Numford
Physics

Atoms and quanta

Light in, electrons out. Orbits, waves and decay.

Photoelectric effect

Pick a metal and a colour of light. Each photon carries hν; it frees an electron only if hν is more than the work function φ. Brighter light means more photons, not bigger ones. Values use h = 6.63 × 10⁻³⁴ J s and e = 1.6 × 10⁻¹⁹ C.

Metal
Light
Photon energy2.49 eV3.98 × 10⁻¹⁹ J
Threshold5.16 × 10¹⁴ Hzλ₀ = 581 nm
K max0.35 eVfastest at 3.49 × 10⁵ m/s
Stopping potential0.35 Vstops even the fastest
Photons per second5.0 × 10¹⁵at 2 mW

Cs plate in light of 500 nm: electrons leave with up to 0.35 eV.

2.557.5-2ν (10¹⁴ Hz)K max (eV)ν₀ = 5.16-φ = -2.14 eVslope h
K max against ν
-10123V (volt)Current2 mW1 mW-V₀
Current against voltage
Show the working
  1. E = hν = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.98 × 10⁻¹⁹ J = 3.98 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 2.49 eV
  2. ν₀ = φ / h = 2.14 × 1.6 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 5.16 × 10¹⁴ Hz, so λ₀ = c / ν₀ = 581 nm
  3. K max = hν - φ = 2.486 - 2.14 = 0.35 eV
  4. eV₀ = K max, so V₀ = 0.35 V
  5. v max = √(2K / m) = √(2 × 5.54 × 10⁻²⁰ / 9.11 × 10⁻³¹) = 3.49 × 10⁵ m/s
  6. Photons per second = P / E = 2.0 × 10⁻³ W / 3.98 × 10⁻¹⁹ J = 5.0 × 10¹⁵

de Broglie wavelength

Every moving particle has a wave with λ = h / p. Give the speed, the kinetic energy, or the voltage a charged particle was accelerated through. Speeds are taken well below the speed of light.

Particle
Given
Wavelength λ0.123 nm1.23 × 10⁻¹⁰ m
Momentum5.4 × 10⁻²⁴kg m/s
Kinetic energy100 eV1.6 × 10⁻¹⁷ J
Speed5.93 × 10⁶ m/sElectron

Electron: de Broglie wavelength 0.123 nm.

Show the working
  1. Given: electron, m = 9.11 × 10⁻³¹ kg, accelerated through V = 100 V.
  2. K = eV = 100 eV = 100 × 1.6 × 10⁻¹⁹ = 1.6 × 10⁻¹⁷ J
  3. p = √(2mK) = √(2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁷) = 5.4 × 10⁻²⁴ kg m/s
  4. λ = h / p = 6.63 × 10⁻³⁴ / 5.4 × 10⁻²⁴ = 1.23 × 10⁻¹⁰ m = 0.123 nm
  5. Shortcut for an electron: λ = 1.227 / √V nm = 1.227 / √100 = 0.123 nm

Bohr model

Orbits grow as n². Energies are -13.6 Z² / n² eV. Wavelengths use the Rydberg constant of each nucleus (1.0968 × 10⁷ per metre for hydrogen); above 200 nm they are given in air, as spectrum tables list them.

Atom
Level n
Jump from n₂
down to n₁
1234
∞0n=1-13.6n=2-3.4n=3-1.51eVLymanBalmerPaschenBrackettPfund
Radius r₂2.12 Å2.12 × 10⁻¹⁰ m
Speed v₂1.09 × 10⁶m/s
Energy E₂-3.4 eV10.2 eV above ground
Photon, 3 to 21.89 eVBalmer α
Wavelength656.3 nmvisible

Hydrogen: the jump from n = 3 to n = 2 gives a 656.3 nm photon in the Balmer series.

Show the working
  1. r₂ = 0.529 n² / Z Å = 0.529 × 4 / 1 = 2.12 Å = 2.12 × 10⁻¹⁰ m
  2. v₂ = 2.19 × 10⁶ × Z / n = 1.09 × 10⁶ m/s
  3. E₂ = -13.6 Z² / n² eV = -13.6 / 4 = -3.4 eV
  4. Jump 3 to 2: ΔE = 13.6 × (1/4 - 1/9) = 1.89 eV
  5. 1/λ = R Z² (1/n₁² - 1/n₂²) = 1.0968 × 10⁷ × (1/4 - 1/9) = 1.523 × 10⁶ per m
  6. λ = 656.5 nm in vacuum, 656.3 nm in air: the Balmer series, visible.

Radioactive decay

N = N₀ e-λt, T½ = 0.693 / λ, mean life τ = 1 / λ and activity R = λN in becquerel. The same law fits a first order reaction, with k in place of λ. The dots are 1000 nuclei.

125 of 1000 left
1T½2T½3T½4T½1½¼⅛N / N₀
Half-life5 dλ = 0.1386 per d
Mean life τ7.213 d1/λ = 1.44 T½
Left after 15 d12.5%N = 125 of 1,000
Activity now2.01 × 10⁻⁴ Bq5.42 × 10⁻¹⁵ Ci
Time to 25% left10 d2 half-lives

After 15 d, 12.5% is left: 125 of 1000 nuclei.

Show the working
  1. λ = 0.693 / T½ = 0.693 / 5 = 0.1386 per d
  2. Mean life τ = 1 / λ = 1.44 T½ = 7.213 d
  3. N = N₀ e-λt = N₀ (½)n with n = t / T½ = 15 / 5 = 3
  4. N / N₀ = (½)n = 0.125, so N = 125 of 1,000
  5. Time to 25% left: t = ln(N₀ / N) / λ = ln(4) / 0.1386 = 10 d
Worked examples

Pick a topic

Photoelectric

de Broglie

Bohr atom

Half-life

Same subject: dimensional formula, unit converter, significant figures, error analysis, vectors, mirrors and lenses, projectile motion, circuits.

Questions students ask

Photons, orbits and half-lives

What is Einstein's photoelectric equation?

K max = hν - φ. One photon of energy hν gives all its energy to one electron; the work function φ is spent getting out of the metal and the rest is the largest kinetic energy an electron can have. The stopping potential obeys eV₀ = K max.

Why does brighter light not give faster electrons?

Brighter light means more photons per second, not bigger photons. More photons free more electrons, so the current rises, but each electron still gets one photon's energy, so K max and the stopping potential stay the same.

How do I find the de Broglie wavelength of an electron?

Use λ = h / p. For an electron accelerated through V volts, p = √(2meV), which gives the shortcut λ = 1.227 / √V nm. Through 100 V that is 0.123 nm, about the size of an atom.

What are the energy levels of hydrogen in the Bohr model?

Eₙ = -13.6 / n² eV: -13.6 eV for n = 1, -3.4 eV for n = 2 and -1.51 eV for n = 3. For a hydrogen-like ion multiply by Z². The orbit radius is 0.529 n² / Z Å.

Which series of the hydrogen spectrum is visible?

The Balmer series, where the electron falls to n = 2. Its first line, H alpha from n = 3, is red at 656.3 nm. Lyman lines (to n = 1) are ultraviolet; Paschen, Brackett and Pfund lines are infrared.

How much of a radioactive sample is left after three half-lives?

One eighth. Each half-life halves what is left: 1/2, then 1/4, then 1/8. In general N = N₀ (1/2)ⁿ with n = t / T½, and T½ = 0.693 / λ.