15.5 The Photoelectric Effect

Photoelectric Effect Interactions

Electron Emission Fundamentals

The photoelectric effect occurs when electromagnetic radiation strikes a photoactive material, causing electrons to be emitted from the material. This phenomenon provides compelling evidence for the particle-like nature of light, as the energy of incident photons is directly transferred to electrons in the material.

• Photoactive materials include certain metals like sodium and potassium, which have relatively low work functions
• Semiconductors such as silicon and germanium also exhibit the photoelectric effect
• The effect forms the basis for many modern technologies including solar cells, light meters, and night vision devices

Threshold Frequency

For electron emission to occur, the incident light must have a minimum frequency known as the threshold frequency. This critical concept helps explain why light behaves as discrete packets of energy rather than continuous waves.

• When light frequency equals or exceeds the threshold frequency, electrons will be emitted regardless of light intensity
• If light frequency is below the threshold, no electrons will be emitted even if the light is extremely intense
• This observation contradicted classical wave theory, which predicted that sufficient light intensity of any frequency should eventually cause electron emission

For example, if a metal has a threshold frequency of $5 \times 10^{14} \text{ Hz}$, light with a frequency of $6 \times 10^{14} \text{ Hz}$ will cause electron emission, while light with a frequency of $4 \times 10^{14} \text{ Hz}$ will not produce any photoelectrons, no matter how bright the light source.

Maximum Kinetic Energy and Frequency

The maximum kinetic energy of emitted electrons depends on both the frequency of the incident light and the material's work function. This relationship reveals important insights about energy conservation in quantum interactions.

• The work function ($\phi$) represents the minimum energy needed to remove an electron from a material
• The relationship between maximum kinetic energy, light frequency, and work function is given by: $K_{max} = hf - \phi$ where $h$ is Planck's constant ($6.63 \times 10^{-34} \text{ J\cdot s}$) and $f$ is the light frequency

In a typical photoelectric effect experiment:

1. Two metal plates are placed in a vacuum chamber and connected to a variable voltage source
2. One plate is illuminated with monochromatic light, causing electrons to be ejected
3. The potential difference between plates is adjusted until the current drops to zero
4. This stopping potential directly relates to the maximum kinetic energy of the emitted electrons

Work Function of Materials

The work function is a characteristic property that represents the minimum energy required to remove an electron from a material's surface. It varies significantly between different materials and affects how they respond to light.

• Materials with lower work functions emit electrons more easily when exposed to light
• The work function determines the threshold frequency through the relationship: $f_{threshold} = \phi/h$
• Some approximate work function values:
  • Sodium: $2.3 \text{ eV}$
  • Potassium: $2.3 \text{ eV}$
  • Copper: $4.7 \text{ eV}$
  • Zinc: $4.3 \text{ eV}$

Practice Problem 1: Threshold Frequency

Solution

To find the threshold frequency, we need to use the relationship between work function and threshold frequency:

$f_{threshold} = \phi/h$

First, let's convert the work function from eV to joules: $\phi = 2.0 \text{ eV} \times 1.60 \times 10^{-19} \text{ J/eV} = 3.20 \times 10^{-19} \text{ J}$

Now we can calculate the threshold frequency: $f_{threshold} = \frac{3.20 \times 10^{-19} \text{ J}}{6.63 \times 10^{-34} \text{ J·s}} = 4.83 \times 10^{14} \text{ Hz}$

Therefore, light with a frequency below 4.83 × 10^14 Hz will not cause electron emission from this metal.

Practice Problem 2: Maximum Kinetic Energy

Solution

We can use the photoelectric effect equation to find the maximum kinetic energy: $K_{max} = hf - \phi$

First, let's calculate the energy of the incident photons: $hf = 6.63 \times 10^{-34} \text{ J\cdot s} \times 7.0 \times 10^{14} \text{ Hz} = 4.64 \times 10^{-19} \text{ J}$

Converting to eV: $hf = 4.64 \times 10^{-19} \text{ J} \div 1.60 \times 10^{-19} \text{ J/eV} = 2.9 \text{ eV}$

Now we can calculate the maximum kinetic energy: $K_{max} = 2.9 \text{ eV} - 1.8 \text{ eV} = 1.1 \text{ eV}$

Therefore, the maximum kinetic energy of the emitted electrons is 1.1 eV.