Photon Energy: E = hf

Introduction to Photons

• Light exhibits wave-particle duality, meaning it can behave as both a wave and a particle.
• The particle-like aspect of light is described by photons, which are discrete packets of energy.
• Classical physics failed to explain phenomena like the photoelectric effect, leading to the concept of quantized energy.

KEY TAKEAWAY: Light has both wave and particle properties. Photons are the ‘particles’ of light.

Quantized Energy of Photons

• Quantization of energy: Energy exists in discrete packets called quanta.
• Photon energy (E): The energy of a single photon is directly proportional to its frequency (f).
• Planck’s constant (h): The constant of proportionality relating energy and frequency, $h = 6.63 \times 10^{-34} \text{ Js}$.

The Formula: E = hf

• The energy of a photon is given by the equation:

  $$E = hf$$

  Where:
  * $E$ is the energy of the photon (in Joules, J)
  * $h$ is Planck’s constant (\$6.63 \times 10^{-34} \text{ Js}$)
  * $f$ is the frequency of the light (in Hertz, Hz)

• Since $f = \frac{c}{\lambda}$, where $c$ is the speed of light and $\lambda$ is the wavelength, the equation can also be written as:

  $$E = \frac{hc}{\lambda}$$

  Where:
  * $c$ is the speed of light (\$3.0 \times 10^8 \text{ m/s}$)
  * $\lambda$ is the wavelength of the light (in meters, m)

REMEMBER: $E = hf$ is the core equation for photon energy. Relate frequency and wavelength using $c = f\lambda$.

Energy Units: Joules (J) and Electron-Volts (eV)

• The standard unit of energy is the Joule (J).
• For atomic-scale phenomena, the electron-volt (eV) is often more convenient.
• Definition of electron-volt: The amount of energy gained (or lost) by a single electron when it moves through an electric potential difference of 1 volt.
• Conversion: \$1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$

Converting Between Joules and Electron-Volts

• J to eV: Divide the energy in Joules by \$1.602 \times 10^{-19}$.
• eV to J: Multiply the energy in electron-volts by \$1.602 \times 10^{-19}$.

EXAM TIP: Pay close attention to units! Convert to Joules before using $E=hf$ if energy is given in eV or vice versa.

Applications of Photon Energy

• Photoelectric effect: Photons with sufficient energy can eject electrons from a metal surface.
• Spectroscopy: Analyzing the wavelengths of light emitted or absorbed by substances to determine their composition and structure.
• Medical imaging: X-rays (high-energy photons) are used to create images of the inside of the body.
• Laser technology: Lasers use stimulated emission of photons to produce coherent light beams.

APPLICATION: The photoelectric effect demonstrates the particle nature of light and is used in light sensors and solar cells.

Relationship Between Photon Energy, Frequency, and Wavelength

• Higher frequency light has higher energy photons.
• Shorter wavelength light has higher energy photons.
• Example: UV light has a higher frequency and shorter wavelength than visible light, therefore UV photons have more energy than visible light photons.

COMMON MISTAKE: Forgetting the inverse relationship between wavelength and energy. Higher energy means SHORTER wavelength.

Examples

1. Calculate the energy of a photon of blue light with a frequency of \$6.5 \times 10^{14} \text{ Hz}$.

   $E = hf = (6.63 \times 10^{-34} \text{ Js})(6.5 \times 10^{14} \text{ Hz}) = 4.31 \times 10^{-19} \text{ J}$

2. Calculate the energy in electron-volts of a photon with a wavelength of \$500 \text{ nm}$.

   First, find the energy in Joules:
   $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34} \text{ Js})(3.0 \times 10^8 \text{ m/s})}{500 \times 10^{-9} \text{ m}} = 3.98 \times 10^{-19} \text{ J}$

   Then, convert to electron-volts:
   $E = \frac{3.98 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} = 2.48 \text{ eV}$

STUDY HINT: Practice converting between Joules and electron-volts to become comfortable with the different units.

Limitations of the Wave Model in Explaining Photon Energy

• The wave model of light could not explain the photoelectric effect.
• The wave model predicts that the energy of emitted electrons should depend on the intensity of the light, but experiments showed it depends on the frequency.
• The wave model could not explain the instantaneous emission of electrons in the photoelectric effect, regardless of intensity.
• Einstein’s explanation, using the photon model ($E=hf$), successfully explained these observations.

VCAA FOCUS: Be prepared to discuss the experimental evidence that supports the photon model and contradicts the wave model of light.