E.2 Quantum physics HL

Read the observations

When monochromatic light illuminates a metal, electrons may be emitted. Increasing intensity increases the emission rate, but for fixed frequency it does not increase the maximum kinetic energy of the emitted electrons.

Use the photon model

Light transfers energy in individual photons. One photon interacts with one electron, so photon frequency sets the energy available per interaction, while intensity changes the number of photons arriving per second.

Identify the evidence

The intensity–energy distinction and the existence of a threshold frequency cannot be explained by a simple continuous wave-energy model. They support the particle nature of light.

Common trap

Do not say that brighter light makes each photoelectron more energetic. At fixed frequency, it produces more emitted electrons, not a larger maximum kinetic energy.

Worked example — threshold frequency

For $\Phi=2.70\,\mathrm{eV}=(2.70)(1.60\times10^{-19})=4.32\times10^{-19}\,\mathrm{J}$, $f_0=\Phi/h=(4.32\times10^{-19})/(6.63\times10^{-34})=6.52\times10^{14}\,\mathrm{Hz}$. Brighter light below this frequency still ejects no electrons.

Explain the intensity result

Below $f_0$, each photon has too little energy to overcome the work function. Increasing intensity supplies more low-energy photons, but it does not make any one photon energetic enough, so no electrons are emitted.

Keep the metal fixed

Threshold frequency depends on the metal’s work function. Two metals illuminated by the same radiation can behave differently because their electron-binding energies differ.

Common trap

Do not explain the threshold in terms of total light energy accumulated over time. The interaction is photon-by-photon.

Worked example — 420 nm light

For $\lambda=420\,\mathrm{nm}$, photon energy is $hc/\lambda=2.96\,\mathrm{eV}$. With $\Phi=2.0\,\mathrm{eV}$, $E_{k,\max}=2.96-2.0=0.96\,\mathrm{eV}$. The result is positive, so emission occurs; a negative calculated remainder would mean no emission.

Use wavelength when given

Because $f=c/\lambda$, write $E_{k,max}=\frac{hc}{\lambda}-\Phi$. Keep $hc/\lambda$ and $\Phi$ in the same energy unit before subtracting.

Find maximum speed

Once $E_{k,max}$ is known, use $E_{k,max}=\frac12m_ev_{max}^2$, so $v_{max}=\sqrt{2E_{k,max}/m_e}$.

Common trap

Do not add the work function to the kinetic energy. The work function is the energy already spent escaping the surface.

Read the diffraction pattern

A beam of particles can produce diffraction or interference patterns after passing through a suitable crystal or narrow structure. The pattern is evidence that the particles have wave-like behaviour.

Use the experiment as evidence

Electron-diffraction experiments demonstrate wave properties of electrons. This complements the photon evidence from the photoelectric effect: matter and radiation can each show both particle-like and wave-like behaviour.

Connect to wavelength

The wave description is quantified by the de Broglie wavelength $\lambda=h/p$. A shorter wavelength generally requires a larger momentum.

Common trap

Rutherford alpha scattering is evidence for the nuclear structure of the atom, not the clearest evidence for matter waves. Use particle diffraction or interference when the question asks for wave properties of electrons.

Hold both descriptions

Quantum objects can show particle-like and wave-like behaviour. The observed property depends on the experiment: photoelectric emission and Compton scattering reveal particle-like transfers, while diffraction reveals wave-like behaviour.

Do not combine classical pictures blindly

Wave-particle duality is not a claim that an object is simultaneously a classical wave and a classical particle. It is a quantum description in which different measurements reveal complementary aspects.

Use scale carefully

For macroscopic objects the de Broglie wavelength is extremely small because momentum is large, so wave effects are not normally detectable. This is a practical limit, not a loss of the relation $\lambda=h/p$.

Common trap

Do not use “wave-particle duality” as an explanation without naming the observation it explains. Match the experiment to the property it demonstrates.

Worked example — moving electron

For $m_e=9.11\times10^{-31}\,\mathrm{kg}$ and $v=5.0\times10^6\,\mathrm{m\,s^{-1}}$, $p=mv=4.56\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}$. Hence $\lambda=h/p=(6.63\times10^{-34})/(4.56\times10^{-24})=1.5\times10^{-10}\,\mathrm{m}$, comparable with atomic spacing.

Choose the momentum form

For non-relativistic motion, use $p=mv$, so $\lambda=h/(mv)$. If kinetic energy is given, use $E_k=p^2/(2m)$ and $p=\sqrt{2mE_k}$.

Scale with accelerating voltage

For an electron accelerated from rest through potential $V$, $eV=E_k$, so $p\propto\sqrt V$ and $\lambda\propto1/\sqrt V$. Quadrupling $V$ halves the wavelength.

Common trap

Do not use $h/E_k$ as the wavelength. The denominator is momentum, not kinetic energy.

Model the collision

In Compton scattering, a photon transfers energy and momentum to an electron. The scattered photon has a changed direction and wavelength, while the electron recoils.

Use the evidence

The measured wavelength shift is evidence that photons carry momentum as well as energy. A wave-only model does not account for the collision-like transfer in the same way.

Track conservation laws

Analyse the photon–electron event using conservation of energy and momentum. The photon’s lost energy becomes kinetic energy of the recoiling electron, with the remaining photon energy determining its new wavelength.

Common trap

Do not describe Compton scattering as simple reflection. The photon transfers energy and momentum, so its wavelength generally changes.

Follow the energy transfer

After Compton scattering, the photon has transferred energy to the electron. Its final energy is lower than its initial energy.

Convert energy to wavelength

Since $E=hc/\lambda$, lower photon energy means larger wavelength. Therefore the scattered photon has a longer wavelength than the incident photon.

Keep the direction of change

The wavelength shift is zero only for no energy transfer. A stronger transfer to the electron produces a larger positive $\Delta\lambda$, subject to the scattering geometry.

Common trap

Do not infer a shorter wavelength from a lower photon energy. Energy and wavelength are inversely related.

Worked example — $30^\circ$ scattering

Using $h/(m_ec)=2.426\times10^{-12}\,\mathrm{m}$, $\Delta\lambda=(2.426\times10^{-12})(1-\cos30^\circ)=3.25\times10^{-13}\,\mathrm{m}$. The shift is positive and depends on angle, not on the incident wavelength.

Check the limits

For $\theta=0^\circ$, $\Delta\lambda=0$. For back-scattering $\theta=180^\circ$, the shift is maximal at $2h/(m_ec)$. The shift is always non-negative for the usual scattering geometry.

Solve for angle

If $\Delta\lambda$ is given, rearrange to $\cos\theta=1-\frac{m_ec\Delta\lambda}{h}$, then take the inverse cosine and check that the result is physically allowed.

Common trap

Do not use the scattered wavelength itself as $\Delta\lambda$. The equation requires the difference $\lambda_f-\lambda_i$ and the scattering angle.