Wave-Particle Duality

• Planck's Relation: The energy of a photon is directly proportional to its frequency, expressed as $E = hf$. This demonstrates that energy is quantized rather than continuous.

• De Broglie Hypothesis: Any particle with momentum $p$ has an associated wavelength $\lambda$ given by $\lambda = \frac{h}{p}$. This relates a wave property (wavelength) to a particle property (momentum).

• Quantization: The principle that certain physical properties, like energy or angular momentum, can only take on discrete values (quanta) rather than any value in a continuous range.

Calculating Photon Energy

• To find the energy of a photon from its wavelength, use the combined wave-energy equation: $E = \frac{hc}{\lambda}$
• Here, $h$ is Planck's constant ($6.63 \times 10^{-34}$ Js), $c$ is the speed of light ($3.00 \times 10^8$ m/s), and $\lambda$ is the wavelength in meters.

Determining Matter Wavelength

• For a particle of mass $m$ moving at velocity $v$, the de Broglie wavelength is calculated as: $\lambda = \frac{h}{mv}$
• This calculation is vital for determining the resolution of electron microscopes, where smaller wavelengths allow for higher precision imaging.

• Unit Consistency: Always convert energy from electronvolts (eV) to Joules (J) before using Planck's constant in calculations ($1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$).

• Scale Awareness: When calculating the de Broglie wavelength for macroscopic objects (like a ball), the resulting $\lambda$ will be extremely small (e.g., $10^{-35}$ m). In exams, explain that this is why wave effects are not observable in daily life.

• Proportionality: Remember that $E \propto f$ and $E \propto \frac{1}{\lambda}$. If the wavelength is halved, the energy of the photon doubles.

• Simultaneity Error: A common misconception is that an entity acts as a wave and a particle at the exact same moment in the same observation. In reality, the experimental setup determines which nature is manifested.

• Massless Photons: Students often try to calculate a photon's momentum using $p = mv$. Since photons are massless, you must use $p = \frac{E}{c}$ or $p = \frac{h}{\lambda}$ instead.

• Intensity vs. Energy: Increasing light intensity increases the quantity of photons (and thus the number of emitted electrons) but does not increase the energy of individual photons or the kinetic energy of emitted electrons.