E.2 Quantum physics · IB DP Physics

      
      IB DP Physics · Theme E · Nuclear and quantum physics
Quantum physics.E.2 · Photoelectric effect, matter waves, Compton scattering
Higher level only: there is no standard level content in E.2

              Additional HL · 8 hours
              E.1 Structure of the atom ↗
            
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      The journey
Three experiments.
Part 01
The photoelectric effectLight behaves as particles
Part 02
Electron diffractionParticles behave as waves
Part 03
Compton scatteringPhotons carry momentum
Together: light and matter both show wave–particle duality.
E.2 QUANTUM PHYSICS · HL ONLY


      
      Part 01
The photoelectric effect.How can light be used to create an electric current?
E.2 QUANTUM PHYSICS · HL ONLY


      From E.1
Photons: a reminder.Light arrives as photons
Energy of each depends on frequency only
Intensity: how many arrive each second
\(E = hf = \dfrac{hc}{\lambda}\)

            RadiationWavelengthPhoton energy

            Red light700 nm1.77 eV
Violet light400 nm3.10 eV
Ultraviolet250 nm4.96 eV

          
E.2 QUANTUM PHYSICS · HL ONLY
The shortcut used all through this topicThe data booklet gives \(hc = 1.24 \times 10^{-6} \,\mathrm{eV} \,\mathrm{m}\). Divide it by the wavelength in meters and the photon energy comes out in eV: \(\frac{1.24 \times 10^{-6}}{400 \times 10^{-9}} = 3.10 \,\mathrm{eV}\).
Two separate thingsFrequency sets how much energy each photon has. Intensity sets how many photons arrive. A dim ultraviolet lamp sends few photons, but each one carries more energy than any photon from the brightest red lamp.
Work in eV in this topic. Work functions and electron energies are quoted in eV, so converting to joules is usually wasted effort.


      The effect
Light knocks electrons out of a metal.Shine light on a clean metal
Electrons may be emitted
They are called photoelectrons
But only if the frequency is high enough.
PHOTOELECTRIC EMISSIONNOT TO SCALE
E.2 QUANTUM PHYSICS · HL ONLY
A demonstrationPut a clean zinc plate on a negatively charged electroscope. Under a bright filament lamp the leaf stays up. Under a weak ultraviolet lamp it falls at once: electrons are leaving the zinc. Put a sheet of glass in the way, which absorbs ultraviolet, and it stops.
What needs explainingThe weak lamp works and the bright one does not. Whatever decides whether electrons escape, it is not how much energy arrives each second.
A photoelectron is an ordinary electron. The name only says how it was released.
“Clean” matters: an oxide layer on the surface changes the energy needed to escape.


      The evidence
Four things the experiments show.01 · Threshold
No emission below a certain frequencyThe threshold frequency \(f_{0}\), different for each metal
02 · No delay
Emission starts at onceEven when the light is very dim
03 · Intensity
Brighter light: more electronsBut their maximum kinetic energy does not change
04 · Frequency
Higher frequency: faster electronsMaximum kinetic energy rises with \(f\)
E.2 QUANTUM PHYSICS · HL ONLY
Why “maximum” kinetic energy?The electrons come out with a range of energies. One at the very surface loses the least on the way out and leaves with the most. One from deeper in loses more in collisions. Only the maximum is a clean, repeatable number.
How these were measuredLight of one frequency falls on a metal plate in a vacuum, and the current of photoelectrons is measured. Changing the frequency, the intensity and the metal one at a time gives the four results.
Learn these four as a set. Almost every extended question on the photoelectric effect asks you to explain two or three of them.


      Classical wave theory against the evidence
Why the wave theory fails.
          
            A wave would…What happens

            FrequencyWork at any frequency, if bright enoughNothing below \(f_{0}\), however bright
TimeNeed time to build up energy in dim lightEmission is immediate
IntensityGive each electron more energy when brighterSame maximum energy: only more electrons
EnergyNot depend on frequencyMaximum energy rises with frequency

          
        
E.2 QUANTUM PHYSICS · HL ONLY
The wave pictureA wave spreads its energy evenly over the whole surface, and a brighter wave delivers energy faster. So any electron should be able to soak up energy until it has enough to leave. A dim wave would just take longer, and a bright one of any color would do it.
Putting a number on the delayFor very dim light, spreading the energy over all the atoms of the surface gives each electron so little that it would need minutes or hours to collect enough. The measured delay is less than a billionth of a second.
For each feature, write what the wave theory predicts, what is seen, and that they disagree. Naming the feature alone does not earn the mark.
Saying the wave theory predicts no emission. It predicts emission; it gets the conditions wrong.


      The story · Bern, 1905
Background story · not assessed
A patent clerk’s revolutionary idea.
            Albert Einstein at the patent office in Bern, about 1905
            The journal’s contents, June 1905. His paper is sixth on the list
          
Einstein, aged 26, checks patents for a living and does physics in his spare time
His paper says light itself comes in packets of energy \(hf\): light quanta
Almost nobody believes it. Light was known to be a wave
1921: the Nobel Prize, for this and not for relativity
He called it the one revolutionary thing he had done.
Photo: Lucien Chavan, public domain. Contents page: Nerd271, CC0.
E.2 QUANTUM PHYSICS · HL ONLY


      Einstein (1905)
One photon, one electron.
            Light is a stream of photons, each of energy \(hf\)
One photon gives all its energy to one electron
Work function \(\Phi\): the least energy needed to leave the metal
What is left over is kinetic energy

          
WHERE THE PHOTON’S ENERGY GOES
E.2 QUANTUM PHYSICS · HL ONLY
How this explains all four resultsThresholdif \(hf\) is less than \(\Phi\), no single photon can free an electron, and photons do not team up. More photons do not help.
No delaythe energy arrives in one lump, so the electron has enough the moment a photon reaches it.
Intensitybrighter light is more photons each second, so more electrons each second. Each photon is no bigger, so the maximum energy is the same.
Frequencya higher frequency means more energy in each photon and more left over after paying \(\Phi\).
Einstein’s Nobel Prize (1921) was for this explanation, not for relativity.
The work function belongs to the metal. It is the minimum: most electrons need more than \(\Phi\) to get out.


      Einstein’s equation
The photoelectric equation.\(E_{\mathrm{max}} = hf - \Phi\)
\(E_{\mathrm{max}}\): maximum kinetic energy of the photoelectrons
\(\Phi\): work function of the metal
At the threshold \(E_{\mathrm{max}} = 0\), so \(hf_{0} = \Phi\)

            Threshold for zinc · \(\Phi = 4.30 \,\mathrm{eV}\)
            \(f_{0} = \dfrac{\Phi}{h} = \dfrac{4.30 \times 1.60 \times 10^{-19}}{6.63 \times 10^{-34}}\)
              \(= \class{res}{1.04 \times 10^{15} \,\mathrm{Hz}}\)
            \(\lambda _{0} = \dfrac{c}{f_{0}} = \class{res}{2.88 \times 10^{-7} \,\mathrm{m}}\)
            288 nm: ultraviolet
          
E.2 QUANTUM PHYSICS · HL ONLY
It is conservation of energyEnergy in (one photon, \(hf\)) = energy to escape \((\Phi )\) + kinetic energy left \((E_{\mathrm{max}})\). Nothing else is going on.
Why zinc ignores visible lightIts threshold wavelength is 288 nm. Visible light runs from 400 to 700 nm, so every visible photon has too little energy, which is what the electroscope showed. Metals such as cesium and sodium have work functions near 2 eV and do respond to visible light.
Mixing units: \(hf\) worked out in joules with \(\Phi\) left in eV. Put both in the same unit before subtracting.
A negative answer for \(E_{\mathrm{max}}\) means the frequency is below the threshold: no electrons are emitted.


      Einstein’s equation as a graph
One straight line, three readings.\(E_{\mathrm{max}} = hf - \Phi\) has the form \(y = mx + c\)
Gradient = \(h\), the same for every metal
Intercept on the \(f\) axis = \(f_{0}\)
Intercept on the energy axis = \(-\Phi\)
Another metal: a parallel line.
MAXIMUM KINETIC ENERGY AGAINST FREQUENCY
E.2 QUANTUM PHYSICS · HL ONLY
Reading the graphBelow \(f_{0}\) there are no electrons, so there are no points; the dashed part is the line extended back to find \(-\Phi\). Above \(f_{0}\), each extra hertz adds the same extra energy, \(h\).
Why every metal gives the same gradientThe gradient is the Planck constant, which belongs to light and not to the metal. Changing the metal changes only \(\Phi\), sliding the line sideways.
Robert Millikan spent years measuring this line, expecting to prove Einstein wrong. His results, published in 1916, fitted the equation and gave an accurate value of \(h\).
If the energy axis is in eV, the gradient is \(h\) in eV s: multiply by \(1.60 \times 10^{-19}\) for J s.


      The story · Chicago, 1912–1915
Background story · not assessed
Millikan sets out to prove Einstein wrong.
            Robert Millikan, pictured in 1923, the year of his Nobel Prize
            His graph, published in 1916: stopping potential against frequency, with the gradient used to find \(h\)
          
Millikan was sure light was a wave. He set out to show that Einstein’s equation failed
He shaved his metal surfaces clean inside the vacuum, so that nothing spoiled the readings
Every point fell on Einstein’s straight line, and its gradient gave an accurate value of \(h\)
Portrait: Nobel Foundation, public domain. Graph: R. A. Millikan, Physical Review, 1916, public domain.
E.2 QUANTUM PHYSICS · HL ONLY


      Measuring the maximum kinetic energy
Stopping the fastest electrons.Make the collector negative
Raise the p.d. until the current is zero
That p.d. is the stopping potential \(V_{\mathrm{s}}\)
\(eV_{\mathrm{s}} = E_{\mathrm{max}}\)
PHOTOCELLNOT TO SCALE
E.2 QUANTUM PHYSICS · HL ONLY
Why it worksAn electron crossing to a collector that is negative by \(V\) loses kinetic energy \(eV\). Slow electrons are turned back first. When even the fastest just fail to arrive, the work done against the field equals their kinetic energy: \(eV_{\mathrm{s}} = E_{\mathrm{max}}\).
The convenient partAn electron stopped by 0.66 V had 0.66 eV. The stopping potential in volts is the maximum kinetic energy in electronvolts.
Brighter light of the same frequency gives a bigger current but the same stopping potential. Higher frequency gives a bigger stopping potential.


      Worked example
250 nm ultraviolet on zinc.
            \(\Phi = 4.30 \,\mathrm{eV}\) · \(hc = 1.24 \times 10^{-6} \,\mathrm{eV} \,\mathrm{m}\)
            Photon: \(E = \dfrac{hc}{\lambda} = \dfrac{1.24 \times 10^{-6}}{250 \times 10^{-9}} = \class{res}{4.96 \,\mathrm{eV}}\)
            \(E_{\mathrm{max}} = hf - \Phi = 4.96 - 4.30 = \class{res}{0.66 \,\mathrm{eV}}\)
              \(= 0.66 \times 1.60 \times 10^{-19} = \class{res}{1.1 \times 10^{-19} \,\mathrm{J}}\)
            Stopping potential \(V_{\mathrm{s}} = \dfrac{E_{\mathrm{max}}}{e} = \class{res}{0.66 \,\mathrm{V}}\)
          

            Double the intensity: still 0.66 eV, twice the current
Use 300 nm instead: 4.13 eV per photon, no emission

          
E.2 QUANTUM PHYSICS · HL ONLY
The order to work inPhoton energy first, then compare it with the work function. If the photon has less, stop: there is no emission and nothing more to calculate. If it has more, the difference is \(E_{\mathrm{max}}\).
Finding the speed\(E_{\mathrm{max}} = \tfrac{1}{2}mv^{2}\) with the energy in joules: \(v = \sqrt{\frac{2 \times 1.06 \times 10^{-19}}{9.11 \times 10^{-31}}} = 4.8 \times 10^{5} \,\mathrm{m} \,\mathrm{s}^{-1}\).
Doubling the intensity does not double \(E_{\mathrm{max}}\). It doubles the number of photons, so the number of electrons.


      Quick check
The photoelectric effect.01What is the threshold frequency?
Lowest frequency that gives emission
02What is the work function?
Least energy needed to free an electron
03Photons of 5.0 eV, metal with \(\Phi = 2.3 \,\mathrm{eV}\). \(E_{\mathrm{max}}\)?
\(5.0 - 2.3 = 2.7 \,\mathrm{eV}\)
04Intensity doubles, same frequency. What changes?
More electrons. Same \(E_{\mathrm{max}}\)
05Why can waves not explain the threshold?
Any frequency should work if bright enough
06What is the gradient of \(E_{\mathrm{max}}\) against \(f\)?
The Planck constant, \(h\)
PRESS → TO REVEAL EACH ANSWER


      
      Part 02
Matter waves.How can particles diffract?
E.2 QUANTUM PHYSICS · HL ONLY


      The story · Paris, 1924
Background story · not assessed
A prince’s thesis.
            Louis de Broglie, pictured in 1929, the year of his Nobel Prize
          
Louis de Broglie, a French prince, studied history before he turned to physics
1924: his doctoral thesis proposes that electrons have a wavelength
His examiners were unsure. One sent a copy to Einstein, who backed it
There was no experimental evidence yet. It came three years later
The prediction came first. The evidence came afterwards.
Photo: Agence Rol, public domain.
E.2 QUANTUM PHYSICS · HL ONLY


      de Broglie (1924)
If waves can be particles…?Light, a wave, also behaves as particles
de Broglie: particles should also behave as waves
Every moving particle has a wavelength
\(\lambda = \dfrac{h}{p}\)
\(p = mv\), the momentum. More momentum → shorter wavelength.
THE DE BROGLIE WAVELENGTH
E.2 QUANTUM PHYSICS · HL ONLY
Where the idea came fromA photon of wavelength \(\lambda\) has momentum \(p = h/\lambda\). de Broglie proposed, with no experimental evidence at the time, that the same relation holds for electrons and every other particle, read the other way round: \(\lambda = h/p\).
From energy to wavelengthFor a particle that is not moving close to the speed of light, \(E_{\mathrm{k}} = p^{2}/2m\), so \(p = \sqrt{2mE_{\mathrm{k}}}\). An electron accelerated from rest through a p.d. \(V\) has \(E_{\mathrm{k}} = eV\).
This also answers a question left open in E.1. A whole number of electron wavelengths fitting round a Bohr orbit, \(2\pi r = n\lambda\), gives exactly Bohr’s rule \(mvr = nh/2\pi\).


      Worked examples
An electron and a tennis ball.
            Electron accelerated through 150 V
            \(E_{\mathrm{k}} = eV = 1.60 \times 10^{-19} \times 150\)
              \(= 2.40 \times 10^{-17} \,\mathrm{J}\)
            \(p = \sqrt{2mE_{\mathrm{k}}}\)
              \(= \sqrt{2 \times 9.11 \times 10^{-31} \times 2.40 \times 10^{-17}}\)
              \(= 6.61 \times 10^{-24} \,\mathrm{kg} \,\mathrm{m} \,\mathrm{s}^{-1}\)
            \(\lambda = \dfrac{h}{p} = \class{res}{1.0 \times 10^{-10} \,\mathrm{m}}\)
            About the spacing of atoms in a crystal
          

            Tennis ball · 0.058 kg at \(50 \,\mathrm{m} \,\mathrm{s}^{-1}\)
            \(p = mv = 0.058 \times 50 = 2.9 \,\mathrm{kg} \,\mathrm{m} \,\mathrm{s}^{-1}\)
            \(\lambda = \dfrac{h}{p} = \dfrac{6.63 \times 10^{-34}}{2.9}\)
              \(= \class{res}{2.3 \times 10^{-34} \,\mathrm{m}}\)
            Far smaller than a nucleus
          
E.2 QUANTUM PHYSICS · HL ONLY
Why we never see a ball diffractDiffraction is noticeable only when the gap is about as wide as the wavelength. An electron’s wavelength matches the gaps between atoms in a crystal, so a crystal diffracts it. The ball’s wavelength is about \(\frac{1}{10^{19}}\) of the diameter of a nucleus, and nothing has gaps that small.
Changing the wavelengthFor the electron, \(\lambda = h/\sqrt{2meV}\), so \(\lambda \propto 1/\sqrt{V}\). Four times the accelerating p.d. halves the wavelength.
Using \(\lambda = hc/E\) for an electron. That is for photons only. For a particle with mass use \(\lambda = h/p\).


      The story · New York and Aberdeen, 1927
Background story · not assessed
An accident, and a father and son.
            Clinton Davisson (left) and Lester Germer with their electron tube, 1927
            The diffraction camera G. P. Thomson built in Aberdeen
            One of Thomson’s electron diffraction photographs
          
Davisson and Germer fired electrons at nickel. A 1925 accident left it as a few large crystals: the electrons now came off in sharp peaks
In Aberdeen, G. P. Thomson sent electrons through thin metal films and photographed rings
His father, J. J. Thomson, showed the electron is a particle. The son showed it is a wave. Davisson and Thomson shared the 1937 Nobel Prize
Photo: unknown photographer, public domain. Camera and diffraction photograph: Science Museum Group and Science Museum Group, CC BY 4.0.
E.2 QUANTUM PHYSICS · HL ONLY


      The test · Davisson & Germer, G. P. Thomson (1927)
Electron diffraction.Fire electrons through thin graphite
The screen shows rings, not one spot
Rings are a diffraction pattern
Only waves diffract
ELECTRON DIFFRACTION TUBENOT TO SCALE
E.2 QUANTUM PHYSICS · HL ONLY
Why graphite?Its layers of carbon atoms are spaced about \(10^{-10} \,\mathrm{m}\) apart, close to the wavelength of electrons accelerated through a few thousand volts. The regular layers act as a diffraction grating. The tiny crystals in the film point in every direction, so each diffracted beam is spread round into a ring.
Why the rings are evidenceParticles passing through a thin film would scatter into a blur that is brightest in the middle and fades steadily outwards. Bright rings with dark gaps between them mean that at some angles the electrons reinforce and at others they cancel: interference, which only waves do.
The same pattern builds up even when electrons are sent through one at a time. Each lands at one point, like a particle; the points pile up into rings, like a wave.


      Evidence → conclusion
Testing \(\lambda = h/p\).OBSERVATIONCONCLUSION
Electrons form rings on the screen
→Electrons diffract: they have a wave nature
Higher accelerating p.d. → smaller rings
→More momentum → shorter wavelength
Ring size gives a wavelength
→It equals \(h/p\)
Diffraction of particles is the evidence for the wave nature of matter.
E.2 QUANTUM PHYSICS · HL ONLY
Why do the rings shrink?Raising the p.d. gives each electron more kinetic energy and so more momentum. By \(\lambda = h/p\) the wavelength falls. A shorter wavelength is diffracted through a smaller angle, as it is for light at a grating, so the rings close in.
A prediction that could have failedde Broglie’s equation says exactly what the wavelength should be for a given p.d. The wavelength measured from the rings, using the known spacing of the atoms, agreed.
Asked for evidence that matter has wave properties: name the experiment, say what is seen, and say that diffraction is something only waves do.


      Diffraction of particles
Where the first minimum falls.Particles scattered by an obstacle of width \(D\)
Counted at each angle \(\theta\)
Fewest arrive at the first minimum
\(\sin \theta \approx \dfrac{\lambda}{D}\)with \(\lambda = h/p\)
Measure \(\theta\) → find \(D\). This is how nuclei are measured.
SCATTERED PARTICLES AGAINST ANGLESKETCH
E.2 QUANTUM PHYSICS · HL ONLY
The same rule as for lightIn C.3, light of wavelength \(\lambda\) passing a slit of width \(b\) has its first minimum at \(\theta = \frac{\lambda}{b}\). Particles do the same with their de Broglie wavelength, and an obstacle diffracts like a gap of the same width.
Worked exampleElectrons with a de Broglie wavelength of 2.9 fm are scattered by carbon-12 nuclei, about 5.5 fm across. \(\sin \theta \approx \frac{2.9}{5.5} = 0.53\), so the first minimum is near 32°. The rule gives an estimate: the exact angle depends on how the charge is spread through the nucleus, which is what such experiments measure.
Why very fast electrons?To give a clear minimum the wavelength must be about the size of the nucleus, a few femtometers. That needs a very large momentum, so the electrons come from an accelerator.
Shorter wavelength (more momentum) or a bigger obstacle moves the minimum to a smaller angle.
Electrons are used because they do not feel the strong nuclear force, so the pattern shows the size of the nucleus cleanly, unlike the alpha particles in E.1.


      Putting it together
Wave–particle duality.
            Behaves as a waveBehaves as a particle

            LightDiffraction, interferencePhotoelectric effect, Compton scattering
ElectronsDiffraction by crystalsEach arrives at one point, with charge and mass

          Neither picture alone is complete. Which one shows up depends on the experiment.
E.2 QUANTUM PHYSICS · HL ONLY
What duality does and does not sayIt does not say an electron is sometimes a little ball and sometimes a ripple. It says that “wave” and “particle” are both pictures taken from everyday objects, and that something as small as an electron or a photon fits neither one completely.
A useful guideHow it travels (spreading through gaps, forming patterns) is described by a wave. How it arrives or exchanges energy (in one place, in one lump) is described by a particle.
The two equations that tie the pictures together: \(E = hf\) and \(p = h/\lambda\). In each, a particle quantity on the left equals \(h\) times a wave quantity.


      Quick check
Matter waves.01State the de Broglie relation.
\(\lambda = h/p\)
02An electron’s speed doubles. Its wavelength?
Halves
03What is the evidence that electrons have a wave nature?
They diffract: rings from a thin crystal
04The accelerating p.d. is raised. What happens to the rings?
They get smaller: shorter wavelength
05Why does a tennis ball not diffract?
Its wavelength is far smaller than any gap
06Where is the first minimum for an obstacle of width \(D\)?
\(\sin \theta \approx \lambda /D\)
PRESS → TO REVEAL EACH ANSWER


      
      Part 03
Compton scattering.A photon collides with an electron
E.2 QUANTUM PHYSICS · HL ONLY


      The story · St Louis, 1922
Background story · not assessed
Compton plays billiards with light.
            Arthur Compton, pictured in 1927, the year of his Nobel Prize
          
Compton, aged 30, scatters X-rays off graphite at Washington University
The scattered X-rays have a longer wavelength. No wave theory allowed that
He treats it as a collision between one photon and one electron. The numbers fit
1927: the Nobel Prize, shared with C. T. R. Wilson, whose cloud chamber showed the recoiling electrons
After this, most physicists accepted the photon.
Photo: unknown photographer, public domain.
E.2 QUANTUM PHYSICS · HL ONLY


      Compton (1923)
X-rays that change wavelength.X-rays are scattered by electrons
The scattered X-rays have a longer wavelength
The bigger the angle, the bigger the change
The electron recoils
A collision between two particles.
ONE PHOTON, ONE ELECTRONNOT TO SCALE
~ Photon− Electron
E.2 QUANTUM PHYSICS · HL ONLY
Why a longer wavelength?The photon gives some of its energy to the electron. A photon cannot slow down: it always travels at \(c\). The only way it can have less energy is a lower frequency, \(E = hf\), and so a longer wavelength.
What the wave theory expectedA wave shaking an electron makes it oscillate at the wave’s own frequency, and the electron then radiates at that same frequency. The scattered X-rays should have had exactly the wavelength of those going in.
Compton used X-rays on graphite. The outer electrons of carbon are held so weakly compared with an X-ray photon’s energy that they behave as free electrons at rest.


      The Compton formula
The shift in wavelength.\(\lambda _{\mathrm{f}} - \lambda _{\mathrm{i}} = \dfrac{h}{m_{\mathrm{e}}c}(1 - \cos \theta )\)
\(\theta\): angle the photon is turned through
\(h/m_{\mathrm{e}}c = \class{res}{2.43 \times 10^{-12} \,\mathrm{m}}\)
The shift depends only on the angle

            Angle \(\theta\)\(1 - \cos \theta\)Shift \(\Delta \lambda\)

            0°00
90°12.43 pm
180°24.85 pm

          
E.2 QUANTUM PHYSICS · HL ONLY
Reading the formulaA photon that carries straight on \((\theta = 0)\) has not interacted, so nothing changes. One that bounces straight back \((\theta = 180^{\circ})\) has hit the electron hardest and gives it the most energy, so it has the largest shift.
What is not in the formulaNeither the starting wavelength nor the material appears. Every photon scattered through 90° by a free electron gains 2.43 pm, whatever it started with.
Why X-rays are neededThe shift is at most 4.85 pm. For X-rays of about 70 pm that is a few percent and can be measured. For visible light of 500 000 pm it is far too small to notice.
The formula is in the data booklet and its derivation is not required. \(1 \,\mathrm{pm} = 10^{-12} \,\mathrm{m}\).
Leaving the calculator in radians when the angles are in degrees. Check with \(\cos 90^{\circ} = 0\).


      Worked example
An X-ray scattered through 60°.
            Incoming wavelength 71.0 pm
            \(\Delta \lambda = \dfrac{h}{m_{\mathrm{e}}c}(1 - \cos \theta )\)
              \(= 2.43 \times (1 - \cos 60^{\circ}) = \class{res}{1.2 \,\mathrm{pm}}\)
            \(\lambda _{\mathrm{f}} = \lambda _{\mathrm{i}} + \Delta \lambda = 71.0 + 1.2 = \class{res}{72.2 \,\mathrm{pm}}\)
            Before: \(\dfrac{hc}{\lambda _{\mathrm{i}}} = \dfrac{1.24 \times 10^{-6}}{71.0 \times 10^{-12}} = 17.46 \,\mathrm{keV}\)
            After: \(\dfrac{hc}{\lambda _{\mathrm{f}}} = \dfrac{1.24 \times 10^{-6}}{72.2 \times 10^{-12}} = 17.17 \,\mathrm{keV}\)
          

            The photon has lost about 0.29 keV
The electron has gained it as kinetic energy

          
E.2 QUANTUM PHYSICS · HL ONLY
Two steps, never oneThe formula gives the change in wavelength. Add it to the starting wavelength to get the final one. The energy the electron receives is then the difference between the two photon energies.
Keep the extra figuresThe two photon energies differ by under 2%, so rounding them early would wipe out the answer. Carry four significant figures until the subtraction is done.
Giving 1.2 pm as the final wavelength. It is the shift.


      Nature of science
Like two balls colliding?THE SAMEDIFFERENT
Energy is conserved
≠The photon has no mass
Momentum is conserved
≠It never slows: always at \(c\)
The target recoils, the projectile is deflected
≠Losing energy means a longer wavelength
A photon has momentum, \(p = h/\lambda\). Only a particle model predicts the shift.
E.2 QUANTUM PHYSICS · HL ONLY
Why is this stronger evidence than the photoelectric effect?The photoelectric effect shows that light delivers energy in lumps. Compton scattering shows more: the photon also has momentum, and energy and momentum are both conserved in a collision with a single electron, exactly as for two particles. The measured shift at every angle matches that calculation.
And it is seen directlyHere the photon is still there afterwards, traveling in a new direction with a new wavelength that can be measured. In the photoelectric effect the photon disappears.
For a photon, \(p = h/\lambda = E/c\). It is the same relation as de Broglie’s, now applied to light.


      Summary
Three experiments, one picture.Photoelectric effect
Light as particlesThreshold frequency, no delay
\(E_{\mathrm{max}} = hf - \Phi\)
Electron diffraction
Matter as wavesRings from a crystal
\(\lambda = h/p\)
Compton scattering
Photons have momentumWavelength shift with angle
\(\Delta \lambda = (h/m_{\mathrm{e}}c)(1 - \cos \theta )\)
Light and matter both show wave–particle duality.
E.2 QUANTUM PHYSICS · HL ONLY


      Quick check
Compton scattering.01What happens to the wavelength of a scattered photon?
It increases
02Why does it increase?
The photon gives energy to the electron: lower \(f\)
03At which angle is the shift greatest?
180°: 4.85 pm
04Which two quantities are conserved?
Energy and momentum
05Shift for scattering through 90°?
2.43 pm
06What would the wave theory predict?
No change in wavelength
PRESS → TO REVEAL EACH ANSWER


      Summary
Key words.Photoelectric effectLight frees electrons from a metal
Threshold frequency\(f_{0}\): lowest that gives emission\(hf_{0} = \Phi\)
Work function\(\Phi\): least energy to leave the metal
Photoelectric equation\(E_{\mathrm{max}} = hf - \Phi\)
Stopping potential\(eV_{\mathrm{s}} = E_{\mathrm{max}}\)
de Broglie wavelength\(\lambda = h/p\)
Electron diffractionEvidence that matter has a wave nature
Compton scatteringPhoton hits electronLonger wavelength
Wave–particle dualityLight and matter show both behaviors
E.2 QUANTUM PHYSICS · HL ONLY


    
Radiation	Wavelength	Photon energy
Red light	700 nm	1.77 eV
Violet light	400 nm	3.10 eV
Ultraviolet	250 nm	4.96 eV
	A wave would…	What happens
Frequency	Work at any frequency, if bright enough	Nothing below \(f_{0}\), however bright
Time	Need time to build up energy in dim light	Emission is immediate
Intensity	Give each electron more energy when brighter	Same maximum energy: only more electrons
Energy	Not depend on frequency	Maximum energy rises with frequency
	Behaves as a wave	Behaves as a particle
Light	Diffraction, interference	Photoelectric effect, Compton scattering
Electrons	Diffraction by crystals	Each arrives at one point, with charge and mass