recall and use hf = Φ + 21mvmax2

Energy and Momentum of a Photon

1. Key Concepts

In the quantum description of light, a photon is a particle that carries both energy and momentum. The relationships are:

• Energy: $E = hf$ where $h$ is Planck’s constant ($6.626\times10^{-34}\,\text{J·s}$) and $f$ is the frequency.
• Momentum: $p = \dfrac{E}{c} = \dfrac{h}{\lambda}$ where $c$ is the speed of light in vacuum ($3.00\times10^{8}\,\text{m·s}^{-1}$) and $\lambda$ is the wavelength.

2. Photoelectric Effect – Objective

Recall and use the photoelectric equation:

$$hf = \Phi + \frac{1}{2} m v_{\max }^{2}$$

where:

• $\Phi$ is the work function of the metal (minimum energy required to liberate an electron).
• $v_{\max}$ is the maximum speed of the emitted electrons.

3. Derivation of Photon Momentum

Starting from the energy–frequency relation and the wave relation $c = f\lambda$:

$$p = \frac{E}{c} = \frac{hf}{c} = \frac{h}{\lambda}$$

This shows that a photon’s momentum is inversely proportional to its wavelength.

4. Numerical Example – Photon Energy and Momentum

Calculate the energy and momentum of a photon of wavelength $500\,\text{nm}$ (green light).

1. Convert wavelength to meters: $\lambda = 500\times10^{-9}\,\text{m}$.
2. Find frequency: $f = \dfrac{c}{\lambda} = \dfrac{3.00\times10^{8}}{5.00\times10^{-7}} = 6.00\times10^{14}\,\text{Hz}$.
3. Energy: $E = hf = (6.626\times10^{-34})(6.00\times10^{14}) = 3.98\times10^{-19}\,\text{J}$ (≈ $2.48\,\text{eV}$).
4. Momentum: $p = \dfrac{h}{\lambda} = \dfrac{6.626\times10^{-34}}{5.00\times10^{-7}} = 1.33\times10^{-27}\,\text{kg·m·s}^{-1}$.

5. Photon Energy Table for Common Wavelengths

Colour / Source	Wavelength $\lambda$ (nm)	Frequency $f$ (THz)	Energy $E$ (eV)	Momentum $p$ ($10^{-27}$ kg·m·s$^{-1}$)
Red (He‑Ne laser)	632.8	474	1.96	1.05
Green (LED)	530	566	2.34	1.25
Blue (Laser)	450	667	2.76	1.47
Ultraviolet (254 nm)	254	1180	4.89	2.61

6. Applying the Photoelectric Equation

Example: A metal with work function $\Phi = 2.2\,\text{eV}$ is illuminated with light of wavelength $400\,\text{nm}$. Find $v_{\max}$ of the emitted electrons.

1. Calculate photon energy: $E = \dfrac{hc}{\lambda} = \dfrac{1240}{400}\,\text{eV} = 3.10\,\text{eV}$.
2. Insert into the photoelectric equation: $3.10 = 2.2 + \dfrac{1}{2}mv_{\max}^{2}$.
3. Solve for kinetic energy: $K_{\max}=0.90\,\text{eV}=0.90\times1.602\times10^{-19}\,\text{J}=1.44\times10^{-19}\,\text{J}$.
4. Find $v_{\max}$: $v_{\max}= \sqrt{\dfrac{2K_{\max}}{m}} = \sqrt{\dfrac{2(1.44\times10^{-19})}{9.11\times10^{-31}}}= 5.6\times10^{5}\,\text{m·s}^{-1}$.

7. Summary

• Photon energy is directly proportional to frequency: $E = hf$.
• Photon momentum is $p = h/\lambda = E/c$.
• The photoelectric effect links photon energy to electron emission via $hf = \Phi + \tfrac12 mv_{\max}^2$.
• These relations are fundamental for understanding phenomena such as solar cells, photomultiplier tubes, and radiation pressure.

8. Practice Questions

1. Calculate the momentum of a $1.0\,\mu\text{m}$ infrared photon.
2. A metal surface has $\Phi = 1.8\,\text{eV}$. Light of wavelength $250\,\text{nm}$ shines on it. Determine the maximum kinetic energy of the emitted electrons in electron‑volts.
3. Explain why increasing the intensity of light (while keeping wavelength constant) increases the number of emitted electrons but not their $v_{\max}$.