Specification: AQA A-level Physics (7408) 3.2.2.1 | Paper: 1 | Data and formulae booklet: given in the exam
🛒 All the equations in one place: the AQA A-level Every Key Equation PowerPoint has 127 equations, 381 practice questions and 15 derivations, for £12.
📌 KEY DEFINITIONS – learn them word for word
Work function φ: the minimum energy needed to remove an electron from the surface of a metal.
Threshold frequency f0: the minimum frequency of electromagnetic radiation that will cause photoelectrons to be emitted from the surface, f0 = φ ÷ h. Below f0 no electrons are emitted, however intense the light.
Stopping potential Vs: the minimum potential difference needed to stop the most energetic photoelectrons reaching the collector, so eVs = Ek(max). Increasing the intensity increases the number of photons per second, so more electrons are emitted per second, but it does not change the maximum kinetic energy.
The equation
photon energy = work function + maximum kinetic energy hf = φ + Ek(max)
| Quantity | Symbol | Unit |
|---|---|---|
| Planck constant | h | joule seconds (J s) |
| frequency of the radiation | f | hertz (Hz) |
| work function | φ | joules (J), or electronvolts (eV) |
| maximum kinetic energy of the photoelectrons | Ek(max) | joules (J), or electronvolts (eV) |
| threshold frequency | f0 | hertz (Hz) |
| stopping potential | Vs | volts (V) |
Rearranged: Ek(max) = hf − φ | φ = hf − Ek(max) | f0 = φ ÷ h | Ek(max) = eVs | vmax = √(2Ek(max) ÷ me) | Put every energy in the same unit before you add or subtract. A graph of Ek(max) against f is a straight line of gradient h, crossing the f axis at f0 and the Ek axis at −φ.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) The work function of a metal is 3.2 × 10−19 J. Calculate its threshold frequency, to 2 significant figures. (h = 6.63 × 10−34 J s)
Show solution
F: hf0 = φ
I: 6.63 × 10−34 × f0 = 3.2 × 10−19
F: f0 = 3.2 × 10−19 ÷ 6.63 × 10−34
A: 4.82… × 1014 Hz = 4.8 × 1014 Hz (2 s.f.)
Q2 (Core) Ultraviolet radiation of frequency 1.2 × 1015 Hz falls on zinc, which has a work function of 4.3 eV. Calculate the maximum kinetic energy of the photoelectrons in joules, to 2 significant figures. (h = 6.63 × 10−34 J s; 1 eV = 1.60 × 10−19 J)
Show solution
Convert first: φ = 4.3 × 1.60 × 10−19 = 6.88 × 10−19 J
F: hf = φ + Ek(max)
I: 6.63 × 10−34 × 1.2 × 1015 = 6.88 × 10−19 + Ek(max), so 7.956 × 10−19 = 6.88 × 10−19 + Ek(max)
F: Ek(max) = 7.956 × 10−19 − 6.88 × 10−19
A: 1.076 × 10−19 J = 1.1 × 10−19 J (2 s.f.)
Q3 (Core) Light of frequency 9.0 × 1014 Hz falls on a metal surface. The stopping potential is 1.5 V. Calculate the work function of the metal in eV, to 2 significant figures. (h = 6.63 × 10−34 J s; 1 eV = 1.60 × 10−19 J)
Show solution
Convert first: Ek(max) = eVs = 1.60 × 10−19 × 1.5 = 2.4 × 10−19 J
F: hf = φ + Ek(max)
I: 6.63 × 10−34 × 9.0 × 1014 = φ + 2.4 × 10−19, so 5.967 × 10−19 = φ + 2.4 × 10−19
F: φ = 5.967 × 10−19 − 2.4 × 10−19 = 3.567 × 10−19 J
A: 3.567 × 10−19 ÷ 1.60 × 10−19 = 2.229… eV = 2.2 eV (2 s.f.)
Q4 (Core) Light of wavelength 400 nm shines on sodium, which has a work function of 2.28 eV. Calculate the stopping potential, to 2 significant figures. (h = 6.63 × 10−34 J s; c = 3.00 × 108 m s⁻¹; 1 eV = 1.60 × 10−19 J)
Show solution
Convert first: 400 nm = 4.00 × 10−7 m
Step 1 – photon energy in eV
E = hc ÷ λ = 6.63 × 10−34 × 3.00 × 108 ÷ 4.00 × 10−7 = 4.9725 × 10−19 J = 4.9725 × 10−19 ÷ 1.60 × 10−19 = 3.107… eV
Step 2 – maximum kinetic energy
F: hf = φ + Ek(max)
I: 3.107… = 2.28 + Ek(max)
F: Ek(max) = 3.107… − 2.28 = 0.827… eV
A: eVs = Ek(max), so a kinetic energy of 0.827… eV needs Vs = 0.83 V (2 s.f.)
Watch out: working in eV makes the stopping potential easy (an energy of x eV needs x volts), but only if every energy is in eV. Do not subtract eV from J.
Q5 (Exam) Zinc has a work function of 4.3 eV. (a) Very bright red light of wavelength 650 nm shines on zinc. Explain why no photoelectrons are emitted. (b) Dim ultraviolet of wavelength 250 nm is used instead. Calculate the maximum kinetic energy of the photoelectrons in eV, to 2 significant figures. (h = 6.63 × 10−34 J s; c = 3.00 × 108 m s⁻¹; 1 eV = 1.60 × 10−19 J)
Show solution
Convert first: 650 nm = 6.50 × 10−7 m; 250 nm = 2.50 × 10−7 m
Step 1 – (a) red light
E = hc ÷ λ = 1.989 × 10−25 ÷ 6.50 × 10−7 = 3.06 × 10−19 J = 1.91… eV
each photon has less energy than the work function (1.91 eV < 4.3 eV) and one electron absorbs one photon, so no electron can escape. Brighter light only means more photons per second, each still with too little energy.
Step 2 – (b) ultraviolet
F: hf = φ + Ek(max)
I: E = 1.989 × 10−25 ÷ 2.50 × 10−7 = 7.956 × 10−19 J = 4.9725 eV, so 4.9725 = 4.3 + Ek(max)
F: Ek(max) = 4.9725 − 4.3
A: 0.6725 eV = 0.67 eV (2 s.f.)
Watch out: intensity changes the number of photoelectrons per second, not their energy. Examiners regularly penalise answers saying brighter light gives the electrons more energy.
Q6 (Exam) Potassium has a work function of 2.30 eV. (a) Calculate the threshold wavelength, to 3 significant figures. (b) State whether photoelectrons are emitted when the surface is lit by light of wavelength 450 nm and of wavelength 600 nm. (h = 6.63 × 10−34 J s; c = 3.00 × 108 m s⁻¹; 1 eV = 1.60 × 10−19 J)
Show solution
Convert first: φ = 2.30 × 1.60 × 10−19 = 3.68 × 10−19 J
F: φ = hf0 = hc ÷ λ0
I: 3.68 × 10−19 = 6.63 × 10−34 × 3.00 × 108 ÷ λ0
F: λ0 = 1.989 × 10−25 ÷ 3.68 × 10−19
A: (a) 5.404… × 10−7 m = 540 nm (3 s.f.)
(b) 450 nm is shorter than λ0, so each photon has more energy than φ: emitted. 600 nm is longer than λ0: not emitted.
Watch out: the threshold is a maximum wavelength but a minimum frequency. Shorter wavelengths, not longer ones, release electrons.
Q7 (Exam) A 5.0 mW ultraviolet laser of wavelength 250 nm shines on a metal of work function 2.30 eV. (a) Calculate the maximum kinetic energy of the photoelectrons in eV, to 3 significant figures. (b) Only 1 in 1000 photons releases an electron. Calculate the photoelectric current, to 2 significant figures. (c) State what happens to your answers to (a) and (b) when the laser power is doubled. (h = 6.63 × 10−34 J s; c = 3.00 × 108 m s⁻¹; 1 eV = 1.60 × 10−19 J; e = 1.60 × 10−19 C)
Show solution
Convert first: 250 nm = 2.50 × 10−7 m; 5.0 mW = 5.0 × 10−3 W
Step 1 – (a) maximum kinetic energy
E = hc ÷ λ = 7.956 × 10−19 J = 4.9725 eV
F: Ek(max) = hf − φ
I: Ek(max) = 4.9725 − 2.30
A: 2.6725 eV = 2.67 eV (3 s.f.)
Step 2 – (b) current
photons per second N = P ÷ E = 5.0 × 10−3 ÷ 7.956 × 10−19 = 6.284… × 1015 s⁻¹
electrons per second = 6.284… × 1015 ÷ 1000 = 6.284… × 1012 s⁻¹
I = charge per second = 6.284… × 1012 × 1.60 × 10−19 = 1.005… × 10−6 A
A: 1.0 × 10−6 A (1.0 µA, 2 s.f.)
(c) Ek(max) stays at 2.67 eV (same photon energy); the current doubles to 2.0 µA (twice as many photons per second)
Q8 (Exam) In an experiment, the maximum kinetic energy of photoelectrons is measured for different frequencies. The results lie on a straight line passing through (6.0 × 1014 Hz, 0.50 eV) and (9.0 × 1014 Hz, 1.74 eV) on a graph of Ek(max) in eV against f. Calculate (a) the value of the Planck constant from the graph, to 3 significant figures, (b) the threshold frequency, to 3 significant figures, and (c) the work function in eV, to 3 significant figures. (1 eV = 1.60 × 10−19 J)
Show solution
Step 1 – (a) gradient
gradient = ΔEk ÷ Δf = (1.74 − 0.50) ÷ (9.0 × 1014 − 6.0 × 1014) = 1.24 ÷ 3.0 × 1014 = 4.133… × 10−15 eV s
the gradient is h, so convert to J s: 4.133… × 10−15 × 1.60 × 10−19 = 6.613… × 10−34 J s
A: h = 6.61 × 10−34 J s (3 s.f.)
Step 2 – (b) threshold frequency (where Ek(max) = 0)
f0 = 6.0 × 1014 − 0.50 ÷ 4.133… × 10−15 = 6.0 × 1014 − 1.209… × 1014 = 4.790… × 1014 Hz
A: 4.79 × 1014 Hz (3 s.f.)
Step 3 – (c) work function
F: φ = hf0 (or −1 × the intercept on the Ek axis)
I: φ = 4.133… × 10−15 × 4.790… × 1014
A: 1.98 eV (3 s.f.) (check: at 9.0 × 1014 Hz, 4.133… × 10−15 × 9.0 × 1014 − 1.98 = 1.74 eV)
Watch out: with Ek in eV the gradient is in eV s. Multiply by 1.60 × 10−19 to get h in J s.
Q9 (Stretch) Ultraviolet radiation of wavelength 200 nm falls on a metal with a work function of 4.7 eV. Calculate the maximum speed of the emitted photoelectrons, to 3 significant figures. (h = 6.63 × 10−34 J s; c = 3.00 × 108 m s⁻¹; 1 eV = 1.60 × 10−19 J; me = 9.11 × 10−31 kg)
Show solution
Convert first: 200 nm = 2.00 × 10−7 m; φ = 4.7 × 1.60 × 10−19 = 7.52 × 10−19 J
Step 1 – photon energy
E = hc ÷ λ = 1.989 × 10−25 ÷ 2.00 × 10−7 = 9.945 × 10−19 J
Step 2 – maximum kinetic energy
F: hf = φ + Ek(max)
I: 9.945 × 10−19 = 7.52 × 10−19 + Ek(max)
F: Ek(max) = 2.425 × 10−19 J
Step 3 – maximum speed
½mevmax² = 2.425 × 10−19, so vmax = √(2 × 2.425 × 10−19 ÷ 9.11 × 10−31) = √(5.323… × 1011)
A: 7.296… × 105 m s⁻¹ = 7.30 × 105 m s⁻¹ (3 s.f.)
Watch out: Ek must be in joules (not eV) and the mass is the electron mass, before you use ½mv².
Q10 (Stretch) Light of wavelength 300 nm releases photoelectrons from a metal surface with a maximum speed of 6.0 × 105 m s⁻¹. Calculate (a) the work function in eV, to 3 significant figures, (b) the threshold wavelength in nm, to 3 significant figures, and (c) the stopping potential, to 3 significant figures. (h = 6.63 × 10−34 J s; c = 3.00 × 108 m s⁻¹; 1 eV = 1.60 × 10−19 J; e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg)
Show solution
Convert first: 300 nm = 3.00 × 10−7 m
Step 1 – maximum kinetic energy
Ek(max) = ½mevmax² = 0.5 × 9.11 × 10−31 × (6.0 × 105)² = 1.6398 × 10−19 J
Step 2 – (a) work function
F: hf = φ + Ek(max)
I: hc ÷ λ = 1.989 × 10−25 ÷ 3.00 × 10−7 = 6.63 × 10−19 J, so 6.63 × 10−19 = φ + 1.6398 × 10−19
F: φ = 4.9902 × 10−19 J = 4.9902 × 10−19 ÷ 1.60 × 10−19 = 3.118… eV
A: 3.12 eV (3 s.f.)
Step 3 – (b) threshold wavelength
λ0 = hc ÷ φ = 1.989 × 10−25 ÷ 4.9902 × 10−19 = 3.985… × 10−7 m
A: 399 nm (3 s.f.), just at the violet end of the visible spectrum
Step 4 – (c) stopping potential
eVs = Ek(max), so Vs = 1.6398 × 10−19 ÷ 1.60 × 10−19 = 1.024… V
A: 1.02 V (3 s.f.)
Watch out: vmax applies only to electrons from the surface. Electrons from deeper in the metal lose energy on the way out, so most photoelectrons are slower.
🧮 Want every A-level equation in one place?
The AQA A-level Every Key Equation PowerPoint has 127 equations, 381 practice questions and 15 derivations, covering the core and all five options, for £12. For the rest of the specification, the AQA A-level Physics Focus Questions turn every point of the spec into a question with the answer underneath, for £24.
← All Particles and radiation equations · AQA A-level Physics revision