A-Level — physicsHard3 min read

The Photoelectric Effect and Photon Model

E = hf; work function; threshold frequency; electron volt; LED and photon energy

Tutor AI Team
Tutor AI Team

# The Photoelectric Effect and Photon Model — A-Level Physics

The photoelectric effect was one of the key experiments that established quantum physics. It showed that light behaves as discrete packets of energy — photons — rather than continuous waves.

1. Photon Energy

E=hf=hcλ\boxed{E = hf = \frac{hc}{\lambda}}

Where:

  • h=6.63×10−34h = 6.63 \times 10^{-34} J·s (Planck's constant)
  • ff = frequency (Hz)
  • λ\lambda = wavelength (m)

Electron Volt

11 eV =1.6×10−19= 1.6 \times 10^{-19} J — the energy gained by an electron accelerated through 1 V.

2. The Photoelectric Effect

When light shines on a metal surface, electrons may be emitted.

Key Observations

  1. Threshold frequency (f0f_0): No electrons emitted below this frequency, regardless of intensity
  2. Instantaneous emission: Electrons emitted immediately, even at low intensity
  3. Intensity: Increasing intensity increases the number of electrons but NOT their maximum KE
  4. Frequency: Increasing frequency increases the maximum KE of emitted electrons

Einstein's Equation

hf=ϕ+12mvmax⁡2\boxed{hf = \phi + \frac{1}{2}mv_{\max}^2}

Or: 12mvmax⁡2=hf−ϕ\frac{1}{2}mv_{\max}^2 = hf - \phi

Where ϕ=hf0\phi = hf_0 = work function (minimum energy to release an electron).

Why Wave Theory Fails

  • Wave theory predicts: any frequency works if bright enough (wrong)
  • Wave theory predicts: time delay for weak light (wrong)
  • Photon model explains both threshold frequency and instantaneous emission

3. Stopping Potential

The voltage needed to stop the most energetic electrons: eVs=hf−ϕeV_s = hf - \phi Vs=hef−ϕeV_s = \frac{h}{e}f - \frac{\phi}{e}

Plotting VsV_s vs ff: gradient = h/eh/e; y-intercept = −ϕ/e-\phi/e; x-intercept = f0f_0.

4. Line Spectra and Energy Levels

Atoms have discrete energy levels. When an electron transitions: ΔE=hf=E2−E1\Delta E = hf = E_2 - E_1

Emission spectrum: electron drops to lower level → photon emitted Absorption spectrum: photon absorbed → electron excited to higher level

Each element has a unique spectrum → used for identification.

Worked Example: Photon Energy

Problem

Find the energy of a photon of wavelength 550 nm in eV.

E=hc/λ=6.63×10−34×3×108/550×10−9=3.62×10−19E = hc/\lambda = 6.63 \times 10^{-34} \times 3 \times 10^8/550 \times 10^{-9} = 3.62 \times 10^{-19} J =3.62×10−19/1.6×10−19=2.26= 3.62 \times 10^{-19}/1.6 \times 10^{-19} = 2.26 eV

Solution

Worked Example: Photoelectric Effect

Problem

Sodium has ϕ=2.28\phi = 2.28 eV. Light of wavelength 400 nm shines on it. Find the max KE.

E=hc/λ=6.63×10−34×3×108/400×10−9=4.97×10−19E = hc/\lambda = 6.63 \times 10^{-34} \times 3 \times 10^8/400 \times 10^{-9} = 4.97 \times 10^{-19} J = 3.11 eV KEmax⁡=3.11−2.28=0.83KE_{\max} = 3.11 - 2.28 = 0.83 eV = 1.33×10−191.33 \times 10^{-19} J

Solution

Worked Example: Threshold Frequency

Problem

A metal has work function 4.5 eV. Find the threshold frequency and wavelength.

ϕ=hf0\phi = hf_0 → f0=ϕ/h=4.5×1.6×10−19/6.63×10−34=1.085×1015f_0 = \phi/h = 4.5 \times 1.6 \times 10^{-19}/6.63 \times 10^{-34} = 1.085 \times 10^{15} Hz λ0=c/f0=3×108/1.085×1015=276\lambda_0 = c/f_0 = 3 \times 10^8/1.085 \times 10^{15} = 276 nm (UV)

Solution

6. Practice Questions

    1. A photon has energy 5.0 eV. Calculate its frequency and wavelength. (3 marks)
    1. The work function of zinc is 4.3 eV. Will light of wavelength 200 nm cause photoemission? Calculate the max KE if yes. (4 marks)
    1. Explain why increasing the intensity of light below the threshold frequency produces no photoelectrons. (3 marks)
    1. An electron in a hydrogen atom drops from −1.5 eV to −13.6 eV. Find the wavelength of the emitted photon. (3 marks)

    Answers

    1. f=E/h=5.0×1.6×10−19/6.63×10−34=1.21×1015f = E/h = 5.0 \times 1.6 \times 10^{-19}/6.63 \times 10^{-34} = 1.21 \times 10^{15} Hz. λ=c/f=248\lambda = c/f = 248 nm.

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Summary

  • E=hf=hc/λE = hf = hc/\lambda; 1 eV = 1.6×10−191.6 \times 10^{-19} J
  • Photoelectric: hf=ϕ+12mvmax⁡2hf = \phi + \frac{1}{2}mv_{\max}^2
  • Threshold frequency: f0=ϕ/hf_0 = \phi/h
  • Intensity affects number of electrons, not max KE
  • Energy levels: ΔE=hf\Delta E = hf
Tags:
#a-level#physics#quantum#photoelectric-effect#photons#work-function

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