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

where

• $u$ 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 $u_{0}$ and a threshold wavelength $\lambda_{0}$.

$$\phi = hu_{0} = \frac{hc}{\lambda_{0}}$$

Thus:

• Threshold frequency: $\displaystyle u_{0} = \frac{\phi}{h}$
• 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:

$$K_{\text{max}} = hu - \phi$$

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

$$0 = hu_{0} - \phi \quad\Rightarrow\quad hu_{0} = \phi$$

From this, the expressions for $u_{0}$ 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: $$u_{0} = \frac{\phi}{h} = \frac{4.005\times10^{-19}}{6.626\times10^{-34}} \approx 6.04\times10^{14}\ \text{Hz}$$
4. Calculate threshold wavelength: $$\lambda_{0} = \frac{c}{u_{0}} = \frac{3.00\times10^{8}}{6.04\times10^{14}} \approx 5.0\times10^{-7}\ \text{m} = 500\ \text{nm}$$

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

Metal	Work Function $\phi$ (eV)	Threshold Wavelength $\lambda_{0}$ (nm)	Threshold Frequency $u_{0}$ ($\times10^{14}$ Hz)
Cesium (Cs)	1.95	637	4.71
Sodium (Na)	2.28	544	5.51
Aluminium (Al)	4.28	290	10.3
Platinum (Pt)	5.65	220	13.6

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 $u_{0}$ 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 $u_{0}$ 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 $hu_{0} = \phi$ 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.