Quantum theory — Question 272

ANSWER

(a) Parallel positive-slope lines; A begins at $4.53\times10^{14}\,\mathrm{Hz}$ and B at $6.04\times10^{14}\,\mathrm{Hz}$. A lies above B for frequencies where both emit. (b) Neither moves; intensity changes emission rate, not the energy per photon or maximum electron energy.

Worked solution

(a) $E_{k,\max}=hf-W$ has common slope $h$. The intercepts with $E_k=0$ are $W_A/h=4.53\times10^{14}\,\mathrm{Hz}$ and $W_B/h=6.04\times10^{14}\,\mathrm{Hz}$. Below threshold there is no photoemission, not a negative photoelectron kinetic energy.

(b) At fixed frequency photon energy is unchanged, so the maximum kinetic energy at each frequency is unchanged.

Equivalent physically justified methods, alternative valid sketches and consistent units accepted.

Accept sensible significant figures and consistent rounding from stated constants.

Part a: Draw positive, equal-gradient lines with A above B in their common domain.

[1 mark]

Part a: Label the two correct threshold frequencies.

[1 mark]

Part a: Restrict the drawn emission lines to nonnegative kinetic energy.

[1 mark]

Part b: State no movement and justify using unchanged photon energy.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.