Photoelectric effect answers should name the classical failure and the photon explanation. More intensity at too-low frequency does not eventually eject electrons, because each photon still lacks enough energy.

When light shines on a metal surface, electrons are emitted only if the light frequency exceeds a threshold frequency.

Below threshold frequency, no photoelectrons are emitted regardless of intensity.

Above threshold frequency, emission is effectively immediate.

Increasing intensity increases the number of emitted photoelectrons, not their maximum kinetic energy.

Einstein explained the observations by treating light as photons with energy E = hf.

One photon transfers energy to one electron; if photon energy exceeds the work function, the electron can escape.

Interpret the photoelectric-effect observations.

Graph

A photoelectric experiment compares three cases: high intensity below threshold gives no current, low intensity above threshold gives immediate current, and high intensity above threshold gives larger current.

1identify the threshold-frequency observation

2separate intensity from maximum kinetic energy

3state photon energy E = hf

Which observation cannot be explained by classical wave theory, and what photon idea explains it?

Explain why the photoelectric effect is evidence for photons.

Only saying electrons are emitted when light shines on metal.

Explain why the photoelectric effect is evidence for photons.

Choose

Photoelectric graphs are full of intercepts. Read the x-intercept as threshold frequency, the y-intercept as negative work function, and the gradient as Planck’s constant.

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

The threshold frequency is f0 = φ/h.

Below f0, no photoelectrons are emitted regardless of intensity.

For a graph of maximum kinetic energy against frequency, the gradient is h, the y-intercept is -φ, and the x-intercept is f0.

Stopping voltage Vs measures maximum kinetic energy using e Vs = E_k(max).

A current-voltage graph reaches zero current at the stopping potential for the fastest electrons.

Interpret photoelectric graph features.

Graph

A graph of E_k(max) against frequency is a straight line crossing the frequency axis at f0 and the energy axis at -φ. A current-voltage graph shows current falling to zero at -Vs.

1read gradient and intercepts

2link stopping voltage to kinetic energy

3distinguish intensity effects from kinetic-energy effects

How do you identify h, φ, f0, and E_k(max) from the graphs?

Interpret a graph of maximum kinetic energy against frequency for the photoelectric effect.

Using the graph height to identify intensity rather than kinetic energy.

Interpret a graph of maximum kinetic energy against frequency for the photoelectric effect.

Choose

Einstein’s equation is an energy budget for one electron absorbing one photon. Photon energy first pays the work function; the remainder is kinetic energy of the fastest emitted electron.

A photon incident on a metal has energy hf.

The work function φ is the minimum energy required to release an electron from the surface.

Any photon energy beyond the work function becomes maximum kinetic energy: E_k(max) = hf - φ.

The stopping voltage relation is eV_s = E_k(max).

Increasing frequency increases maximum kinetic energy; increasing intensity above threshold increases photoelectron number but not E_k(max).

Assemble the photoelectric equation.

Formula

Target formula E_k(max) = hf - φ

E_k(max)

maximum kinetic energy of emitted photoelectrons

J or eV

h

Planck constant

J s

f

light frequency

Hz

φ

work function

J or eV

V_s

stopping voltage

V

e

elementary charge

C

1Write photon energy.E = hf

2Subtract the work function to get maximum kinetic energy.E_k(max) = hf - φ

3At threshold, kinetic energy is zero.hf0 = φ

4Link stopping voltage to kinetic energy.eV_s = E_k(max)

Build the formula

Solve a photoelectric-effect problem using Einstein’s equation.

Putting intensity into E_k(max)=hf-φ.

Solve a photoelectric-effect problem using Einstein’s equation.

Choose

Electron diffraction is the key matter-wave experiment. Do not just say “electrons make rings”; explain why rings are wave evidence and how changing voltage changes wavelength.

Accelerated electrons passing through a thin graphite film produce diffraction rings on a screen.

Diffraction is a wave behaviour, so the pattern is evidence for the wave nature of electrons.

The pattern is similar in character to X-ray diffraction by crystal planes.

Increasing the accelerating voltage increases electron momentum.

Higher momentum means shorter de Broglie wavelength and a smaller diffraction angle or ring diameter.

Repair the electron-diffraction claims.

Spot Errors

Describe how electron diffraction provides evidence for wave-particle duality.

Saying diffraction rings are just a particle spray pattern.

Describe how electron diffraction provides evidence for wave-particle duality.

Choose

Wave-particle duality is not vague “sometimes wave, sometimes particle” language. It is a summary of experimental evidence: diffraction/interference for waves, photoelectric and Compton effects for particle-like photons, and electron diffraction for matter waves.

Light shows wave behaviour in interference and diffraction.

Light shows particle behaviour in the photoelectric effect and Compton scattering.

Matter such as electrons shows particle behaviour in tracks and localized detection.

Matter shows wave behaviour in diffraction experiments.

De Broglie proposed that particles have wavelength λ = h/p.

Wave-particle duality means quantum objects cannot be described fully by only a classical wave model or only a classical particle model.

Match each evidence cue to the wave-particle duality claim.

Match

Reasons

0/5

Discuss evidence for wave-particle duality of light and matter.

Giving only definitions without experimental evidence.

Discuss evidence for wave-particle duality of light and matter.

Choose

De Broglie wavelength is an inverse-momentum relation. In electron diffraction, increasing accelerating voltage increases momentum, so wavelength decreases and diffraction angle decreases.

De Broglie proposed that particles have wavelength λ = h/p.

For a non-relativistic particle, p = mv, so λ = h/(mv).

Greater momentum gives shorter de Broglie wavelength.

For an electron accelerated through potential difference V, kinetic energy is eV = p^2/(2m).

Therefore λ = h/sqrt(2meV) for a non-relativistic accelerated electron.

Assemble de Broglie wavelength equations.

Formula

Target formula λ = h/p

λ

de Broglie wavelength

m

h

Planck constant

J s

p

particle momentum

kg m s^-1

m

particle mass

kg

v

particle speed

m s^-1

V

accelerating potential difference

V

1Start from de Broglie’s relation.λ = h/p

2For non-relativistic motion use momentum p=mv.λ = h/(mv)

3For an accelerated electron, relate voltage to kinetic energy.eV = p^2/(2m)

4Substitute momentum.λ = h/sqrt(2meV)

Build the formula

Calculate the de Broglie wavelength of a particle or accelerated electron.

Using photon wavelength equations instead of λ = h/p.

Calculate the de Broglie wavelength of a particle or accelerated electron.

Choose

Compton scattering is the collision evidence for photons. The wavelength shift is not just “redder light”; it shows the photon lost energy and momentum to an electron, so photons carry momentum.

Compton scattering involves X-ray photons scattering from electrons.

The scattered photons are observed with longer wavelength than the incident photons.

A longer wavelength means lower photon energy.

The result is explained by treating photons as particles with momentum p = h/λ.

During the collision, the photon transfers energy and momentum to the electron.

Classical wave theory did not predict the observed wavelength shift.

Connect Compton observations to photon momentum.

Formula

Target formula p = h/λ

p

photon momentum

kg m s^-1

h

Planck constant

J s

λ

photon wavelength

m

E

photon energy

J

θ

scattering angle

degrees or radians

1Assign photon momentum.p = h/λ

2Relate photon energy to wavelength.E = hc/λ

3Describe collision transfer.photon transfers energy and momentum to electron

4Interpret the scattered photon.λ_scattered > λ_incident -> lower photon energy

Build the formula

Explain how Compton scattering supports the particle nature of light.

Saying the photon wave simply bends with no energy transfer.

Explain how Compton scattering supports the particle nature of light.

Choose

This card is the qualitative story behind the formula. Photon loses energy to the electron; lower energy means larger wavelength. The scattering angle controls how much shift occurs.

In Compton scattering, an incident photon collides with an electron.

The electron recoils and gains kinetic energy.

The scattered photon therefore has less energy than the incident photon.

Since photon energy is E = hc/λ, lower energy means longer wavelength.

The amount of wavelength increase depends on scattering angle.

The photon is not absorbed and re-emitted as an atomic spectral line in this model; it is scattered in a collision-like interaction.

Put the Compton wavelength-increase explanation in order.

Order

1

Electron recoils with kinetic energy.

2

Incident X-ray photon approaches an electron.

3

Scattered photon has lower energy than before.

4

Because E = hc/λ, the photon wavelength is longer.

5

Photon scatters from the electron and transfers energy and momentum.

Explain why the wavelength of a photon increases in Compton scattering.

Saying the photon slows down.

Explain why the wavelength of a photon increases in Compton scattering.

Choose

The formula tells you how much the scattered photon wavelength increases. It is an angle formula: no deflection gives no shift, backscatter gives the largest shift.

The Compton wavelength shift is Δλ = λ_f - λ_i = h/(m_ec)(1 - cosθ).

m_e is the electron mass and θ is the photon scattering angle.

The derivation is not required for IB; use the formula and interpret its angle dependence.

At θ = 0°, cosθ = 1, so Δλ = 0.

At θ = 180°, cosθ = -1, so the shift is maximum: Δλ = 2h/(m_ec).

Assemble the Compton shift formula.

Formula

Target formula Δλ = h/(m_ec)(1 - cosθ)

Δλ

increase in photon wavelength, λ_f - λ_i

m

h

Planck constant

J s

m_e

electron mass

kg

c

speed of light

m s^-1

θ

scattering angle of photon

degrees or radians

1Use the Compton shift equation.Δλ = h/(m_ec)(1 - cosθ)

2Check no-deflection case.θ = 0° -> Δλ = 0

3Check backscatter case.θ = 180° -> Δλ = 2h/(m_ec)

4Find final wavelength if needed.λ_f = λ_i + Δλ

Build the formula

Calculate or interpret the Compton wavelength shift for a scattered photon.

Forgetting that Δλ is λ_f - λ_i and must be added to the incident wavelength.

Calculate or interpret the Compton wavelength shift for a scattered photon.

Choose

E.2 is an evidence map, not a formula list. The photoelectric effect and Compton scattering support particle-like photons; electron diffraction supports matter waves; de Broglie links particle momentum to wavelength.

Photons are quanta with energy E = hf.

The photoelectric effect shows particle-like light: emission requires f ≥ f0, where f0 = φ/h, and E_k(max) = hf - φ = eV_s.

Photoelectric graphs give gradient h, y-intercept -φ, and x-intercept f0.

Matter has wave-like properties: de Broglie wavelength λ = h/p and electron diffraction through graphite gives diffraction rings.

For accelerated electrons, λ = h/sqrt(2meV) in the non-relativistic model.

Compton scattering shows photon momentum p = h/λ and wavelength shift Δλ = h/(m_ec)(1 - cosθ).

Match each E.2 cue to the quantum model it retrieves.

Match

Reasons

0/8

Summarize the E.2 quantum physics evidence and equations.

Listing equations without connecting them to experiments.

Summarize the E.2 quantum physics evidence and equations.

Choose

IB Physics HL E.2 Quantum physics Notes

Interpret Photoelectric effect

Interpret Threshold frequency

Interpret Photoelectric equation

Model Particle diffraction

Interpret Matter wave-particle duality

Interpret de Broglie wavelength

Interpret Compton scattering

Explain Photon wavelength increase

Interpret Compton wavelength shift

Retrieve the E.2 Quantum physics Model