understand and use the terms threshold frequency and threshold wavelength

Energy and Momentum of a Photon

In the quantum description of light, a photon is a particle that carries both energy and momentum despite having no rest mass. The relationships are derived from Planck’s constant \$h\$ and the speed of light \$c\$.

Key Equations

• Energy: \$E = h\nu = \dfrac{hc}{\lambda}\$
• Momentum: \$p = \dfrac{E}{c} = \dfrac{h}{\lambda} = \dfrac{h\nu}{c}\$

where

• \$\nu\$ is the frequency of the photon (Hz),
• \$\lambda\$ is the wavelength (m),
• \$h = 6.626 \times 10^{-34}\ \text{J·s}\$,
• \$c = 3.00 \times 10^{8}\ \text{m·s}^{-1}\$.

Threshold Frequency and Threshold Wavelength

When a photon strikes a metal surface, it can liberate an electron if its energy exceeds the work function \$\phi\$ of the metal. The minimum photon energy required is called the threshold energy, corresponding to a threshold frequency \$\nu{0}\$ and a threshold wavelength \$\lambda{0}\$.

\$\phi = h\nu{0} = \frac{hc}{\lambda{0}}\$

Thus:

• Threshold frequency: \$\displaystyle \nu_{0} = \frac{\phi}{h}\$
• Threshold wavelength: \$\displaystyle \lambda_{0} = \frac{hc}{\phi}\$

Derivation from the Photoelectric Equation

The kinetic energy \$K_{\text{max}}\$ of the most energetic emitted electrons is given by Einstein’s photoelectric equation:

\$K_{\text{max}} = h\nu - \phi\$

Setting \$K_{\text{max}} = 0\$ defines the threshold condition:

\$0 = h\nu{0} - \phi \quad\Rightarrow\quad h\nu{0} = \phi\$

From this, the expressions for \$\nu{0}\$ and \$\lambda{0}\$ follow directly.

Worked Example

1. Given: work function \$\phi = 2.5\ \text{eV}\$ for a metal. (Recall \$1\ \text{eV}=1.602\times10^{-19}\ \text{J}\$.)
2. Convert \$\phi\$ to joules:

   \$\phi = 2.5\ \text{eV} \times 1.602\times10^{-19}\ \frac{\text{J}}{\text{eV}} = 4.005\times10^{-19}\ \text{J}\$

3. Calculate threshold frequency:

   \$\nu_{0} = \frac{\phi}{h} = \frac{4.005\times10^{-19}}{6.626\times10^{-34}} \approx 6.04\times10^{14}\ \text{Hz}\$

4. Calculate threshold wavelength:

   \$\lambda{0} = \frac{c}{\nu{0}} = \frac{3.00\times10^{8}}{6.04\times10^{14}} \approx 5.0\times10^{-7}\ \text{m} = 500\ \text{nm}\$

Therefore, photons with wavelength shorter than \$500\ \text{nm}\$ (or frequency higher than \$6.0\times10^{14}\ \text{Hz}\$) will eject electrons from this metal.

Table: Typical Threshold Wavelengths for Common Metals

Conceptual Checks

• If a photon’s wavelength is longer than \$\lambda_{0}\$, can it cause photoemission? No – its energy is insufficient.
• Does increasing the intensity of light below \$\nu_{0}\$ cause electrons to be emitted? No – intensity changes the number of photons, not their individual energy.
• How does the momentum of a photon relate to its threshold wavelength?

  \$p{0} = \frac{h}{\lambda{0}}\$

  Shorter \$\lambda_{0}\$ means larger momentum.

Suggested Diagram

Summary

Understanding threshold frequency \$\nu{0}\$ and threshold wavelength \$\lambda{0}\$ is essential for applying the photoelectric effect in A‑Level physics. They are directly linked to the work function of a material through the simple relations \$h\nu{0} = \phi\$ and \$\lambda{0}=hc/\phi\$. Mastery of these concepts enables accurate prediction of whether a given light source can liberate electrons from a particular metal surface.