Numford
Photoelectric

Photoelectric effect

Light of 6.0 × 10¹⁴ Hz has photons of 3.98 × 10⁻¹⁹ J = 2.49 eV. On caesium (φ = 2.14 eV) it frees electrons with K max = 0.35 eV, so the stopping potential is 0.35 V. A 2.0 mW beam carries 5.0 × 10¹⁵ photons per second.

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
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

Photoelectric effect: quick answers

What is the photoelectric effect?

When light of high enough frequency falls on a metal, electrons are given off at once. Einstein explained it with photons: each photon of energy hν frees at most one electron, which leaves with K max = hν - φ.

What happens if the light on caesium is made twice as bright?

Twice the photons arrive each second, so twice the electrons leave and the current doubles. Each electron still gets one photon, so K max stays 0.35 eV and the stopping potential stays 0.35 V.

What happens below the threshold frequency?

No electrons leave at all, however bright the light. For caesium the threshold is ν₀ = φ / h = 5.16 × 10¹⁴ Hz; below it a photon has less energy than the work function.

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