22. Quantum physics A2

Use the model that explains the observation; neither classical wave nor classical particle language alone covers every experiment.

A diffraction pattern supports wave behaviour, while the photoelectric effect requires discrete energy transfers.

Photons travel at c in vacuum and are absorbed or emitted as whole quanta in interactions.

A higher-frequency ultraviolet photon carries more energy than a visible red photon.

For λ=630 nm, E=hc/λ=(6.63×10⁻³⁴)(3.00×10⁸)/(630×10⁻⁹)=3.16×10⁻¹⁹ J=1.97 eV.

For photon energy 74 keV, first convert E=74×10³×1.60×10⁻¹⁹ J, then λ=hc/E=1.68×10⁻¹¹ m.

A 1.0×10⁻² W beam of 1.97 eV photons emits (1.0×10⁻²)/(1.97×1.60×10⁻¹⁹)=3.2×10¹⁶ photons s⁻¹.

Use eV for particle-scale energies and convert to joules when applying SI equations or comparing macroscopic work.

A 5.0 eV electron has energy about 8.0×10⁻¹⁹ J.

For E=3.11×10⁻¹⁹ J, p=E/c=(3.11×10⁻¹⁹)/(3.00×10⁸)=1.04×10⁻²⁷ kg m s⁻¹ (or N s).

Increasing intensity above threshold increases the number of emitted electrons, while frequency controls each photon’s energy and maximum electron speed.

Dim ultraviolet light can eject electrons while bright red light may not if red photon energy is below threshold.

Below f₀ or above λ₀ no electrons are emitted regardless of intensity in the ideal model.

A metal with threshold frequency 6.0×10¹⁴ Hz has threshold wavelength about 500 nm.

Use Φ as the minimum energy needed to escape the surface and connect it to threshold frequency via Φ=hf₀.

Increasing frequency above threshold increases maximum photoelectron kinetic energy by the excess photon energy.

One photon supplies hf. The maximum-energy electrons spend Φ escaping and keep the remainder Kmax; a stopping potential Vs removes that maximum kinetic energy because Kmax=eVs.

For f=1.10×10¹⁵ Hz and Φ=5.80×10⁻¹⁹ J: Kmax=hf−Φ=1.49×10⁻¹⁹ J, so vmax=√(2Kmax/me)=5.72×10⁵ m s⁻¹ and Vs=Kmax/e=0.933 V.

Different metals give parallel Kmax–f lines because h is universal. A larger work function shifts the threshold to a larger frequency and the extrapolated intercept to a more negative value.

At a fixed frequency above threshold, raising intensity sends more photons per second, so the photocurrent can rise, but each photon still supplies the same energy and K_max is unchanged.

Doubling the intensity of ultraviolet light doubles the possible emission rate in an ideal experiment, not the stopping potential.

Photoelectric effect → particulate evidence: energy arrives in photons E=hf. A threshold frequency, immediate emission and Kmax depending on frequency rather than intensity follow from one-electron/one-photon transfer.

Interference and diffraction → wave evidence: overlapping alternatives produce stable maxima and minima through constructive and destructive superposition.

Dim radiation above threshold can eject electrons immediately, while bright sub-threshold radiation cannot; yet the same electromagnetic radiation can form interference fringes or diffract through an aperture.

The pattern is explained by a wavelength associated with the moving electrons; changing electron momentum changes the spacing of maxima and minima.

A ring pattern from a polycrystalline film is evidence of constructive interference from many crystal orientations.

For a non-relativistic electron p=mv; use the momentum actually delivered by the apparatus, not simply the particle’s rest mass.

Accelerating an electron through a larger potential difference increases its speed and reduces λ, changing the diffraction scale.

If speed is known and v is far below c, calculate p=mv first; keep units consistent before comparing λ with a slit or lattice spacing.

For a 9.11×10⁻³¹ kg electron moving at 2.0×10⁶ m s⁻¹, λ is about 0.36 nm, comparable with atomic spacings.

An electron changes level by absorbing or emitting a photon whose energy equals the gap: ΔE=hf. Larger gaps correspond to higher-frequency, shorter-wavelength lines.

A hydrogen emission spectrum contains separate lines because only certain transitions are allowed between its levels.

Excited electrons later fall and re-emit photons, but in random directions. Along the original viewing direction, fewer photons remain at the absorbed wavelengths, so the dark lines persist.

For the same element, absorption and emission lines occur at the same characteristic wavelengths because both use the same energy-level gaps.

Each line has one frequency f and photon energy hf. Discrete line frequencies therefore imply discrete photon energies, discrete energy gaps and hence discrete atomic electron levels.

Downward transition: one photon is emitted. Upward transition: one matching photon is absorbed. In both cases photon energy is the positive magnitude of the level difference.

For hydrogen E3=−1.51 eV and E2=−3.40 eV, the n=3→2 gap is (−1.51)−(−3.40)=1.89 eV=3.02×10⁻¹⁹ J.

Then f=ΔE/h=(3.02×10⁻¹⁹)/(6.63×10⁻³⁴)=4.56×10¹⁴ Hz and λ=c/f=6.58×10⁻⁷ m=658 nm.

To find an unknown upper level from an emitted photon, use Eupper=Elower+hf. Keep signed level energies until after the subtraction.