Photon Energy and the Photoelectric Effect

Light is not only a wave — it also behaves as a stream of discrete energy packets called photons. The energy of each photon depends on its frequency (or equivalently, its wavelength).

Photon Energy

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

Where:

• $h = 6.626 \times 10^{-34}\,\text{J·s}$ is Planck's constant
• $f$ is the frequency in Hz
• $c = 3 \times 10^8\,\text{m/s}$ is the speed of light
• $\lambda$ is the wavelength in meters

In electron-volts ($1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J}$):

$E_{\text{eV}} = \frac{hc}{\lambda \cdot e}$

Visible light energies:

The Photoelectric Effect

Einstein's 1905 explanation of the photoelectric effect earned him the Nobel Prize: when light shines on a metal, electrons are ejected only if the photon energy exceeds the metal's work function $\phi$.

The maximum kinetic energy of ejected electrons is:

$KE_{\max} = hf - \phi = \frac{hc}{\lambda} - \phi$

Key observations:

• No electrons are emitted below a threshold frequency, regardless of intensity.
• Above threshold, $KE_{\max}$ depends only on frequency, not intensity.
• Intensity determines the number of ejected electrons, not their energy.

Example: UV light at $\lambda = 200\,\text{nm}$ on sodium ($\phi = 2.3\,\text{eV}$):

$E = \frac{hc}{\lambda} \approx 6.2\,\text{eV}$ $KE_{\max} = 6.2 - 2.3 = 3.9\,\text{eV}$

Your Task

Implement the two functions. Use $h = 6.626 \times 10^{-34}\,\text{J·s}$, $c = 3 \times 10^8\,\text{m/s}$, $e = 1.602 \times 10^{-19}\,\text{J/eV}$.