Atomic Structure
1. Discovery of Sub-atomic Particles
Thomson's cathode-ray experiments established the electron and measured its charge-to-mass ratio ($e/m_e = 1.758820 \times 10^{11}\ C\,kg^{-1}$). Millikan's oil-drop experiment then fixed the charge, $e = 1.602 \times 10^{-19}\ C$, giving $m_e = 9.1094 \times 10^{-31}\ kg$. Rutherford's $\alpha$-particle scattering showed that almost all the mass and all the positive charge sit in a tiny nucleus, but his model could not explain why an orbiting (accelerating) electron does not radiate energy and spiral into the nucleus.
- Atomic number ($Z$): number of protons = number of electrons in a neutral atom.
- Mass number ($A$): protons + neutrons. Isotopes have the same $Z$, different $A$; isobars have the same $A$, different $Z$.
2. Dual Nature of Radiation and Matter
Planck's quantum theory: energy is emitted or absorbed in packets, $E = h\nu = hc/\lambda$. In the photoelectric effect, $h\nu = h\nu_0 + \frac{1}{2}m_ev^2$; below the threshold frequency $\nu_0$ no electrons are ejected, however intense the light.
de Broglie extended this to matter: $\lambda = \frac{h}{mv} = \frac{h}{p}$. For an electron accelerated through a potential $V$, $\lambda = \frac{h}{\sqrt{2meV}}$.
Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$. Exact position and momentum cannot both be known, so fixed Bohr orbits are replaced by orbitals (probability regions).
3. Bohr Model of the Hydrogen Atom
Valid only for one-electron species (H, He+, Li2+):
The wavenumber of an emitted line is given by the Rydberg equation:
- Lyman ($n_1 = 1$, UV), Balmer ($n_1 = 2$, visible), Paschen ($n_1 = 3$), Brackett ($n_1 = 4$), Pfund ($n_1 = 5$) — the last three are infrared.
- Maximum number of spectral lines when an electron falls from level $n$ to the ground state: $\frac{n(n-1)}{2}$.
Interactive diagram: open this page in a browser to use it.
4. Quantum Numbers and Orbitals
- Principal ($n$): shell and size; $n = 1, 2, 3 \ldots$
- Azimuthal ($l$): subshell and shape; $l = 0$ to $n-1$ (s, p, d, f).
- Magnetic ($m_l$): orientation; $-l$ to $+l$, giving $2l+1$ orbitals.
- Spin ($m_s$): $+\frac{1}{2}$ or $-\frac{1}{2}$.
Nodes: radial nodes $= n - l - 1$; angular nodes $= l$; total nodes $= n - 1$. Orbital angular momentum $= \sqrt{l(l+1)}\,\frac{h}{2\pi}$.
5. Filling Electrons
- Aufbau principle and the $(n+l)$ rule: lower $(n+l)$ fills first; if equal, lower $n$ fills first. So 4s (4+0) fills before 3d (3+2).
- Pauli exclusion principle: no two electrons in an atom share all four quantum numbers, so an orbital holds at most two electrons of opposite spin.
- Hund's rule: electrons occupy degenerate orbitals singly, with parallel spins, before pairing.
⚠️ Exceptions to Learn
Cr is $[Ar]\,3d^5 4s^1$ and Cu is $[Ar]\,3d^{10} 4s^1$, not $3d^4 4s^2$ and $3d^9 4s^2$. Half-filled and fully filled d subshells are extra stable because of symmetry and greater exchange energy.
🎯 JEE/NEET Corner
Spin-only magnetic moment $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. Expect it in atomic structure, d-block and coordination questions alike.
Practice questions
The number of radial nodes in a 3p orbital is:
- 0
- 1
- 2
- 3
Answer
B. Radial nodes = n - l - 1 = 3 - 1 - 1 = 1.
Which electronic configuration of chromium (Z = 24) is correct?
- [Ar] 3d⁴ 4s²
- [Ar] 3d⁵ 4s¹
- [Ar] 3d⁶
- [Ar] 4s² 4p⁴
Answer
B. A half-filled 3d subshell is extra stable, so one 4s electron moves into 3d: [Ar] 3d⁵ 4s¹.
The maximum number of spectral lines when an electron in a hydrogen atom falls from n = 5 to the ground state is:
- 4
- 5
- 10
- 15
Answer
C. n(n-1)/2 = 5 × 4/2 = 10.
The radius of the first Bohr orbit of He⁺ is:
- 52.9 pm
- 26.45 pm
- 105.8 pm
- 13.2 pm
Answer
B. r = 52.9 n²/Z = 52.9 × ½ = 26.45 pm.
Which set of quantum numbers is NOT allowed?
- n=2, l=1, mₗ=0
- n=3, l=2, mₗ=-2
- n=2, l=2, mₗ=1
- n=4, l=0, mₗ=0
Answer
C. l can only take values up to n-1. For n = 2, l = 2 is impossible.
The visible lines of the hydrogen spectrum belong to the:
- Lyman series
- Balmer series
- Paschen series
- Pfund series
Answer
B. Balmer lines end on n₁ = 2 and fall in the visible region. Lyman is UV; Paschen, Brackett and Pfund are infrared.