Numford
Photoelectric

Threshold frequency and threshold wavelength

For caesium (φ = 2.14 eV) the threshold frequency is ν₀ = φ / h = 5.16 × 10¹⁴ Hz and the threshold wavelength is λ₀ = c / ν₀ = 581 nm. Light of lower frequency, or longer wavelength, frees no electrons.

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
Working

Step by step

  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
Ten metals

Threshold frequency and wavelength

Metalφν₀λ₀
Caesium (Cs)2.14 eV5.16 × 10¹⁴ Hz581 nm
Potassium (K)2.3 eV5.55 × 10¹⁴ Hz540 nm
Sodium (Na)2.75 eV6.64 × 10¹⁴ Hz452 nm
Calcium (Ca)3.2 eV7.72 × 10¹⁴ Hz388 nm
Molybdenum (Mo)4.17 eV1.01 × 10¹⁵ Hz298 nm
Aluminium (Al)4.28 eV1.03 × 10¹⁵ Hz290 nm
Copper (Cu)4.65 eV1.12 × 10¹⁵ Hz267 nm
Silver (Ag)4.7 eV1.13 × 10¹⁵ Hz264 nm
Nickel (Ni)5.15 eV1.24 × 10¹⁵ Hz241 nm
Platinum (Pt)5.65 eV1.36 × 10¹⁵ Hz220 nm
Derivation

Why it works

  1. Light arrives in packets, photons, each of energy E = hν.
  2. One photon is absorbed by one electron, all at once, so there is no delay however dim the light.
  3. The electron must spend at least the work function φ to leave the metal.
  4. The fastest electrons keep the rest: K max = hν - φ. This is Einstein's photoelectric equation.
  5. A retarding voltage V₀ stops even these, so eV₀ = K max. When hν = φ, K max = 0: that frequency is the threshold ν₀ = φ / h.
Questions

Threshold frequency: quick answers

What is the threshold frequency?

The lowest frequency of light that can free electrons from a metal. At ν₀ a photon has just enough energy to pay the work function, so hν₀ = φ.

Why does caesium respond to visible light?

Its work function is small, 2.14 eV, so its threshold wavelength 581 nm lies in the visible. Most metals need ultraviolet light.

How are the threshold frequency and wavelength related?

λ₀ = c / ν₀ = hc / φ. A larger work function means a higher threshold frequency and a shorter threshold wavelength.

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