Generated ERQ

✓ passed E.2 Quantum physics (HL) × C.3 Wave phenomena 14 marks HL 3 passes 123.86s $0.8186
## ERQ · 14 marks · Topics: E.2 Quantum physics (HL) + C.3 Wave phenomena · Archetype: experimental_analysis **Integration:** primary=E.2 Quantum physics (HL), secondary=C.3 Wave phenomena (strength: supporting) **Stem.** A student investigates the photoelectric effect using a vacuum phototube whose cathode is coated with caesium. Light from a mercury discharge lamp passes through a transmission diffraction grating of slit spacing d = 1.67 × 10⁻⁶ m, and a narrow slit selects one spectral line at a time before the light strikes the cathode. The anode is held at a variable negative potential and the stopping potential V_s is measured for each spectral line. The collected data are: | Spectral line | Frequency f / 10¹⁴ Hz | Stopping potential V_s / V | |---|---|---| | Yellow | 5.19 | 0.27 | | Green | 5.49 | 0.39 | | Blue | 6.88 | 0.96 | | Violet | 7.41 | 1.18 | | UV | 8.20 | 1.50 | A graph of V_s against f is straight, with gradient (4.10 ± 0.15) × 10⁻¹⁵ V s and y-intercept (−1.86 ± 0.08) V. The student wishes to determine the Planck constant h and the work function φ of caesium, and later uses the grating to investigate the role of intensity. ### Part (a) State [2 marks] · AO1 · Topic: E.2 State the Einstein photoelectric equation and identify how the gradient and the y-intercept of the V_s–f graph relate to h and φ. ### Part (b)(i) Determine [3 marks] · AO2 · Topic: E.2 Using the gradient of the graph, determine an experimental value for the Planck constant, and state its absolute uncertainty. ### Part (b)(ii) Calculate [4 marks] · AO2|AO3 · Topic: E.2 Using the y-intercept, calculate the work function of caesium in electronvolts and its absolute uncertainty. Comment on whether the experimental value is consistent with the accepted value φ = 2.14 eV. ### Part (c) Derive [3 marks] · AO2|AO3 · Topic: E.2 + C.3 The yellow spectral line is selected by setting the grating at a first-order diffraction angle θ. Using the grating equation and the data for the yellow line, derive an expression for θ in terms of d, c and f, substitute the numerical values, and calculate θ in degrees. ### Part (d) Explain [2 marks] · AO3 · Topic: E.2 + C.3 · ASSUMPTIONS DISCRIMINATOR The student now widens the slit illuminating the grating, increasing the intensity of the yellow line falling on the cathode without changing the grating angle. Explain why the measured stopping potential for the yellow line is unchanged, while the photocurrent increases. --- ## Mark Scheme ### Part (a) [2 marks] — State - M1: Einstein equation stated as eV_s = hf − φ (or E_k,max = hf − φ with eV_s = E_k,max). [no ECF] - M2: Gradient = h/e AND y-intercept = −φ/e (both required). [no ECF] ### Part (b)(i) [3 marks] — Determine - M1: h = gradient × e = (4.10 × 10⁻¹⁵) × (1.60 × 10⁻¹⁹). [no ECF] - M2: h = 6.56 × 10⁻³⁴ J s (accept 6.5–6.6 × 10⁻³⁴). [ECF from (a)] - M3: Δh = (0.15/4.10) × h = 0.24 × 10⁻³⁴ J s ≈ 0.2 × 10⁻³⁴ J s; quote as (6.6 ± 0.2) × 10⁻³⁴ J s. [ECF from M2] ### Part (b)(ii) [4 marks] — Calculate + Comment - M1: φ = |y-intercept| × e = 1.86 eV (accept use of e directly since intercept is in volts → φ in eV numerically equals 1.86). [no ECF] - M2: Δφ = 0.08 eV, so φ = (1.86 ± 0.08) eV. [ECF from M1] - M3: Difference from accepted value = 2.14 − 1.86 = 0.28 eV, which exceeds the uncertainty 0.08 eV (≈ 3.5σ). [ECF from M2] - M4: Comment: result is NOT consistent with the accepted value — systematic error likely (e.g. contact potential between cathode and anode, surface oxidation lowering apparent φ, or stray light). [ECF from M3] ### Part (c) [3 marks] — Derive - M1: Grating equation d sin θ = mλ with m = 1 and λ = c/f → sin θ = c/(d f). [no ECF] - M2: Substitution: sin θ = (3.00 × 10⁸)/[(1.67 × 10⁻⁶)(5.19 × 10¹⁴)] = 0.3463. [ECF from M1] - M3: θ = arcsin(0.3463) = 20.3° (accept 20.2°–20.4°). [ECF from M2] ### Part (d) [2 marks] — Explain (causal chain per §4.4.1) - M1: At fixed grating angle θ and spacing d, the grating equation d sin θ = mλ fixes the wavelength λ (and hence the photon frequency f = c/λ) of the light reaching the cathode, regardless of how wide the input slit is opened. - M2: Therefore widening the slit increases the number of photons per second arriving but does not change each photon's energy hf, so by V_s = hf/e − φ/e the stopping potential is unchanged, while the rate of electron ejection — and hence the photocurrent — rises in proportion to the photon flux. [linking marking point: fixed λ → fixed hf → fixed V_s; higher flux → higher current] ### Marker notes - Part (b)(i): accept fractional uncertainty propagated as (Δgradient/gradient) applied to h; do not penalise rounding 0.24 → 0.2. - Part (b)(ii): accept any plausible systematic error in M4 (contact potential, anode work function, oxidised cathode, finite anode work function, residual gas). Reject "random error" or "human error" alone. - Part (c): alternative — students who compute λ = c/f = 578 nm first, then sin θ = λ/d = 0.346, then θ = 20.3°, earn M1+M2+M3 by equivalent route. - Part (d): do NOT award M2 unless the candidate explicitly links fixed frequency to unchanged V_s. Stating only "intensity doesn't affect V_s" without invoking the grating-fixed wavelength scores M2 only if M1 already establishes the wavelength selection mechanism.