Why doesn’t more intense light overcome the threshold?

Intensity increases the number of photons, not the energy of each. An electron absorbs one photon at a time, so if one photon cannot free it, more photons do not help.

Is light a wave or a particle?

Both models are needed. Interference and diffraction show wave behaviour; the photoelectric effect shows particle behaviour. Which appears depends on the experiment.

Why is it the maximum kinetic energy?

Electrons deeper in the metal lose some energy in collisions before escaping. Only those at the surface leave with the full hf -.

What does the stopping potential measure?

The maximum kinetic energy in electronvolts: the reverse voltage that just halts the fastest emitted electrons.

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90–95WavesStretch

The Photoelectric Effect

Shine ultraviolet light on a clean zinc plate and electrons are ejected. Shine bright red light for as long as you like and nothing happens. That observation — that the $\textit{colour}$ matters and the brightness does not — cannot be explained if light is purely a wave, and it forced physics to accept that light also comes in $\textbf{photons}$ .

40 min Topic 5 · Waves and the Particle Nature of Light

What you'll be able to do

Use $E = hf$ to find photon energy and convert between joules and electronvolts
Describe the observations of the photoelectric effect
Explain why the wave model fails and how the photon model succeeds
Apply Einstein’s equation $hf = \phi + \tfrac12 mv_{\max}^{2}$

Photons

In the photon model, light is emitted and absorbed in discrete packets of energy called $\textbf{photons}$ . Each photon’s energy is proportional to the frequency: $E = hf$ , where $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ is the Planck constant.

Combining with $c = f\lambda$ gives $E = hc/\lambda$ — shorter wavelengths carry more energy per photon, which is why ultraviolet can damage skin and visible light cannot.

Photon energies are tiny in joules, so the $\textbf{electronvolt}$ is used: $1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}$ , the energy gained by an electron accelerated through 1 V.

E = hf = \dfrac{hc}{\lambda}

Photon energy.

1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}

Electronvolt.

Example 1Find the energy of a photon of wavelength 450 nm, in joules and electronvolts.

E = \dfrac{(6.63\times10^{-34})(3.00\times10^{8})}{450\times10^{-9}} = 4.42\times10^{-19}

2In eV:

\dfrac{4.42\times10^{-19}}{1.60\times10^{-19}}

Answer

4.42\times10^{-19}\ \mathrm{J} = 2.76\ \mathrm{eV}

Tip — To convert joules to eV, divide by $1.60\times10^{-19}$ ; to go back, multiply. Mixing the two units in one equation is a common error.

What is observed

For a given metal there is a $\textbf{threshold frequency}$ $f_{0}$ . Below it, no electrons are emitted however intense the light. Above it, emission happens.

Emission is $\textbf{instantaneous}$ — there is no delay even for very dim light.

The $\textbf{maximum kinetic energy}$ of emitted electrons depends on the frequency, not the intensity. Increasing intensity increases only the $\textit{number}$ of electrons emitted per second.

Why waves fail and photons succeed

The wave model says energy arrives continuously and spread across the surface. So brighter light, of any frequency, should eventually give electrons enough energy to escape — there should be no threshold. And dim light should need time to build up enough energy — there should be a delay. Neither is seen.

The photon model explains all three. An electron absorbs $\textbf{one photon}$ , all of its energy at once. If that single photon’s energy $hf$ is less than the energy needed to escape, the electron cannot leave — no matter how many photons arrive, since it cannot combine them. That is the threshold.

Absorption of one photon is immediate, so there is no delay. Higher intensity means more photons per second, so more electrons — but each photon still has energy $hf$ , so the maximum kinetic energy is unchanged.

Tip — The phrase that earns the mark is "one photon is absorbed by one electron". It is the single idea that explains the threshold frequency.

Einstein’s photoelectric equation

The $\textbf{work function}$ $\phi$ is the minimum energy needed to free an electron from the metal surface. Energy conservation for one photon then gives $hf = \phi + E_{k,\max}$ : the photon’s energy pays the escape cost, and any excess becomes kinetic energy.

At threshold, $E_{k,\max} = 0$ , so $hf_{0} = \phi$ .

The maximum kinetic energy is measured using a $\textbf{stopping potential}$ $V_{s}$ — the reverse voltage just large enough to stop the fastest electrons: $eV_{s} = E_{k,\max}$ . A graph of $E_{k,\max}$ against $f$ is a straight line with gradient $h$ and intercept $-\phi$ .

hf = \phi + \tfrac{1}{2}mv_{\max}^{2}

Einstein’s equation.

\phi = hf_{0}

Work function and threshold.

eV_{s} = E_{k,\max}

Stopping potential.

Example 1Light of frequency

1.2\times10^{15}

Hz falls on a metal with work function 2.3 eV. Find the maximum kinetic energy of the electrons in eV, and the stopping potential.

hf = 6.63\times10^{-34} \times 1.2\times10^{15} = 7.96\times10^{-19}

= 4.97

eV.

E_{k,\max} = 4.97 - 2.3 = 2.67

eV.

3Stopping potential in volts equals the energy in eV.

Answer

E_{k,\max} = 2.7\ \mathrm{eV}

;

V_{s} = 2.7\ \mathrm{V}

Equation recap

E = hf = hc/\lambda

Photon energy.

hf = \phi + E_{k,\max}

Einstein’s equation.

\phi = hf_{0}

Threshold frequency.

eV_{s} = E_{k,\max}

Stopping potential.

Common mistakes to avoid

Saying brighter light gives faster electrons.

Brighter light gives more electrons; maximum speed depends on frequency.

Mixing eV and J in Einstein’s equation.

Convert to one unit before subtracting.

Explaining the threshold by saying low-frequency light is too weak.

Each photon has too little energy, and an electron absorbs only one photon.

Calling

E_{k,\max}

the kinetic energy of every electron.

It is the maximum; electrons below the surface lose some energy escaping.

Key takeaways

Light delivers energy in photons with $E = hf = hc/\lambda$ .
There is a threshold frequency, emission is instant, and intensity only changes electron number.
The wave model cannot explain these; one-photon-one-electron absorption can.
Einstein: $hf = \phi + E_{k,\max}$ , with $\phi = hf_{0}$ .
The stopping potential gives $E_{k,\max} = eV_{s}$ ; the $E_{k}$ – $f$ graph has gradient $h$ .

Test yourself

Answer a few questions, then review anything you missed.

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