Photons, photoelectric effect and line spectra

What you'll learn

How the photon model links light frequency to energy using $E = h f$ .
Why the photoelectric effect is strong evidence that light can behave like particles.
How to use $hf=ϕ+Ek,max⁡hf=\phi+E_{k,\max}$ and stopping potential graphs.
Why atoms produce line spectra rather than a continuous spread of colours.

Before we start: waves, energy and charge

You already know that electromagnetic waves include radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. In a vacuum, they all travel at the same speed:

\times 10^8\ \text{m s}^{-1}

For any wave,

v=fλv=f\lambda

so for light in a vacuum,

c=fλc=f\lambda

where $f$ is frequency in hertz, and $λ\lambda$ is wavelength in metres. Higher frequency means shorter wavelength.

Definition

Electronvolt

An electronvolt, eV, is a unit of energy. One electronvolt is the energy transferred when an electron moves through a potential difference of 1 volt: $J1\ \text{eV}=1.60\times10^{-19}\ \text{J}$ .

The electronvolt is useful in atomic physics because photon energies and electron energy-level changes are often very small in joules.

Photons: light as packets of energy

The photon model says that electromagnetic radiation is emitted, absorbed and transferred in discrete packets called photons.

Definition

Photon

A photon is a discrete packet, or quantum, of electromagnetic radiation. The energy of one photon is $E = h f$ , where $h$ is the Planck constant.

The Planck constant is:

\times 10^{-34}\ \text{J s}

A photon’s energy depends on its frequency:

E = h f

Using $c=fλc=f\lambda$ , you can also write:

E=hcλE=\frac{hc}{\lambda}

So high-frequency ultraviolet photons have more energy per photon than visible light photons. Increasing the intensity of light at a fixed frequency means more photons arrive each second, not that each photon has more energy.

Key Idea

Frequency decides photon energy

For one photon, energy depends on frequency only: $E = h f$ . Intensity affects the number of photons arriving per second.

Example

Finding the energy of an ultraviolet photon

An ultraviolet photon has wavelength 250 nm. Calculate its energy in joules and electronvolts.

Use $E=hcλE=\frac{hc}{\lambda}$ because the wavelength is given, and convert the wavelength: $m250\ \text{nm}=250\times10^{-9}\ \text{m}$ .
Substitute the constants:
$E=(6.63×10−34)(3.00×108)250×10−9E=\frac{(6.63\times10^{-34})(3.00\times10^8)}{250\times10^{-9}}$ $JE=7.96\times10^{-19}\ \text{J}$
Convert to electronvolts by dividing by $eV−11.60\times10^{-19}\ \text{J eV}^{-1}$ :
$eVE=\frac{7.96\times10^{-19}}{1.60\times10^{-19}}=4.98\ \text{eV}$

The photoelectric effect

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency is incident on the metal.

Labelled photoelectric effect apparatus showing photons striking a metal cathode, emitted electrons, photocurrent and stopping potential graph

Definition

Photoelectron

A photoelectron is an electron emitted from a material by the photoelectric effect.

The key observations are:

Below a certain frequency, no electrons are emitted, no matter how intense the light is.
Above that frequency, emission starts almost immediately.
Increasing intensity increases the number of emitted electrons per second.
Increasing frequency increases the maximum kinetic energy of the emitted electrons.

The minimum frequency needed to release electrons from a particular metal is called the threshold frequency, $f_0$ .

Definition

Threshold frequency

The threshold frequency, $f_0$ , is the minimum frequency of incident radiation needed to emit photoelectrons from a metal surface.

These observations are difficult to explain using a simple wave model of light, because a wave model suggests energy could build up gradually. The photon model explains them neatly: one photon transfers its energy to one electron.

Key Idea

One photon, one electron

For photoemission, a single electron absorbs energy from a single photon. If the photon energy is too small, increasing intensity cannot make that individual photon release an electron.

Example

Explaining threshold frequency and intensity

A metal is illuminated with bright red light and no photoelectrons are emitted. The intensity of the red light is then doubled.

Compare photon energy with the energy needed to release an electron. Red light has a relatively low frequency, so each photon has energy $E = h f$ .
If the red light frequency is below the threshold frequency, then $hf<ϕhf<\phi$ , where $ϕ\phi$ is the work function. Each photon has insufficient energy to remove one electron.
Doubling intensity doubles the number of photons arriving per second, but it does not change $f$ or $E = h f$ . So no photoelectrons are emitted.

Work function and the photoelectric equation

The work function is the minimum energy needed to remove an electron from the surface of a metal.

Definition

Work function

The work function, $ϕ\phi$ , is the minimum energy required for an electron to escape from the surface of a material.

If a photon gives energy $h f$ to an electron, some energy is used to overcome the work function. Any remaining energy becomes kinetic energy of the emitted electron.

hf=ϕ+Ek,max⁡hf=\phi+E_{k,\max}

Rearranging:

Ek,max⁡=hf−ϕE_{k,\max}=hf-\phi

At the threshold frequency, emitted electrons have zero maximum kinetic energy, so:

ϕ=hf0\phi=hf_0

and therefore:

E_{k,\max}=h(f-f_0)

Common Mistake

Using intensity in the photoelectric equation

Intensity does not appear in $hf=ϕ+Ek,max⁡hf=\phi+E_{k,\max}$ . Increasing intensity increases the emission rate, but it does not increase the maximum kinetic energy of the photoelectrons.

Example

Calculating maximum kinetic energy and speed

Light of frequency $Hz8.00\times10^{14}\ \text{Hz}$ is incident on a metal with work function 2.30 eV. Calculate the maximum speed of the emitted photoelectrons. Use electron mass $kgm_e=9.11\times10^{-31}\ \text{kg}$ .

Convert the work function into joules:
$J\phi=2.30\times(1.60\times10^{-19})=3.68\times10^{-19}\ \text{J}$
Calculate the photon energy:
$E=hf=(6.63×10−34)(8.00×1014)E=hf=(6.63\times10^{-34})(8.00\times10^{14})$ $JE=5.30\times10^{-19}\ \text{J}$
Find the maximum kinetic energy:
$Ek,max⁡=hf−ϕ=5.30×10−19−3.68×10−19E_{k,\max}=hf-\phi=5.30\times10^{-19}-3.68\times10^{-19}$ $JE_{k,\max}=1.62\times10^{-19}\ \text{J}$
Use $Ek=12mv2E_k=\frac{1}{2}mv^2$ and rearrange:
$v=2Ekmv=\sqrt{\frac{2E_k}{m}}$ $v=2(1.62×10−19)9.11×10−31v=\sqrt{\frac{2(1.62\times10^{-19})}{9.11\times10^{-31}}}$ $s−1v=5.96\times10^5\ \text{m s}^{-1}$

Stopping potential

In photoelectric experiments, a potential difference can be applied to oppose the motion of the photoelectrons.

The stopping potential is the potential difference needed to reduce the photocurrent to zero. It stops even the fastest photoelectrons from reaching the collecting electrode.

Definition

Stopping potential

The stopping potential, $V_s$ , is the magnitude of the potential difference needed to stop the most energetic photoelectrons from reaching the anode.

The electrical work done per electron is $eV_s$ , where $e$ is the elementary charge. At the stopping potential:

E_{k,\max}=eV_s

Combining this with the photoelectric equation:

eVs=hf−ϕeV_s=hf-\phi

Vs=hef−ϕeV_s=\frac{h}{e}f-\frac{\phi}{e}

This is a straight-line relationship between stopping potential and frequency.

Tip

Photoelectric graph clues

For a graph of $E_{k,\max}$ against $f$ , the gradient is $h$ , the y-intercept is $−ϕ-\phi$ , and the x-intercept is $f_0$ .

Example

Finding Planck’s constant from stopping potential data

In an experiment, the stopping potential is 0.40 V at frequency $Hz6.00\times10^{14}\ \text{Hz}$ and 1.23 V at $Hz8.00\times10^{14}\ \text{Hz}$ . Estimate $h$ .

Use the straight-line relationship $Vs=hef−ϕeV_s=\frac{h}{e}f-\frac{\phi}{e}$ , so the gradient of a $V_s$ against $f$ graph is $he\frac{h}{e}$ .
Calculate the gradient:
$gradient=1.23−0.408.00×1014−6.00×1014\text{gradient}=\frac{1.23-0.40}{8.00\times10^{14}-6.00\times10^{14}}$ $s\text{gradient}=4.15\times10^{-15}\ \text{V s}$
Multiply by the elementary charge to find $h$ :
$h=e×gradienth=e\times\text{gradient}$ $h=(1.60×10−19)(4.15×10−15)h=(1.60\times10^{-19})(4.15\times10^{-15})$ $sh=6.64\times10^{-34}\ \text{J s}$

Common Mistake

Sign of stopping potential

The graph may show the current reaching zero at a negative potential, such as $V_s$ . In calculations, $V_s$ is usually used as a positive magnitude.

Line spectra and energy levels

A line spectrum is a set of separate, sharp lines at specific wavelengths or frequencies. It is not a continuous spectrum.

Definition

Line spectrum

A line spectrum is a spectrum containing discrete lines at particular frequencies or wavelengths, caused by transitions between quantised energy levels.

Atoms have quantised energy levels, meaning electrons in atoms can only have certain allowed energies. When an electron moves from a higher energy level to a lower energy level, it emits a photon. The photon energy equals the energy difference between the levels.

Energy-level diagram showing electron transitions producing photons and corresponding emission lines

hf=ΔEhf=\Delta E

So:

f=ΔEhf=\frac{\Delta E}{h}

and:

λ=hcΔE\lambda=\frac{hc}{\Delta E}

Because each element has its own unique set of energy levels, each element produces a unique line spectrum. This is why line spectra can be used to identify elements in stars and gas lamps.

Key Idea

Atoms only make certain photon energies

Line spectra show that atomic energy levels are quantised. Only photons with energies matching differences between allowed levels can be emitted or absorbed.

Emission spectra are produced when excited electrons fall to lower levels and emit photons. Absorption spectra are produced when electrons absorb photons of exactly the right energy to jump to higher levels.

Example

Calculating wavelength from an energy-level transition

An electron in an atom drops between two energy levels separated by 2.55 eV. Calculate the wavelength of the emitted photon.

Convert the energy difference into joules:
$ΔE=2.55×(1.60×10−19)\Delta E=2.55\times(1.60\times10^{-19})$ $J\Delta E=4.08\times10^{-19}\ \text{J}$
Use $f=ΔEhf=\frac{\Delta E}{h}$ :
$f=4.08×10−196.63×10−34f=\frac{4.08\times10^{-19}}{6.63\times10^{-34}}$ $Hzf=6.15\times10^{14}\ \text{Hz}$
Use $c=fλc=f\lambda$ , so $λ=cf\lambda=\frac{c}{f}$ :
$λ=3.00×1086.15×1014\lambda=\frac{3.00\times10^8}{6.15\times10^{14}}$ $m\lambda=4.88\times10^{-7}\ \text{m}$
This is 488 nm, which is in the visible region.

Bringing the ideas together

Photons explain both the photoelectric effect and line spectra.

In the photoelectric effect, a photon transfers energy to an electron in a metal. If $h f$ is greater than the work function, the electron can escape.

In line spectra, photons are emitted or absorbed when electrons in atoms move between allowed energy levels. The photon energy is exactly equal to the energy change.

Exam technique

In the exam

Always check whether an energy is given in joules or electronvolts before substituting into equations.
For photoelectric questions, separate the effects of frequency and intensity: frequency affects photon energy; intensity affects photon number per second.
On graphs, identify the gradient and intercept carefully: $E_{k,\max}$ against $f$ gives gradient $h$ , while $V_s$ against $f$ gives gradient $he\frac{h}{e}$ .

Self review

Check yourself

Why can increasing the intensity of light below the threshold frequency not release photoelectrons?
What does the work function represent in the equation $hf=ϕ+Ek,max⁡hf=\phi+E_{k,\max}$ ?
How does an energy-level diagram explain the sharp lines in an emission spectrum?

Photons, photoelectric effect and line spectra

What you'll learn

Before we start: waves, energy and charge

Photons: light as packets of energy

The photoelectric effect

Work function and the photoelectric equation

Stopping potential

Line spectra and energy levels

Bringing the ideas together

You've reached the end

Photons, photoelectric effect and line spectra Revision Guide