Photoelectric effect

1. Overview

The photoelectric effect is the process by which a metal surface emits electrons, known as photoelectrons, when it is illuminated by electromagnetic radiation of a sufficiently high frequency.

This phenomenon is a cornerstone of modern physics because it cannot be explained by classical wave theory. Classical physics suggests that the energy of a wave is determined by its intensity (amplitude), implying that any frequency of light should eventually cause electron emission if given enough time or intensity. However, experimental observations show that emission depends primarily on frequency.

The photoelectric effect provides definitive evidence for the particulate nature of light, demonstrating that electromagnetic radiation consists of discrete packets of energy called photons. This led to the development of the photon model, where the energy of each photon is directly proportional to its frequency ($E = hf$).

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Key Definitions

To secure full marks in "define" or "explain" questions, use the following precise terminology:

• Photon: A discrete packet (or quantum) of electromagnetic energy.
• Photoelectron: An electron emitted from the surface of a metal due to the absorption of energy from an incident photon.
• Work Function Energy ($\Phi$): The minimum energy required for an electron to escape from the surface of a metal.
• Threshold Frequency ($f_0$): The minimum frequency of incident electromagnetic radiation required to cause the emission of photoelectrons from a metal surface.
• Threshold Wavelength ($\lambda_0$): The maximum wavelength of incident electromagnetic radiation that will cause the emission of photoelectrons.
• Stopping Potential ($V_s$): The minimum negative potential applied to the collecting electrode (anode) that is required to reduce the photoelectric current to zero. It is a measure of the maximum kinetic energy of the emitted electrons ($eV_s = K_{max}$).

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Content

The Photon Model and One-to-One Interaction

In the photoelectric effect, the interaction between electromagnetic radiation and the metal is a one-to-one process.

1. A single incident photon interacts with a single conduction electron in the metal.
2. The electron absorbs the entire energy of the photon ($E = hf$).
3. If the absorbed energy is greater than the work function ($\Phi$), the electron can escape the metal surface.
4. If the absorbed energy is less than the work function, the electron gains some energy but remains trapped within the metal; it eventually loses this extra energy to the metal lattice as heat.

Einstein’s Photoelectric Equation

Einstein applied the Law of Conservation of Energy to this one-to-one interaction. The energy of the incident photon is split into two parts:

1. The energy required to liberate the electron from the metal surface (the work function, $\Phi$).
2. The remaining energy, which appears as the kinetic energy of the emitted electron.

The equation is expressed as: $$hf = \Phi + K_{max}$$

Where:

• $h$ is the Planck constant ($6.63 \times 10^{-34}$ J s).
• $f$ is the frequency of the incident radiation (Hz).
• $\Phi$ is the work function of the metal (J).
• $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons (J).

Since $K_{max} = \frac{1}{2}m_e v_{max}^2$, the equation can also be written as: $$hf = \Phi + \frac{1}{2}m_e v_{max}^2$$

Why is it "Maximum" Kinetic Energy?

The work function $\Phi$ is the minimum energy required to escape the surface.

• Electrons at the very surface of the metal require the least energy to escape. When they absorb a photon, they exit with the maximum possible kinetic energy ($K_{max}$).
• Electrons deeper within the metal lattice lose some of their kinetic energy through collisions with ions or other electrons as they make their way to the surface. Consequently, these electrons are emitted with kinetic energies less than $K_{max}$.
• Therefore, photoelectrons are emitted with a range of kinetic energies from zero up to $K_{max}$.

Threshold Frequency ($f_0$) and Wavelength ($\lambda_0$)

Emission only occurs if the photon energy is at least equal to the work function ($hf \geq \Phi$).

• At the threshold frequency ($f_0$), the photon has just enough energy to liberate the electron, but with zero kinetic energy remaining. $$\Phi = h f_0$$
• Using the wave equation $c = f \lambda$, we can derive the threshold wavelength ($\lambda_0$): $$\lambda_0 = \frac{hc}{\Phi}$$
• If $f < f_0$ (or $\lambda > \lambda_0$), no emission occurs, regardless of the intensity of the light.

The Effect of Intensity vs. Frequency

This is the most common area for conceptual exam questions. You must distinguish between the rate of emission and the energy of emission.

1. Intensity (at constant frequency):

   • Intensity is the power per unit area. In the photon model, increasing intensity means increasing the number of photons hitting the surface per unit area per second.
   • Since the interaction is 1-to-1, more photons result in more electrons being emitted per second.
   • Result: Photoelectric current is directly proportional to intensity.
   • Crucial Note: Intensity has no effect on the maximum kinetic energy of the photoelectrons.

2. Frequency (at constant intensity):

   • Increasing the frequency increases the energy of each individual photon ($E = hf$).
   • According to Einstein's equation, if $f$ increases and $\Phi$ is constant, $K_{max}$ must increase.
   • Result: $K_{max}$ is linearly dependent on frequency.
   • Crucial Note: Frequency has no effect on the photoelectric current (provided $f > f_0$), because the number of photons per second remains the same if intensity is constant (actually, at constant intensity, increasing frequency slightly decreases the number of photons, but for A-Level, we focus on the fact that $K_{max}$ is the variable affected by frequency).

Stopping Potential ($V_s$)

The maximum kinetic energy can be measured experimentally using a vacuum photocell. A negative potential is applied to the collecting anode to repel the incoming photoelectrons. When the potential is high enough that even the fastest electrons (those with $K_{max}$) are turned back, the current drops to zero. At this point: $$K_{max} = e V_s$$ Where $e$ is the elementary charge ($1.60 \times 10^{-19}$ C).

Substituting this into Einstein's equation: $$hf = \Phi + e V_s$$

Evidence Against Classical Wave Theory

The photoelectric effect provided three key observations that classical wave theory could not explain:

Observation	Classical Wave Theory Prediction	Photon Model Explanation
Threshold Frequency	Any frequency should cause emission if intensity is high enough.	One photon gives all its energy to one electron. If $hf < \Phi$, no emission is possible.
Instantaneous Emission	For low intensity, there should be a "time lag" while the electron builds up energy.	Energy is delivered in a single packet. If $hf > \Phi$, emission is immediate.
Max Kinetic Energy	$K_{max}$ should increase with intensity (larger wave amplitude).	$K_{max}$ depends only on the energy of a single photon ($hf$). Intensity only affects the number of electrons.

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Worked Example 1 — Calculating $K_{max}$ and Velocity

A zinc surface has a work function of $4.31$ eV. Ultraviolet light of wavelength $220$ nm is incident on the surface. Calculate the maximum speed of the emitted photoelectrons.

Step 1: Convert the work function to Joules $$\Phi = 4.31 \times 1.60 \times 10^{-19} = 6.896 \times 10^{-19} \text{ J}$$

Step 2: Calculate the energy of the incident photon $$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34}) \times (3.00 \times 10^8)}{220 \times 10^{-9}}$$ $$E = 9.041 \times 10^{-19} \text{ J}$$

Step 3: Use Einstein’s equation to find $K_{max}$ $$hf = \Phi + K_{max}$$ $$K_{max} = E - \Phi = 9.041 \times 10^{-19} - 6.896 \times 10^{-19}$$ $$K_{max} = 2.145 \times 10^{-19} \text{ J}$$

Step 4: Calculate the maximum velocity $$K_{max} = \frac{1}{2} m_e v_{max}^2$$ $$v_{max} = \sqrt{\frac{2 \times K_{max}}{m_e}} = \sqrt{\frac{2 \times 2.145 \times 10^{-19}}{9.11 \times 10^{-31}}}$$ $$v_{max} = 6.86 \times 10^5 \text{ m s}^{-1}$$

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Worked Example 2 — Threshold Calculations

The maximum wavelength of light that can cause photoelectric emission from a particular metal is $550$ nm. Determine the threshold frequency and the work function of the metal in eV.

Step 1: Calculate threshold frequency ($f_0$) $$c = f_0 \lambda_0 \implies f_0 = \frac{c}{\lambda_0}$$ $$f_0 = \frac{3.00 \times 10^8}{550 \times 10^{-9}} = 5.45 \times 10^{14} \text{ Hz}$$

Step 2: Calculate work function ($\Phi$) in Joules $$\Phi = h f_0 = (6.63 \times 10^{-34}) \times (5.45 \times 10^{14})$$ $$\Phi = 3.613 \times 10^{-19} \text{ J}$$

Step 3: Convert to eV $$\Phi_{eV} = \frac{3.613 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.26 \text{ eV}$$

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Key Equations

Equation	Symbols	SI Units	Status
$E = hf$	$E$: photon energy, $f$: frequency	J, Hz	Data Sheet
$E = \frac{hc}{\lambda}$	$\lambda$: wavelength	J, m	Derived
$\Phi = h f_0$	$\Phi$: work function, $f_0$: threshold freq.	J, Hz	Memorise
$hf = \Phi + K_{max}$	$K_{max}$: max kinetic energy	J	Memorise
$K_{max} = eV_s$	$V_s$: stopping potential, $e$: electron charge	J, V	Memorise
$K_{max} = \frac{1}{2}m_e v_{max}^2$	$m_e$: mass of electron ($9.11 \times 10^{-31}$ kg)	J, kg, m/s	Data Sheet

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Common Mistakes to Avoid

• ❌ Wrong: Using the mass of the metal atom or a proton in the kinetic energy formula.
  • ✓ Right: Always use the mass of an electron ($m_e = 9.11 \times 10^{-31}$ kg) because it is the electron that is being emitted.
• ❌ Wrong: Adding the work function to the photon energy ($hf + \Phi = K_{max}$).
  • ✓ Right: Remember the energy balance: Photon Energy = Energy to get out + Energy left over. Thus, $hf = \Phi + K_{max}$.
• ❌ Wrong: Forgetting to convert eV to Joules before using Einstein's equation.
  • ✓ Right: All terms in $hf = \Phi + K_{max}$ must be in Joules. Convert eV to J by multiplying by $1.60 \times 10^{-19}$.
• ❌ Wrong: Thinking that "threshold wavelength" is the minimum wavelength.
  • ✓ Right: Threshold wavelength is the maximum wavelength. Since $E \propto 1/\lambda$, a longer wavelength means lower energy. If the wavelength is too long, the photon won't have enough energy to overcome the work function.
• ❌ Wrong: Stating that intensity increases the speed of photoelectrons.
  • ✓ Right: Intensity increases the number of photoelectrons (current). Only frequency (or wavelength) affects the speed/kinetic energy.

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Exam Tips

1. Graph Analysis: $K_{max}$ vs. Frequency

If you are given a graph of $K_{max}$ (y-axis) against frequency $f$ (x-axis):

• The equation is $K_{max} = hf - \Phi$. This follows the form $y = mx + c$.
• Gradient: The gradient of the line is always the Planck constant ($h$).
• x-intercept: The point where $K_{max} = 0$ is the threshold frequency ($f_0$).
• y-intercept: If the line is extrapolated back to the y-axis, the intercept is negative work function ($-\Phi$).
• Different Metals: If the experiment is repeated with a different metal, the line will be parallel (same gradient $h$) but will have different intercepts.

2. Graph Analysis: Photoelectric Current vs. Voltage

• Saturation Current: The flat part of the graph where all emitted electrons reach the anode. The height of this line is proportional to the intensity of the light.
• Stopping Potential ($V_s$): The x-intercept on the negative side. This value depends only on the frequency of the light and the metal's work function.
• If intensity increases (same frequency): The saturation current increases, but $V_s$ stays the same.
• If frequency increases (same intensity): $V_s$ becomes more negative (moves left), but the saturation current stays roughly the same.

3. Structuring "Evidence" Questions

When asked why the photoelectric effect supports the particle theory of light, structure your answer with these three points:

1. Threshold Frequency: Mention that no electrons are emitted below $f_0$ regardless of intensity, which contradicts wave theory.
2. Instantaneous Emission: Mention that emission occurs without a time delay, even at very low intensities.
3. Max Kinetic Energy: State that $K_{max}$ depends on frequency, not intensity.
4. Conclusion: These observations suggest that light energy is delivered in discrete packets (photons).