Particle-Like Properties of Light Revision Notes for VCE SSCE Physics

Particle-Like Properties of Light

Introduction to quantum physics

Modern technology depends heavily on quantum physics. Devices such as silicon chips, lasers, optical fibres, GPS and MRI scanners all rely on applications of quantum theory.

In the late 19th century, physicists believed they had solved most of the major problems in physics. However, a few puzzling phenomena remained unexplained:

The blackbody problem: How hot objects emit electromagnetic radiation
The photoelectric effect: How light can cause electrons to be ejected from metal surfaces

Classical physics could not explain these observations. Max Planck solved the blackbody problem by proposing that energy comes in discrete quantities called quanta. Albert Einstein then applied this idea to explain the photoelectric effect, marking a crucial step in the development of quantum physics.

The photoelectric effect

What is the photoelectric effect?

The photoelectric effect occurs when light shining onto a metal surface causes electrons to be ejected from that surface. This phenomenon was first observed by Heinrich Hertz in 1887, and Philipp Lenard identified in 1902 that the emitted particles were electrons.

Demonstration with simple apparatus

The photoelectric effect can be demonstrated using basic equipment.

In this demonstration:

A clean zinc plate is placed on top of a negatively charged electroscope (a device that detects electric charge)
When ultraviolet (UV) light shines on the zinc plate, electrons are ejected
The loss of negative charge causes the gold leaf to collapse back towards the stand
This rapid discharge indicates electrons are escaping from the metal surface

infoNote

The electroscope provides a simple visual demonstration of the photoelectric effect. The collapse of the gold leaf directly shows the loss of electrons from the zinc surface when UV light strikes it.

A more sophisticated setup

This apparatus uses potassium as the metal surface. When light strikes the potassium:

Electrons are emitted from the metal
They travel across the evacuated tube to the positive electrode
This completes the circuit, and current flows through the ammeter

Key experimental observations

Early experiments revealed several important features:

chatImportant

Threshold frequency exists: Below a certain frequency of light, no electrons are emitted, regardless of light intensity
Different metals have different thresholds: For example, zinc requires UV light, while potassium emits electrons with visible light
Frequency matters more than intensity: The threshold frequency varies between metals

Measuring photoelectron energy

Experimental apparatus

To measure the energy of emitted electrons (called photoelectrons), more complex equipment is needed.

This apparatus includes:

Light source: A hot filament producing white light
Filter: Allows nearly monochromatic (single-frequency) light to pass
Photocell: An evacuated glass tube protecting the metal plate from oxidation
Metal plate: Surface from which photoelectrons are emitted
Collector electrode: Captures the emitted photoelectrons
Variable DC voltage source: Creates a retarding potential to oppose electron movement
Ammeter: Measures current
Voltmeter: Measures potential difference

infoNote

The photocell is evacuated (contains no air) to allow free movement of photoelectrons and to prevent the metal surface from oxidising. This ensures accurate measurements and maintains the metal's properties throughout the experiment.

Energy units: the electronvolt

Photoelectron energies are conveniently measured in electronvolts (eV):

$1 \text{ eV}$ is the energy an electron gains or loses when moving through a potential difference of $1$ volt
In joules: $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$
Common multiples: $1 \text{ keV} = 10^3 \text{ eV}$ and $1 \text{ MeV} = 10^6 \text{ eV}$
Conversion: $1 \text{ J} = 6.25 \times 10^{18} \text{ eV}$

infoNote

The electronvolt is particularly useful in quantum physics because it provides convenient numbers when dealing with atomic and subatomic energies. Working in joules would require constantly managing very small powers of ten.

Understanding stopping potential

The DC voltage source is connected to oppose the motion of electrons (retarding potential). By gradually increasing this voltage:

Less energetic electrons are turned back
The ammeter current decreases
Eventually the current reaches zero

At this point, the voltage reading gives the maximum kinetic energy of the photoelectrons in eV. A reading of $1.2$ V means the maximum electron kinetic energy is $1.2$ eV.

The stopping potential ( $V_0$ ) is the minimum potential difference needed to just stop the most energetic photoelectrons.

Photocurrent versus potential difference

This graph shows typical behaviour:

At the stopping potential ( $-V_0$ ), current just reaches zero
The stopping potential gives the maximum kinetic energy of photoelectrons in eV
When the potential is attractive (positive), the graph levels off as all photoelectrons are collected
The flat region indicates saturation - all emitted electrons are being captured

Effect of light intensity

Surprising results

When investigating how light intensity affects photoelectron energy, researchers made surprising discoveries:

At constant frequency:

More intense light produces more photoelectrons
However, their kinetic energy does not increase
The stopping potential remains the same
There is no time delay in photoelectron emission, even with very dim light

chatImportant

This graph demonstrates that changing intensity (at fixed frequency) affects the number of photoelectrons emitted (photocurrent) but not their maximum energy (same stopping potential).

These results contradicted the wave model of light, which predicted that more intense light should transfer more energy to electrons.

Effect of light frequency

Millikan's experiments

Robert Millikan conducted careful measurements showing that frequency directly affects photoelectron kinetic energy.

His results for potassium showed:

A linear relationship between frequency and stopping potential
Below $43.9 \times 10^{13}$ Hz, no electrons are ejected
Higher frequencies produce photoelectrons with greater kinetic energy

Comparing different metals

This graph compares four different metals:

All show the same gradient (slope)
Each metal has a different threshold frequency (x-intercept)
Different metals require different minimum energies to release electrons

infoNote

Photoelectrons from Metal D at frequency $6.4 \times 10^{14}$ Hz have much less kinetic energy than those from other metals at the same frequency. This indicates that some metals hold their electrons more tightly than others - a property we'll see is related to the work function.

Einstein's explanation

The photon model

In 1905, Einstein proposed a revolutionary explanation using Planck's idea of energy quantisation. He suggested that light consists of discrete packets of energy called quanta (later renamed photons).

Einstein received the Nobel Prize in 1921 for this work, which was initially controversial as it contradicted the well-established wave model of light.

Energy of a photon

Einstein proposed that each photon carries energy given by:

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

Where:

$E$ = energy of the photon (J or eV)
$h$ = Planck's constant = $6.63 \times 10^{-34}$ J s (or $4.14 \times 10^{-15}$ eV s)
$f$ = frequency of the light (Hz)
$c$ = speed of light = $3.0 \times 10^8 \text{ m} \cdot \text{s}^{-1}$
$\lambda$ = wavelength of the photon (m)

This formula shows that:

Higher frequency light carries more energy per photon
Energy is directly proportional to frequency
Energy is inversely proportional to wavelength

The photoelectric effect equation

Einstein proposed that when a photon strikes a metal surface:

An electron absorbs the entire photon energy
Some energy is used to escape the metal (the work function, $\phi$ )
The remaining energy becomes the electron's kinetic energy

This is expressed mathematically as:

$E_{k \text{ max}} = hf - \phi$

Where:

$E_{k \text{ max}}$ = maximum kinetic energy of emitted photoelectrons (J or eV)
$h$ = Planck's constant
$f$ = frequency of incident light (Hz)
$\phi$ = work function of the metal (J or eV)

Work function

The work function ( $\phi$ ) is:

The minimum energy required to release a photoelectron from a metal surface
A property of the metal, independent of the light source
Different for each type of metal

Threshold frequency

From Einstein's equation, we can find the threshold frequency - the minimum frequency needed to emit photoelectrons:

At threshold: $E_{k \text{ max}} = 0$

Therefore: $hf_{\text{threshold}} = \phi$

$f_{\text{threshold}} = \frac{\phi}{h}$

chatImportant

Below this frequency, photons don't have enough energy to overcome the work function, so no photoelectrons are emitted regardless of the light intensity.

Graphical representation

Einstein's equation can be graphed as a straight line for any metal:

Gradient: equals Planck's constant, $h$ (in J s or eV s)
x-intercept: gives the threshold frequency (Hz)
y-intercept: gives the negative work function, $-\phi$ (J or eV)

infoNote

Different metals produce parallel lines (same gradient) with different intercepts, reflecting their different work functions and threshold frequencies. The fact that all metals show the same gradient provides strong evidence for the universality of Planck's constant.

Worked example: photoelectric calculations

Consider a photoelectric experiment using a caesium metal plate.

lightbulbExample

Worked Example: Photoelectric Calculations with Caesium

Part a: Green light with frequency $5.45 \times 10^{14}$ Hz passes through the filter. Calculate the photon energy in electronvolts.

Solution: Using $E = hf$ with $h = 4.14 \times 10^{-15}$ eV s:

$E = 4.14 \times 10^{-15} \times 5.45 \times 10^{14} = 2.26 \text{ eV}$

Part b: The work function of caesium is $2.10$ eV. Calculate the maximum kinetic energy of photoelectrons in joules.

Solution: Using $E_{k \text{ max}} = hf - \phi$ :

$E_{k \text{ max}} = 2.26 - 2.10 = 0.16 \text{ eV}$

Converting to joules:

$E_{k \text{ max}} = 0.16 \times 1.60 \times 10^{-19} = 2.56 \times 10^{-20} \text{ J}$

Part c: The filter is changed to allow violet photons with energy $2.80$ eV. Calculate the maximum speed of emitted photoelectrons. (Electron mass = $9.1 \times 10^{-31}$ kg)

Solution: First find maximum kinetic energy:

$E_{k \text{ max}} = 2.80 - 2.10 = 0.70 \text{ eV}$

Converting to joules:

$E_{k \text{ max}} = 0.70 \times 1.60 \times 10^{-19} = 1.12 \times 10^{-19} \text{ J}$

Using $E_{k \text{ max}} = \frac{1}{2}mv_{\text{max}}^2$ , rearrange for speed:

$v_{\text{max}} = \sqrt{\frac{2E_{k \text{ max}}}{m}} = \sqrt{\frac{2 \times 1.12 \times 10^{-19}}{9.1 \times 10^{-31}}} = 4.96 \times 10^5 \text{ m} \cdot \text{s}^{-1}$

Limitations of the wave model

Why classical wave theory fails

The photoelectric effect revealed several experimental observations that classical wave theory could not explain:

1. No time delay with dim light

Wave prediction: Dim light has low energy wavefronts, so electrons should need time to accumulate enough energy to escape.

Observation: Photoelectrons are emitted instantly, even with extremely faint light.

Photon explanation: Each electron receives energy from a single photon in one interaction - no accumulation time needed.

2. Existence of threshold frequency

Wave prediction: Sufficiently intense light of any frequency should eventually provide enough energy to liberate electrons.

Observation: Below the threshold frequency, no photoelectrons are emitted, regardless of light intensity.

Photon explanation: Each photon must have energy $hf \geq \phi$ . Low-frequency photons never have enough individual energy.

3. Intensity affects number, not energy

Wave prediction: More intense light should transfer more energy to electrons, increasing their kinetic energy.

Observation: More intense light produces more photoelectrons, but their maximum kinetic energy remains unchanged.

Photon explanation: Intensity depends on the number of photons. More photons mean more interactions, but each interaction transfers the same energy $hf$ .

4. Photoelectron energy depends on frequency

Wave prediction: Wave energy depends on both amplitude and frequency, but the clear linear relationship with frequency alone is difficult to explain.

Observation: Maximum photoelectron kinetic energy increases linearly with frequency above threshold.

Photon explanation: Each photon's energy is $E = hf$ , so higher frequency directly means more energy per photon, giving electrons more kinetic energy after escaping.

chatImportant

Why the wave model fails:

The wave model predicts that energy is distributed continuously across the wavefront, so all of these observations contradict fundamental wave principles. The photon model successfully explains all experimental observations by treating light as discrete packets of energy.

Understanding intensity in the photon model

Intensity is the product of:

The number of photons per second
The energy of each photon ( $hf$ )

Important consequence: If intensity is kept constant while frequency changes:

Increasing frequency → fewer photons (to maintain constant total energy)
Decreasing frequency → more photons (to maintain constant total energy)

infoNote

This relationship explains why changing frequency affects photoelectron energy while changing intensity (at constant frequency) only affects the number of photoelectrons emitted.

bookmarkSummary

Key Points to Remember:

The photoelectric effect occurs when light causes electrons to be ejected from metal surfaces, providing key evidence for the particle nature of light
Threshold frequency is the minimum frequency required to emit photoelectrons; it varies between different metals
Photon energy is given by $E = hf = \frac{hc}{\lambda}$ , showing that energy increases with frequency
Einstein's photoelectric equation is $E_{k \text{ max}} = hf - \phi$ , where $\phi$ is the work function
The work function is the minimum energy needed to release an electron from a specific metal surface
Stopping potential (in volts) numerically equals the maximum photoelectron kinetic energy (in eV)
Increasing light intensity at fixed frequency produces more photoelectrons but does not increase their kinetic energy
The wave model of light cannot explain the threshold frequency, instantaneous emission, or why photoelectron energy depends on frequency rather than intensity

Particle-Like Properties of Light (VCE SSCE Physics): Revision Notes