understand that photoelectrons may be emitted from a metal surface when it is illuminated by electromagnetic radiation

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Energy and Momentum of a Photon

Energy and Momentum of a Photon

Learning Objective

Understand that photoelectrons may be emitted from a metal surface when it is illuminated by electromagnetic radiation.

1. Photons – Quantised Packets of Light

A photon is a quantum of electromagnetic radiation. Its energy \$E\$ and momentum \$p\$ are related to its frequency \$\nu\$ (or wavelength \$\lambda\$) by:

\$\$

E = h\nu = \frac{hc}{\lambda},\qquad p = \frac{E}{c} = \frac{h}{\lambda}

\$\$

  • \$h = 6.626\times10^{-34}\ \text{J·s}\$ – Planck’s constant
  • \$c = 3.00\times10^{8}\ \text{m·s}^{-1}\$ – speed of light in vacuum

2. The Photoelectric Effect

When light of sufficient frequency shines on a clean metal surface, electrons can be ejected. This phenomenon is called the photoelectric effect.

2.1 Key Observations

  1. Electrons are emitted only if the incident light frequency \$\nu\$ exceeds a certain threshold \$\nu_0\$.
  2. The kinetic energy of the emitted electrons depends on the frequency of the light, not its intensity.
  3. Increasing the light intensity (at fixed \$\nu > \nu_0\$) increases the number of emitted electrons but not their maximum kinetic energy.
  4. There is an instantaneous emission – no measurable time lag between illumination and electron ejection.

2.2 Einstein’s Photoelectric Equation

Einstein explained the effect by treating light as a stream of photons. The energy balance for a single photon interacting with an electron is:

\$\$

h\nu = \phi + K_{\text{max}}

\$\$

  • \$\phi\$ – work function of the metal (minimum energy required to liberate an electron)
  • \$K_{\text{max}}\$ – maximum kinetic energy of the emitted electron

The work function is related to the threshold frequency by \$\phi = h\nu_0\$.

2.3 Stopping Potential

In an experimental setup, a reverse (retarding) voltage \$V_s\$ is applied to stop the most energetic electrons. The stopping potential satisfies:

\$\$

eVs = K{\text{max}} = h\nu - \phi

\$\$

  • \$e = 1.602\times10^{-19}\ \text{C}\$ – elementary charge

3. Conditions for Photoelectron Emission

ParameterRequirement for Emission
Frequency \$\nu\$\$\nu > \nu_0 = \dfrac{\phi}{h}\$ (must exceed threshold)
Photon Energy \$E = h\nu\$Must be greater than the work function \$\phi\$
IntensityControls number of photons per unit time → controls number of emitted electrons, not \$K_{\text{max}}\$

4. Momentum Transfer

When a photon ejects an electron, conservation of momentum must be considered. The photon’s momentum \$p = h/\lambda\$ is transferred to the electron and the metal lattice. In most A‑Level treatments, the recoil of the metal is negligible, but the concept reinforces the particle nature of light.

5. Example Calculation

Determine the maximum kinetic energy of electrons emitted from a sodium surface (\$\phi = 2.28\ \text{eV}\$) when illuminated with light of wavelength \$400\ \text{nm}\$.

  1. Convert work function to joules: \$\phi = 2.28\ \text{eV} \times 1.602\times10^{-19}\ \text{J/eV} = 3.65\times10^{-19}\ \text{J}\$.
  2. Photon energy: \$E = \dfrac{hc}{\lambda} = \dfrac{6.626\times10^{-34}\times3.00\times10^{8}}{4.00\times10^{-7}} = 4.97\times10^{-19}\ \text{J}\$.
  3. Maximum kinetic energy: \$K_{\text{max}} = E - \phi = 4.97\times10^{-19} - 3.65\times10^{-19} = 1.32\times10^{-19}\ \text{J}\$.
  4. Convert back to electron‑volts: \$K_{\text{max}} = \dfrac{1.32\times10^{-19}}{1.602\times10^{-19}}\ \text{eV} \approx 0.82\ \text{eV}\$.

6. Experimental Setup (Suggested Diagram)

Suggested diagram: Schematic of a photoelectric experiment showing a light source, metal photocathode, anode, and variable stopping potential \$V_s\$.

7. Summary

  • Photons carry quantised energy \$E = h\nu\$ and momentum \$p = h/\lambda\$.
  • The photoelectric effect demonstrates the particle nature of light.
  • Emission occurs only when \$h\nu > \phi\$; the excess energy becomes kinetic energy of the electron.
  • The stopping potential provides a direct measurement of \$K_{\text{max}}\$ and allows determination of \$\phi\$.
  • Intensity affects the number of emitted electrons, not their kinetic energy.

8. Quick Revision Questions

  1. What is the threshold frequency for a metal with work function \$4.5\ \text{eV}\$?
  2. If light of wavelength \$250\ \text{nm}\$ shines on a metal with \$\phi = 3.0\ \text{eV}\$, calculate the stopping potential required to just stop the most energetic electrons.
  3. Explain why increasing the intensity of light of frequency below the threshold does not cause electron emission.