnetic moment.
Show answer & solution
Concept: radial = n − l − 1, angular = l; μ = √(n(n + 2)).
- (a) 4d: n = 4, l = 2 ⇒ radial = 1, angular = 2 (total 3 = n − 1).
- (b) Fe = [Ar]3d⁶4s²; Fe³⁺ loses both 4s and one 3d electron ⇒ [Ar]3d⁵.
- 3d⁵ has 5 unpaired electrons: μ = √(5 × 7) = √35 ≈ 5.92 BM.
Answer: (a) 1 radial, 2 angular; (b) [Ar]3d⁵, about 5.92 BM
14. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
Which set of quantum numbers is NOT possible?
- n = 2, l = 1, ml = 0
- n = 3, l = 2, ml = −2
- n = 2, l = 2, ml = 1
- n = 4, l = 0, ml = 0
Show solution
l can only be 0 to n − 1. For n = 2, l cannot be 2, so set (C) is impossible. The other three sets follow all the rules.
Answer: (C)
The electronic configuration of copper (Z = 29) is:
- [Ar]3d⁹4s²
- [Ar]3d¹⁰4s¹
- [Ar]3d¹⁰4s²
- [Ar]3d⁸4s²4p¹
Show solution
A completely filled 3d¹⁰ subshell is extra stable, so one 4s electron moves into 3d. (A) is the configuration the Aufbau order would predict, but it is wrong. (C) has 30 electrons.
Answer: (B)
The number of radial nodes in a 3s orbital and a 2p orbital, respectively, is:
- 2, 0
- 0, 2
- 1, 1
- 2, 1
Show solution
Radial nodes = n − l − 1. For 3s: 3 − 0 − 1 = 2. For 2p: 2 − 1 − 1 = 0. The 2p orbital has one angular node, but no radial node.
Answer: (A)
According to the (n + l) rule, which orbital is filled first?
- 3d
- 4s
- 4p
- 5s
Show solution
The (n + l) values are: 3d = 5, 4s = 4, 4p = 5 and 5s = 5. The 4s orbital has the lowest value, so it fills first.
Answer: (B)
The maximum number of spectral lines when an electron in hydrogen falls from n = 5 to n = 1 is:
- 4
- 5
- 10
- 15
Show solution
n(n − 1)/2 = 5 × 4/2 = 10. The answer 4 counts only the single jumps from level to level, and 15 uses n(n + 1)/2 by mistake.
Answer: (C)
Which of these species are isoelectronic?
- Na⁺, Mg²⁺, F⁻
- Na, K, Rb
- Cl⁻, Br⁻, I⁻
- O, S, Se
Show solution
Na⁺, Mg²⁺ and F⁻ all have 10 electrons, the same as neon. The other options are members of the same group, and they have different numbers of electrons.
Answer: (A)
The ratio of the radius of the second Bohr orbit of He⁺ to that of the first Bohr orbit of hydrogen is:
- 1 : 1
- 2 : 1
- 4 : 1
- 1 : 2
Show solution
r ∝ n²/Z. For He⁺ with n = 2: 4/2 = 2. For H with n = 1: 1/1 = 1. The ratio is 2 : 1. (C) forgets to divide by Z.
Answer: (B)
The number of unpaired electrons in Mn²⁺ (Z = 25) and its spin-only magnetic moment are:
- 5, 5.92 BM
- 4, 4.90 BM
- 3, 3.87 BM
- 5, 4.90 BM
Show solution
Mn is [Ar]3d⁵4s². Mn²⁺ loses the two 4s electrons, leaving [Ar]3d⁵ with 5 unpaired electrons. So μ = √35 ≈ 5.92 BM. (B) would be correct for 3d⁴ or 3d⁶, and (D) pairs 5 unpaired electrons with the wrong magnetic moment.
Answer: (A)
15. Quick Revision Summary
- E = hν = hc/λ; hν = hν₀ + KE.
- Bohr: En = −13.6Z²/n² eV; rn = 52.9n²/Z pm; lines n(n − 1)/2.
- λ = h/mv; Δx·Δp ≥ h/4π.
- n, l (0 to n − 1), ml (−l to +l), ms (±½); nodes n − l − 1 and l.
- Aufbau (n + l), Pauli, Hund; Cr 3d⁵4s¹, Cu 3d¹⁰4s¹; μ = √(n(n + 2)).