Numford
Photoelectric

Stopping potential

For 300 nm light on sodium (φ = 2.75 eV), hν = 4.14 eV and K max = 1.39 eV, so the stopping potential is V₀ = K max / e = 1.39 V.

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 energy4.14 eV6.63 × 10⁻¹⁹ J
Threshold6.64 × 10¹⁴ Hzλ₀ = 452 nm
K max1.39 eVfastest at 7.0 × 10⁵ m/s
Stopping potential1.39 Vstops even the fastest
Photons per second3.0 × 10¹⁵at 2 mW

Na plate in light of 300 nm: electrons leave with up to 1.4 eV.

510-22ν (10¹⁴ Hz)K max (eV)ν₀ = 6.64-φ = -2.75 eVslope h
K max against ν
-2-10123V (volt)Current2 mW1 mW-V₀
Current against voltage
Show the working
  1. ν = c / λ = 3 × 10⁸ / (300 × 10⁻⁹) = 1.0 × 10¹⁵ Hz
  2. E = hν = 6.63 × 10⁻³⁴ × 1.0 × 10¹⁵ = 6.63 × 10⁻¹⁹ J = 6.63 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 4.14 eV
  3. ν₀ = φ / h = 2.75 × 1.6 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 6.64 × 10¹⁴ Hz, so λ₀ = c / ν₀ = 452 nm
  4. K max = hν - φ = 4.144 - 2.75 = 1.39 eV
  5. eV₀ = K max, so V₀ = 1.39 V
  6. v max = √(2K / m) = √(2 × 2.23 × 10⁻¹⁹ / 9.11 × 10⁻³¹) = 7.0 × 10⁵ m/s
  7. Photons per second = P / E = 2.0 × 10⁻³ W / 6.63 × 10⁻¹⁹ J = 3.0 × 10¹⁵
Working

Step by step

  1. ν = c / λ = 3 × 10⁸ / (300 × 10⁻⁹) = 1.0 × 10¹⁵ Hz
  2. E = hν = 6.63 × 10⁻³⁴ × 1.0 × 10¹⁵ = 6.63 × 10⁻¹⁹ J = 6.63 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 4.14 eV
  3. ν₀ = φ / h = 2.75 × 1.6 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 6.64 × 10¹⁴ Hz, so λ₀ = c / ν₀ = 452 nm
  4. K max = hν - φ = 4.144 - 2.75 = 1.39 eV
  5. eV₀ = K max, so V₀ = 1.39 V
  6. v max = √(2K / m) = √(2 × 2.23 × 10⁻¹⁹ / 9.11 × 10⁻³¹) = 7.0 × 10⁵ m/s
  7. Photons per second = P / E = 2.0 × 10⁻³ W / 6.63 × 10⁻¹⁹ J = 3.0 × 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

Stopping potential: quick answers

What is the stopping potential?

The negative collector voltage that just stops the fastest photoelectrons, so the current falls to zero. Then eV₀ = K max.

Does the stopping potential depend on the intensity?

No. Brighter light gives more electrons and a larger current, but the same K max, so V₀ is the same. It rises only with the frequency of the light.

How fast are the fastest electrons here?

K max = 1.39 eV = 2.23 × 10⁻¹⁹ J, so v = √(2K / m) = 7.0 × 10⁵ m/s.

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