Wave nature of electromagnetic radiation
Light, radio waves, X-rays and microwaves are all electromagnetic (EM) waves: electric and magnetic fields vibrating at right angles to each other and to the direction of travel. They need no medium and all travel in vacuum at c = 3.0 × 108 m s−1.
- Wavelength λ: distance between two crests (m, nm).
- Frequency ν: waves passing a point per second (Hz = s−1).
- Wavenumber ν̄ = 1/λ: waves per metre (m−1 or cm−1).
- They are linked: c = νλ.
Order by increasing wavelength: gamma rays, X-rays, UV, visible (about 400 nm violet to 750 nm red), IR, microwaves, radio waves. The wave picture explains interference and diffraction.
Particle nature: Planck's quantum theory
Some facts could not be explained by waves: black-body radiation (a hot iron rod turns red, then yellow, then white), the photoelectric effect and line spectra.
Planck (1900): energy is given out or taken in not continuously but in small packets called quanta. For light, a quantum is a photon. Energy of one quantum: E = hν = hc/λ, where h = 6.626 × 10−34 J s (Planck's constant). Energy can only be hν, 2hν, 3hν … never 1.5hν.
So light has a dual behaviour: wave-like in travel (interference), particle-like when it meets matter (photoelectric effect).
Photoelectric effect
When light of suitable frequency falls on a clean metal surface (like potassium, rubidium, caesium), electrons are thrown out at once. Key facts:
- Electrons come out instantly, with no delay.
- Below a threshold frequency ν₀ no electron comes out, however bright the light.
- The number of electrons depends on the intensity (brightness).
- The kinetic energy of electrons depends on the frequency, not the brightness.
Einstein's explanation: one photon hits one electron. Part of the photon's energy (the work function W₀ = hν₀) frees the electron; the rest becomes its kinetic energy:
hν = hν₀ + ½ mev²
Brighter light = more photons = more electrons, but each photon still has the same energy. Waves could not explain the threshold.
Atomic spectra: emission and absorption
A spectrum is the light spread out by wavelength. White light gives a continuous spectrum (all colours merge). When atoms are excited (heated or given electric energy), they give out light only at certain wavelengths: an emission line spectrum (bright lines on dark background). If white light passes through cool atoms, those same wavelengths are missing: an absorption spectrum (dark lines on a bright background).
Each element has its own line pattern, like a fingerprint. Rubidium, caesium, thallium, indium, gallium and scandium were discovered this way, and helium was first seen in the Sun's spectrum.
Line spectrum of hydrogen and the Rydberg formula
Hydrogen's lines fall into series. All are given by one formula (Rydberg):
ν̄ = 1/λ = RH (1/n₁² − 1/n₂²), RH = 109,677 cm−1 (≈ 1.097 × 107 m−1), n₂ > n₁.
| Series | n₁ | n₂ | Region |
|---|---|---|---|
| Lyman | 1 | 2, 3, 4 … | Ultraviolet |
| Balmer | 2 | 3, 4, 5 … | Visible |
| Paschen | 3 | 4, 5 … | Infrared |
| Brackett | 4 | 5, 6 … | Infrared |
| Pfund | 5 | 6, 7 … | Infrared |
Number of lines when electrons fall from level n to the ground state (all routes): n(n − 1)/2.
Bohr's model of the hydrogen atom: postulates
- The electron moves round the nucleus in circular paths of fixed radius and energy, called orbits, stationary states or energy levels (n = 1, 2, 3 …, also K, L, M …).
- In an orbit its energy does not change, so it does not radiate. This solves the stability problem.
- Energy is absorbed or given out only when the electron jumps between orbits: ΔE = E₂ − E₁ = hν (Bohr frequency rule).
- Only orbits with angular momentum a whole multiple of h/2π are allowed: mevr = nh/2π.
Radius and energy of the nth orbit
For hydrogen and hydrogen-like ions (He+, Li2+, Be3+, with atomic number Z):
- Radius: rₙ = 52.9 × n²/Z pm (a₀ = 52.9 pm is the Bohr radius).
- Energy: Eₙ = −RH(Z²/n²) = −2.18 × 10−18 Z²/n² J = −13.6 Z²/n² eV.
- Speed: vₙ = 2.18 × 106 × Z/n m s−1.
Why negative? Zero energy is chosen for an electron free and at rest far from the nucleus (n = ∞). A bound electron has less energy than that, so its energy is negative. More negative = more tightly held. n = 1 is the ground state; higher n are excited states. Ionisation energy of H = +13.6 eV = 2.18 × 10−18 J.
For a jump: ΔE = 2.18 × 10−18(1/n₁² − 1/n₂²) J. Dividing by hc gives exactly the Rydberg formula: this is how Bohr explained the hydrogen spectrum.
Limitations of Bohr's model
- Works only for one-electron species (H, He+, Li2+); fails for multi-electron atoms.
- Cannot explain the fine structure (a line is really a close doublet), or line splitting in a magnetic field (Zeeman effect) and electric field (Stark effect).
- Gives the electron a sharp path with exact position and speed at once, which the uncertainty principle forbids; ignores the electron's wave nature.
- Cannot explain how atoms join to form molecules (chemical bonding).
Try it: predict, then check
In the last story step, first predict: which jump gives a longer wavelength, 3 → 2 or 4 → 2? (Hint: smaller energy gap = longer wavelength.) Then press Jump for both and check. At home: look at a CD/DVD under a tube-light and a sodium street lamp. The tube-light shows a few bright colour bands; that is a line spectrum of mercury.
Board exam corner
Frequent questions: calculate ν, ν̄ or energy of a photon; photoelectric numericals (KE, threshold); wavelength of a hydrogen line with the Rydberg formula; energy/radius of the nth orbit of H or He+; postulates and limitations of Bohr's model (3–5 marks).
Key formulas and definitions
- c = νλ; ν̄ = 1/λ
- E = hν = hc/λ (h = 6.626 × 10⁻³⁴ J s)
- hν = hν₀ + ½mv² (photoelectric)
- 1/λ = R_H (1/n₁² − 1/n₂²), R_H = 1.097 × 10⁷ m⁻¹
- rₙ = 52.9 n²/Z pm
- Eₙ = −2.18 × 10⁻¹⁸ Z²/n² J = −13.6 Z²/n² eV
- mvr = nh/2π
- Lines from level n to ground = n(n − 1)/2
Worked examples
1. Vividh Bharati broadcasts at 1368 kHz. Find the wavelength.
λ = c/ν = 3.0 × 10⁸ ÷ 1.368 × 10⁶ = 219.3 m. It is a radio wave.
2. Yellow light has λ = 580 nm. Find ν and ν̄.
ν = c/λ = 3 × 10⁸ ÷ 580 × 10⁻⁹ = 5.17 × 10¹⁴ Hz. ν̄ = 1/λ = 1 ÷ 580 × 10⁻⁹ = 1.72 × 10⁶ m⁻¹.
3. Find the energy of one photon and of one mole of photons of frequency 5 × 10¹⁴ Hz.
E = hν = 6.626 × 10⁻³⁴ × 5 × 10¹⁴ = 3.31 × 10⁻¹⁹ J. Per mole: 3.31 × 10⁻¹⁹ × 6.022 × 10²³ = 1.99 × 10⁵ J = 199 kJ mol⁻¹.
4. Light of 400 nm falls on a metal with work function 2.13 eV. Find the KE of the ejected electrons.
Photon energy E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) ÷ 400 × 10⁻⁹ = 4.97 × 10⁻¹⁹ J = 3.10 eV. KE = 3.10 − 2.13 = 0.97 eV = 1.55 × 10⁻¹⁹ J.
5. The threshold frequency of a metal is 7.0 × 10¹⁴ s⁻¹. Find the KE of an electron emitted when light of 1.0 × 10¹⁵ s⁻¹ hits it.
KE = h(ν − ν₀) = 6.626 × 10⁻³⁴ × (1.0 × 10¹⁵ − 7.0 × 10¹⁴) = 6.626 × 10⁻³⁴ × 3 × 10¹⁴ = 1.99 × 10⁻¹⁹ J.
6. Find the wavelength of the line when the electron in H falls from n = 3 to n = 2. Which series?
1/λ = 1.097 × 10⁷ (1/4 − 1/9) = 1.097 × 10⁷ × 5/36 = 1.524 × 10⁶ m⁻¹. λ = 656 nm (red). n₁ = 2, so Balmer series.
7. Find the energy of the electron in the 2nd orbit of He⁺ and the radius of that orbit.
Z = 2, n = 2. E = −2.18 × 10⁻¹⁸ × 4/4 = −2.18 × 10⁻¹⁸ J (same as H ground state). r = 52.9 × 4/2 = 105.8 pm.
8. Find the ionisation energy of H in kJ mol⁻¹ and the energy needed to excite H from n = 1 to n = 3.
IE = 2.18 × 10⁻¹⁸ J × 6.022 × 10²³ = 1.313 × 10⁶ J = 1313 kJ mol⁻¹. ΔE(1→3) = 2.18 × 10⁻¹⁸ (1 − 1/9) = 1.94 × 10⁻¹⁸ J.
9. Electrons in a sample of H atoms are in n = 5. How many spectral lines can appear as they fall back to n = 1?
Lines = n(n − 1)/2 = 5 × 4/2 = 10.
Common mistakes
- Forgetting to convert nm to m: 400 nm = 400 × 10⁻⁹ m.
- Thinking brighter light gives faster photoelectrons. Brightness changes the number of electrons; frequency changes their KE.
- Putting n₁ and n₂ the wrong way round. n₁ is the lower (final) orbit for emission, and n₂ > n₁.
- Using Bohr's formulas for He, Li or Na atoms. They work only for one-electron species like H, He⁺, Li²⁺.