Light: Quantum Model

How the glow of hot objects and the electrons knocked out of metals by light forced physics to treat light as a stream of photons, each carrying energy hf.

60 min · Module 7: The Nature of Light

Assumes you have read Special Relativity.

The full teaching. Start here.

Syllabus requirementsModule 7: The Nature of Light

analyse the experimental evidence gathered about black body radiation, including Wien’s Law related to Planck's contribution to a changed model of light (ACSPH137)ACSPH137
$λ_{m a x} = \frac{b}{T}$
investigate the evidence from photoelectric effect investigations that demonstrated inconsistency with the wave model for light (ACSPH087, ACSPH123, ACSPH137)ACSPH087 ACSPH123 ACSPH137
analyse the photoelectric effect $K_{m a x} = h f - ϕ$ as it occurs in metallic elements by applying the law of conservation of energy and the photon model of light, (ACSPH119)ACSPH119

What this actually asks for

Two experiments, black body radiation and the photoelectric effect, each broke the wave model of light, and the syllabus asks you to handle both.

The black body dot point says analyse: you need to link the measured spectrum, Wien's law and Planck's idea of quantised energy into one argument, not list them. The photoelectric dot point says investigate, so expect questions built on experimental results, such as stopping voltage data or a graph of $K_{m a x}$ against frequency. The last dot point asks you to analyse the photoelectric equation with conservation of energy, which is where most of the calculations come from.

The short version

Every hot object glows with a continuous spectrum. For an ideal black body the shape depends only on temperature, and the peak wavelength obeys Wien's law, $λ_{m a x} = b / T$ . Classical wave physics predicted infinite intensity at short wavelengths, the ultraviolet catastrophe. Planck fixed the curve by assuming energy comes in quanta of $E = h f$ .

Einstein took Planck's quanta literally. Light is a stream of photons, each of energy $h f$ , and each photon gives all its energy to one electron. In the photoelectric effect an electron escapes a metal only if $h f$ exceeds the work function $ϕ$ , and leaves with at most

$K_{m a x} = h f - ϕ$

So frequency sets the energy of each electron, intensity sets how many are emitted, and below the threshold frequency $f_{0} = ϕ / h$ nothing is emitted however bright the light. The wave model predicted none of this.

Glowing hot

Everything above absolute zero emits electromagnetic radiation, because its charged particles vibrate and accelerating charges radiate. A warm body emits mostly infrared. Heat it and it glows dull red, then orange, then white.

A black body absorbs all the radiation that falls on it, and so is also the best possible emitter at every wavelength. Its spectrum depends only on its temperature, not on what it is made of. A small hole in a hollow box kept at constant temperature is a very good black body: light entering the hole bounces around inside until it is absorbed, and the radiation leaking out is characteristic of the walls' temperature. Stars are close to black bodies too.

Classical prediction
5800 K

Raise the temperature and the whole curve grows while the peak moves to shorter wavelengths, as Wien’s law says. The dashed classical curve keeps climbing at short wavelengths: the ultraviolet catastrophe.

Temperature5800 K

Peak wavelength, b / T: 500 nm (green)
Peak intensity: 84.4 kW m⁻² nm⁻¹
Total power per square metre: 6.42 × 10⁷ W m⁻²

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Drag the temperature and watch three things happen together. The whole curve rises, the peak shifts to shorter wavelengths, and the area under the curve, the total power radiated per square metre, grows rapidly.

Wien's law

The peak wavelength is inversely proportional to the absolute temperature:

Wien's displacement lawon the NESA formulae sheet

λ_{m a x} = \frac{b}{T}

Symbols: $λ_{m a x}$ wavelength at which a black body's emission is most intensem
$b$ Wien's constant, 2.898 × 10⁻³ m Km K
$T$ absolute temperature of the bodyK
Valid when: A body that emits close to a black body spectrum, such as a star's photosphere, a hot filament or a cavity with a small hole. T must be in kelvin.
Not valid when: The source emits a line spectrum, such as a gas discharge tube or a laser, or the temperature is left in degrees Celsius.
Rearranged: for T: $T = \frac{b}{λ _{m a x}}$
Where it turns up: Estimating the surface temperature of a star from its peak wavelength
Explaining why hotter bodies glow blue and cooler ones red
Finding the peak wavelength of radiation from the human body or the Earth
Where marks go missing: Using the temperature in degrees Celsius
Leaving the wavelength in metres when nanometres were asked for, or the reverse
Reading the peak as the colour the body appears, when the eye sees a mixture of wavelengths

Where this comes from

This is how astronomers measure the surface temperature of stars they will never visit. A red star such as Betelgeuse peaks in the infrared at around $3500$ K; a blue-white star such as Rigel peaks in the ultraviolet at over $10000$ K.

Worked example3 marks

Reading a star's temperature from its colour

The spectrum of the star Sirius peaks at a wavelength of $290$ nm. Calculate its surface temperature, and explain whether it would look redder or bluer than the Sun, whose surface is at $5800$ K.

Convert to metres

Wien's constant is in m K.

$λ_{m a x} = 290 nm = 2.90 \times 1 0^{- 7} m$

Rearrange Wien's law

The unknown is the temperature.

$T = \frac{b}{λ _{m a x}} = \frac{2.898 \times 1 0 ^{- 3}}{2.90 \times 1 0 ^{- 7}} = 1.0 \times 1 0^{4} K$

Compare with the Sun

The question asks for a colour judgement with a reason.

The Sun peaks at $b /5800 = 500$ nm, in the green. Sirius peaks in the ultraviolet, so across the visible band its output rises towards the blue end.

Answer

Sirius has a surface temperature of about $1.0 \times 1 0^{4}$ K. It emits relatively more blue than red light, so it looks bluer than the Sun.

Is that answer sensible?

The answer is higher than the Sun's $5800$ K, which it must be for a shorter peak wavelength, and $λ_{m a x} T = 2.90 \times 1 0^{- 7} \times 1.0 \times 1 0^{4} \approx 2.9 \times 1 0^{- 3}$ m K, which is $b$ .

CheckpointAnswer before reading on.

A black body is heated from $3000$ K to $6000$ K. Which description matches how its intensity against wavelength curve changes?

AThe curve is higher at every wavelength and its peak moves to half the original wavelength.
BThe curve is higher at every wavelength and its peak moves to twice the original wavelength.
CThe peak moves to half the wavelength but stays at the same height.
DThe curve keeps the same peak wavelength and simply doubles in height.

Hint 1Wien's law links the peak wavelength to temperature.

Hint 2 $λ_{m a x} = b / T$ , so doubling $T$ halves $λ_{m a x}$ .

Hint 3A hotter body radiates more at every wavelength.

CheckpointAnswer before reading on.

The Sun's photosphere behaves as a black body at $5800$ K. Calculate the wavelength at which its emission is most intense, in nanometres.

Your answernm

Give it to 3 significant figures.

Hint 1Use Wien's law, $λ_{m a x} = b / T$ .

Hint 2 $b = 2.898 \times 1 0^{- 3}$ m K

Hint 3Convert metres to nanometres by multiplying by $1 0^{9}$ .

CheckpointAnswer before reading on.

Human skin at $37^{\circ} C$ radiates approximately as a black body. Calculate the wavelength at which its emission peaks, in micrometres.

Your answerμm

Give it to 2 significant figures.

Hint 1Convert the temperature to kelvin.

Hint 2 $T = 37 + 273 = 310$ K

Hint 3 $1 μ m = 1 0^{- 6}$ m

The ultraviolet catastrophe

Wien's law described the peak, but nobody could explain the shape of the whole curve.

Where classical physics broke

AnswersWhy did the wave model predict infinite energy?

Inside a hot cavity, radiation forms standing waves between the walls. Each allowed standing wave is a mode, and the shorter the wavelength, the more modes fit: their number grows without limit as the wavelength shrinks.

Classical physics shares the thermal energy equally, giving every mode the same average energy, proportional to the temperature. More modes at short wavelengths then means more energy there, with no upper limit. This is the Rayleigh–Jeans prediction, the dashed curve on the graph above. It matches at long wavelengths but climbs off the top of the plot towards the ultraviolet, predicting that a cup of tea should radiate infinite power.

Measured spectra do the opposite: they peak and fall to zero at short wavelengths. The failure became known as the ultraviolet catastrophe.

CheckpointAnswer before reading on.

What was the ultraviolet catastrophe?

AClassical theory predicted that a black body's intensity would rise without limit at short wavelengths, but measurements showed it falling to zero.
BVery hot objects were found to emit dangerous amounts of ultraviolet radiation.
CWien's law failed to predict the peak wavelength of hot stars.
DUltraviolet light was found to cause no photoelectric emission from any metal.

Hint 1It was a failure of the classical wave model, not a physical event.

Hint 2The classical model shared energy equally among all modes of vibration in a cavity.

Hint 3There are more modes at shorter wavelengths.

Planck's quanta

In 1900 Planck found a formula that matched the measured curves exactly, and then looked for a reason. It required one assumption: an oscillator of frequency $f$ can only hold energy in whole multiples of a basic quantum,

Photon energyon the NESA formulae sheet

E = h f

Symbols: $E$ energy of one photonJ
$h$ Planck's constant, 6.626 × 10⁻³⁴ J sJ s
$f$ frequency of the lightHz
Valid when: Any electromagnetic radiation, one photon at a time. Combine it with c = fλ when the wavelength is given.
Not valid when: It is applied to a whole beam, whose energy is the photon energy times the number of photons, or the energy is wanted in electronvolts and the conversion is skipped.
Rearranged: for f: $f = \frac{E}{h}$ for E: $E = \frac{h c}{λ}$
Where it turns up: The energy carried by one photon of a given colour
The number of photons per second in a beam of known power
Planck's quantised oscillators in black body radiation
Where marks go missing: Substituting a wavelength for f
Mixing joules and electronvolts in one line
Thinking a brighter beam has more energetic photons

Where this comes from

where $h = 6.626 \times 1 0^{- 34}$ J s is now called Planck's constant.

At long wavelengths $h f$ is small compared with the thermal energy available, so energy is effectively continuous and the classical prediction holds. At short wavelengths one quantum is larger than the typical thermal energy, so those modes are rarely excited at all. Their average energy falls towards zero faster than their number grows, and the curve turns down. Between the two regions sits the peak.

Planck saw quantisation as a mathematical device about the oscillators in the walls, not a claim about light. The next experiment made it physical.

CheckpointAnswer before reading on.

How did Planck's hypothesis produce a black body curve that falls to zero at short wavelengths?

AEnergy comes in quanta of size $h f$ , so high-frequency modes need large quanta and are rarely excited.
BThere are fewer modes of vibration at short wavelengths than classical theory assumed.
CShort-wavelength light is absorbed by the walls of the cavity before it can escape.
DEnergy is continuous, but high-frequency waves travel more slowly.

Hint 1Planck restricted the energy an oscillator of frequency $f$ could have.

Hint 2The allowed energies are $0, h f, 2 h f, \dots$

Hint 3What happens when $h f$ is much larger than the typical thermal energy available?

The photoelectric effect

Shine light on a clean metal surface and electrons can be knocked out. The standard apparatus is a vacuum tube with the metal as one electrode and a collector as the other. A variable voltage can speed the electrons towards the collector or push them back.

Lenard, and later Millikan, measured how the current depended on the light. Four results stood out:

Observation	What the wave model predicted
Below a threshold frequency $f_{0}$ , which depends on the metal, no electrons are emitted however bright the light.	Bright enough light of any frequency should eventually free electrons.
The current is proportional to the intensity of the light.	Agrees, but for the wrong reason: more energy per electron, not more electrons.
The maximum kinetic energy of the electrons rises with frequency and is unaffected by intensity.	Brighter light, with larger amplitude, should give faster electrons.
Emission starts as soon as the light is switched on, even when it is very dim.	Dim light spreads its energy over the whole surface, so electrons should take a long time to collect enough.

Speed

Light frequency8.0 × 10¹⁴ Hz

Photons per second6

Work function2.3 eV

Extrapolated
K_max = hf − φ

The same situation drawn as a graph, shown because this page has no JavaScript to animate it.

Below the threshold nothing is emitted, however many photons arrive. Above it, electrons appear from the first photon, with no delay, and brighter light only means more of them. Their speeds depend on the frequency.

Set the frequency below the threshold and turn the intensity up as far as it goes: photons strike, but nothing leaves. Now drop the intensity and raise the frequency just past the threshold, and electrons appear with the very first photons.

Einstein's photons

In 1905 Einstein proposed that light itself is made of quanta, later called photons. Each carries energy $h f$ , and each is absorbed whole by a single electron.

Derivation

The photoelectric equation

Starts from

Conservation of energy for one photon and one electron

Ends at

K_{m a x} = h f - ϕ

Holds only if

One photon is absorbed by one electron
The photon gives up all its energy
The work function is the least energy needed to remove an electron from this metal

The photoelectric equation

4 steps

1
Energy in
The electron gains the whole photon.
The electron's energy increases by $h f$ .
2
Energy spent escaping
The metal holds its conduction electrons with a minimum binding energy.
Escaping costs at least $ϕ$ , the work function. Electrons deeper in the metal lose more on the way out.
3
What is left
Conservation of energy.
The kinetic energy left over is at most
$K_{m a x} = h f - ϕ$
4
The threshold
The electron cannot leave with negative kinetic energy.
Emission needs $h f \geq ϕ$ , so the threshold frequency is $f_{0} = \frac{ϕ}{h}$ .

Photoelectric equationon the NESA formulae sheet

K_{m a x} = h f - ϕ

Symbols: $K_{m a x}$ maximum kinetic energy of an emitted electronJ
$h$ Planck's constant, 6.626 × 10⁻³⁴ J sJ s
$f$ frequency of the incident lightHz
$ϕ$ work function of the metal, the least energy that frees an electronJ
Valid when: Light falling on a clean metal surface, one photon to one electron. The result is the energy of the fastest electrons; most leave with less.
Not valid when: hf is less than φ, where no electrons are emitted and a negative answer has no meaning, or the kinetic energy of a typical electron is wanted rather than the maximum.
Rearranged: for \phi: $ϕ = h f - K_{m a x}$ for f_{0}: $f_{0} = \frac{ϕ}{h}$ for V_{s}: $V_{s} = \frac{K _{m a x}}{e}$
Where it turns up: The stopping voltage for a metal lit at a given frequency
The threshold frequency or wavelength of a metal
Reading h and φ from the gradient and intercept of Millikan's graph
Where marks go missing: Adding the work function instead of subtracting it
Leaving φ in electronvolts while hf is in joules
Expecting a brighter beam to raise K_max

Where this comes from

Every observation now follows. Below $f_{0}$ a single photon cannot free an electron, and an electron cannot bank energy from several photons, so intensity does not help. Above it, doubling the intensity doubles the number of photons and so the number of electrons, but each photon still carries $h f$ , so $K_{m a x}$ is unchanged. And one photon delivers its energy at once, so there is no delay.

Extrapolated
K_max = hf − φ

Below the threshold frequency no electrons are emitted, however bright the light. Above it, the line has gradient h for every metal; only the intercept, set by the work function, changes.

Work function2.3 eV

Light frequency7.5 × 10¹⁴ Hz

Photon energy, hf: 3.10 eV (400 nm, violet)
Work function: 2.3 eV, about that of potassium
Threshold frequency, φ / h: 5.56 × 10¹⁴ Hz
Maximum kinetic energy: 0.80 eV = 1.28 × 10⁻¹⁹ J
Stopping voltage: 0.80 V

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

The line has the same gradient, $h$ , for every metal. Only the intercept changes: the line crosses the frequency axis at $f_{0}$ , and extended back it meets the energy axis at $- ϕ$ .

CheckpointAnswer before reading on.

Light above the threshold frequency falls on a metal. The intensity is doubled while the frequency stays the same. What happens?

ATwice as many electrons are emitted, with the same maximum kinetic energy.
BThe same number of electrons are emitted, each with twice the kinetic energy.
CTwice as many electrons are emitted, each with more kinetic energy.
DNothing changes, because only frequency matters.

Hint 1Intensity measures the number of photons arriving each second.

Hint 2Each photon gives its energy to one electron.

Hint 3What sets the energy each electron receives?

CheckpointAnswer before reading on.

Zinc has a work function of $4.3$ eV. A very bright violet lamp, with photons of $3.10$ eV, shines on a clean zinc plate for an hour. What happens?

ANo electrons are emitted, however bright the lamp or long the exposure.
BElectrons are emitted after a delay while they absorb enough energy.
CElectrons are emitted with $3.10 - 4.3 = - 1.2$ eV of kinetic energy.
DA few electrons are emitted because the lamp is very bright.

Hint 1Compare the photon energy with the work function.

Hint 2An electron absorbs one photon at a time.

Hint 3Can an electron save up energy from several photons?

Common mistake

A brighter light gives the electrons more kinetic energy.

Why it happens: In the wave model, a larger amplitude carries more energy, so it feels natural that brighter light should push electrons harder.

Intensity is the number of photons arriving each second. Each electron still gets exactly one photon's worth, $h f$ . Brighter light means more electrons at the same energies; only raising the frequency makes them faster.

Measuring the fastest electrons: stopping voltage

To measure the fastest electrons, make the collector negative. The field pushes the electrons back, and only those with enough kinetic energy reach the collector. Increase the reverse voltage until the current just stops. At this stopping voltage $V_{s}$ even the fastest electron is turned back, so its kinetic energy equals the work done against it:

$K_{m a x} = e V_{s}$

In electronvolts the two numbers are the same: a stopping voltage of $1.2$ V means $K_{m a x} = 1.2$ eV.

Full intensity
50 %

A negative voltage repels electrons, and the current falls to zero at the stopping voltage. Dimming the light lowers the current but leaves the stopping voltage unchanged; raising the frequency moves it. The model assumes the emitted energies are spread evenly from zero to the maximum.

Light frequency10.0 × 10¹⁴ Hz

Intensity50 %

Work function2.3 eV

Photon energy: 4.14 eV
Stopping voltage: 1.84 V
Saturation current: 50 % of maximum

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Raising the intensity lifts the whole curve but leaves the stopping voltage fixed. Raising the frequency moves the stopping voltage further negative.

CheckpointAnswer before reading on.

In a photoelectric experiment the photocurrent falls to zero when the collector is $1.2$ V negative relative to the metal surface. What is the maximum kinetic energy of the emitted electrons, in joules?

Your answerJ

Give it to 2 significant figures.

Hint 1The fastest electron just fails to reach the collector.

Hint 2Its kinetic energy equals the work the field does against it.

Hint 3 $K_{m a x} = e V_{s}$

Worked example4 marks

Planck's constant from stopping voltages

A student lights a potassium surface with filtered light and measures the stopping voltage at three frequencies.

$f$ ( $\times 1 0^{14}$ Hz)	$V_{s}$ (V)
$7.0$	$0.60$
$9.0$	$1.42$
$11.0$	$2.25$

Use the data to find Planck's constant and the work function of potassium.

Turn the equation into a straight line

Plotting V_s against f is how Millikan analysed his data.

$e V_{s} = h f - ϕ \Rightarrow V_{s} = \frac{h}{e} f - \frac{ϕ}{e}$

A graph of $V_{s}$ against $f$ has gradient $h / e$ .

Gradient

The points lie on a line, so the end points give the gradient.

$gradient = \frac{2.25 - 0.60}{( 11.0 - 7.0 ) \times 1 0 ^{14}} = 4.1 \times 1 0^{- 15} V s$

Planck's constant

Multiply the gradient by e.

$h = 4.1 \times 1 0^{- 15} \times 1.602 \times 1 0^{- 19} = 6.6 \times 1 0^{- 34} J s$

Work function

Rearrange for φ using any point.

At $f = 7.0 \times 1 0^{14}$ Hz, $h f = \frac{6.6 \times 1 0 ^{- 34} \times 7.0 \times 1 0 ^{14}}{1.602 \times 1 0 ^{- 19}} = 2.9$ eV, so

$ϕ = h f - e V_{s} = 2.9 - 0.60 = 2.3 eV$

Answer

$h \approx 6.6 \times 1 0^{- 34}$ J s and $ϕ \approx 2.3$ eV, matching the accepted work function of potassium.

Is that answer sensible?

The threshold frequency is $ϕ / h = \frac{2.3 \times 1.602 \times 1 0 ^{- 19}}{6.6 \times 1 0 ^{- 34}} = 5.6 \times 1 0^{14}$ Hz, below all three frequencies used, so every measurement did show emission. Try it on the graph above with $ϕ = 2.3$ eV.

Millikan set out to disprove Einstein's equation. His careful measurements instead gave a gradient matching Planck's value of $h$ from black body radiation, found by a completely different method, and he conceded that the equation worked.

CheckpointAnswer before reading on.

A student plots stopping voltage against frequency for one metal. The best-fit line has a gradient of $4.1 \times 1 0^{- 15}$ V s and meets the frequency axis at $5.5 \times 1 0^{14}$ Hz.

Which pair of values do these results give?

A $h = 6.6 \times 1 0^{- 34}$ J s and $ϕ = 2.3$ eV
B $h = 4.1 \times 1 0^{- 15}$ J s and $ϕ = 2.3$ eV
C $h = 6.6 \times 1 0^{- 34}$ J s and $ϕ = 5.5$ eV
D $h = 6.6 \times 1 0^{- 34}$ J s and $ϕ = 3.6 \times 1 0^{- 19}$ eV

Hint 1 $e V_{s} = h f - ϕ$ , so $V_{s} = \frac{h}{e} f - \frac{ϕ}{e}$ .

Hint 2The gradient is $h / e$ . Multiply by $e$ to get $h$ .

Hint 3The intercept on the frequency axis is $f_{0}$ , and $ϕ = h f_{0}$ .

Calculating with photons

Worked example3 marks

Ultraviolet on potassium

Ultraviolet light of wavelength $250$ nm falls on potassium, work function $2.3$ eV. Find the maximum kinetic energy of the emitted electrons and the stopping voltage.

Photon energy

The wavelength is given, so combine E = hf with c = fλ.

$h f = \frac{h c}{λ} = \frac{6.626 \times 1 0 ^{- 34} \times 3.00 \times 1 0 ^{8}}{2.50 \times 1 0 ^{- 7}} = 7.95 \times 1 0^{- 19} J$

Put both energies in the same unit

The work function is in eV and the photon energy in J.

$h f = \frac{7.95 \times 1 0 ^{- 19}}{1.602 \times 1 0 ^{- 19}} = 4.96 eV$

Subtract the work function

Conservation of energy.

$K_{m a x} = 4.96 - 2.3 = 2.66 eV = 4.3 \times 1 0^{- 19} J$

Stopping voltage

K_max = eV_s.

$V_{s} = \frac{K _{m a x}}{e} = 2.66 V$

Answer

$K_{m a x} = 4.3 \times 1 0^{- 19}$ J, or $2.66$ eV, and the stopping voltage is $2.66$ V.

Is that answer sensible?

$K_{m a x}$ is positive, so the photon is above the threshold, and it is less than the photon energy, as it must be.

CheckpointAnswer before reading on.

Calculate the energy of one photon of blue light of wavelength $450$ nm, in joules.

Your answerJ

Give it to 3 significant figures.

Hint 1Combine $E = h f$ with $c = f λ$ .

Hint 2 $E = \frac{h c}{λ}$

Hint 3 $λ = 450 \times 1 0^{- 9}$ m

CheckpointAnswer before reading on.

Sodium has a work function of $2.4$ eV. Calculate its threshold frequency.

Your answerHz

Give it to 2 significant figures.

Hint 1At the threshold, the photon has just enough energy to free an electron: $h f_{0} = ϕ$ .

Hint 2Convert the work function to joules: $1$ eV $= 1.602 \times 1 0^{- 19}$ J.

Hint 3 $f_{0} = \frac{ϕ}{h}$

CheckpointAnswer before reading on.

Ultraviolet light of wavelength $250$ nm falls on potassium, which has a work function of $2.3$ eV. Calculate the maximum kinetic energy of the emitted electrons, in joules.

Your answerJ

Give it to 2 significant figures.

Hint 1Find the photon energy with $E = h c / λ$ .

Hint 2Put both energies in the same unit before subtracting.

Hint 3 $K_{m a x} = h f - ϕ$

CheckpointAnswer before reading on.

Ultraviolet light of frequency $1.50 \times 1 0^{15}$ Hz falls on zinc, work function $4.3$ eV. Calculate the stopping voltage.

Your answerV

Give it to 2 significant figures.

Hint 1Find the photon energy in eV: $h f / e$ .

Hint 2 $K_{m a x} = h f - ϕ$

Hint 3 $V_{s}$ in volts equals $K_{m a x}$ in eV.

CheckpointAnswer before reading on.

A red laser pointer emits $5.0$ mW of light at $650$ nm. How many photons does it emit each second?

Your answers⁻¹

Give it to 2 significant figures.

Hint 1Power is energy per second.

Hint 2Find the energy of one photon with $E = h c / λ$ .

Hint 3Number per second $= \frac{P}{E}$

CheckpointAnswer before reading on.

Order these photons from lowest energy to highest: red light, X-ray, microwave, ultraviolet, infrared.

infrared
ultraviolet
X-ray
microwave
red light

Hint 1 $E = h f$ , so energy rises with frequency.

Hint 2Frequency rises as wavelength falls.

Hint 3Recall the order of the electromagnetic spectrum.

Exam question

Harder · about 6 min

4 marks

Describe two results of photoelectric effect investigations and explain why each is inconsistent with the wave model of light.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1Think about threshold frequency, the effect of intensity and the time delay.

Hint 2For each result, state what the wave model predicted.

Hint 3Link each result to the idea that energy arrives in photons of $h f$ .

Where students lose marks on this one

Describing the photon explanation without saying what the wave model predicted.
Why it happens: The question asks why the result conflicts with the wave model.
State the wave prediction and show the result contradicts it.

Written for this site.

Exam question

Standard · about 5 min

3 marks

Use the law of conservation of energy and the photon model of light to explain the equation $K_{m a x} = h f - ϕ$ , and explain why most emitted electrons have less kinetic energy than $K_{m a x}$ .

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1Where does the photon's energy go?

Hint 2What does the work function represent?

Hint 3Electrons deeper in the metal lose energy on their way out.

Where students lose marks on this one

Saying every electron leaves with exactly $hf-\phi$.
Why it happens: The equation has no range in it.
$K_{m a x}$ is the upper limit for electrons at the surface. Others lose extra energy escaping.

Written for this site.

Analysetypically 4 to 7 marks

Demands: Break the situation into its parts and show how they relate and affect each other.
Shape: Identify the components, then the relationships, then what follows from them.
Loses marks: Summarising the stimulus rather than taking it apart.

Through a marker’s eyes

3 marks

Explain why the existence of a threshold frequency in the photoelectric effect supports the particle model of light. (3 marks)

The attempt

1 out of 3

Light is made of particles called photons with energy $E = h f$ . If the frequency is too low the photons do not have enough energy so no electrons come out.

What the marker sees

The photon idea is there, but the answer never says what the wave model predicted, so it does not explain why the observation supports the particle model. It also misses the key point that one photon is absorbed by one electron, which is why intensity cannot make up the shortfall.

The same answer, fixed

3 out of 3

In the wave model the energy delivered depends on intensity, so sufficiently bright light of any frequency should eventually free electrons. Experiments show that below a threshold frequency no electrons are emitted, however intense the light.
In the particle model light is a stream of photons of energy $h f$ , and each electron absorbs a single photon. If $h f$ is less than the work function, no single photon can free an electron, and adding more photons does not help.
The threshold is therefore explained by the particle model and contradicts the wave model.

Earth as a black body

Sunlight delivers about $1360$ W m $^{- 2}$ to Earth. Earth intercepts it over a disc of area $π r^{2}$ but radiates from its whole surface, $4 π r^{2}$ , so a black body Earth would need to radiate $1360/4 = 340$ W m $^{- 2}$ . About a third of the sunlight is reflected, leaving roughly $240$ W m $^{- 2}$ , which by the Stefan–Boltzmann law $P / A = σ T^{4}$ corresponds to about $- 18^{\circ} C$ . The real average is about $15^{\circ} C$ because greenhouse gases absorb the outgoing infrared, which by Wien's law peaks near $10 μ$ m.

Compton scattering

In 1923 Compton fired X-rays at graphite and found the scattered X-rays had longer wavelengths than the incident ones. A wave would make electrons vibrate at its own frequency and re-radiate the same wavelength. Compton explained the shift as a collision between a photon and an electron, with the photon carrying momentum

$p = \frac{h}{λ}$

and losing some energy to the recoiling electron. This gave photons momentum as well as energy and made the particle model hard to deny.

Neither wave nor particle

Light still diffracts and interferes, which only waves do, and the photoelectric effect and Compton scattering need particles. Modern physics treats light as a quantum object that shows either behaviour depending on the experiment. The same turns out to be true of electrons, which the next module explores through de Broglie's matter waves.