AQA A-Level Physics: Turning points in physics (7408) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.12 Turning points in physics of the AQA AS and A-level Physics specification (7407/7408, version 1.4, AS and A-level exams June 2016 onwards), sub-sections 3.12.1 to 3.12.3. Turning points is one of the options (sections 3.9 to 3.13); you study one option, assessed in Paper 3 Section B. All of it is A-level only.

Review the study guide and revision notes first; see also the course hub, checklist and free diagnostics.

Use e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, mass of hydrogen ion = 1.67 × 10⁻²⁷ kg, h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, g = 9.81 N kg⁻¹, μ₀ = 4π × 10⁻⁷ H m⁻¹.

Questions

1. Describe how cathode rays are produced in a discharge tube. [3]

2. An electron gun has an anode pd of 4.5 kV.

(a) Calculate the kinetic energy of an electron leaving the gun. [1] (b) Calculate its speed. [2] (c) Explain why the cathode is heated. [1]

3. An electron beam passes undeflected through crossed fields: plates 40 mm apart at 1200 V, and 1.5 mT. With the electric field off, the beam moves in a circle of radius 7.6 cm.

(a) Calculate the speed of the electrons. [2] (b) Calculate the specific charge of the electron. [2] (c) Explain the significance of this value compared with the hydrogen ion’s specific charge. [2]

4. An oil drop (density 870 kg m⁻³) falls between horizontal plates 8.0 mm apart. Air viscosity is 1.8 × 10⁻⁵ N s m⁻². With no pd, the drop falls at a terminal speed of 8.5 × 10⁻⁵ m s⁻¹.

(a) Show that the radius of the drop is about 9.0 × 10⁻⁷ m. [3] (b) Calculate the mass of the drop. [2] (c) The drop is held stationary when the pd is 324 V. Calculate its charge and express it in terms of e. [3] (d) State what Millikan concluded from many such results. [1]

5. (a) Give two reasons why Newton’s corpuscular theory was preferred to Huygens’ wave theory. [2] (b) Explain, in general terms, how Young’s double slits produce bright and dark fringes. [2]

6. (a) Calculate the permittivity of free space ε₀. [2] (b) In a Hertz-type experiment, stationary radio waves have nodes 0.75 m apart. The frequency is 200 MHz. Calculate the speed of the waves. [2] (c) State what ε₀ relates to. [1]

7. In a Fizeau-type experiment a wheel has 400 teeth and the mirror is 7.5 km away. The returning light is first blocked at 25 revolutions per second.

(a) Calculate the speed of light. [3] (b) State the implication of this result when compared with 1/√(μ₀ε₀). [1]

8. (a) Describe the ultraviolet catastrophe. [2] (b) State Planck’s interpretation. [1] (c) Explain three ways classical wave theory fails to account for photoelectricity. [3] (d) State the significance of Einstein’s explanation. [1]

9. (a) Estimate the anode pd needed to give electrons a de Broglie wavelength of 1.2 × 10⁻¹⁰ m. [3] (b) In low-energy electron diffraction the pd is increased. Explain the effect on the pattern. [2] (c) Explain why a TEM sample must be very thin. [2]

10. Explain how a scanning tunnelling microscope works in constant-current mode. [4]

11. (a) Outline the Michelson-Morley experiment as a means of detecting absolute motion. [3] (b) State the significance of its result. [2] (c) Define an inertial frame of reference. [1] (d) Muons at 0.98c pass detectors 1800 m apart vertically; proper half-life 1.5 μs. 1000 pass the upper detector. Calculate how many reach the lower one with and without time dilation. [4] (e) Show that an observer travelling with the muons predicts the same number. [2]

12. An electron has kinetic energy 3.0 MeV.

(a) Calculate its speed. [3] (b) Calculate its mass. [1] (c) Show that Newtonian mechanics gives an impossible speed. [2] (d) Explain how Bertozzi’s experiment showed that kinetic energy varies with speed as relativity predicts. [2]

Answers

1. Low-pressure gas with a high pd between cathode and anode [1]. Positive ions strike the cathode and release electrons [1]. The electrons accelerate towards the anode; these beams are the cathode rays [1]. [3] Examiner insight: “Electrons come off the cathode” alone does not say why, so scores only one mark.

2. (a) Ek = eV = 7.2 × 10⁻¹⁶ J [1] (b) v = √(2Ek/m) [1] = √(2 × 7.2 × 10⁻¹⁶ / 9.11 × 10⁻³¹) = 3.98 × 10⁷ m s⁻¹ [1] (c) Heating gives conduction electrons enough energy to leave the metal (thermionic emission) [1]. Examiner insight: A wrong (a) carried into a correct method in (b) is usually allowed as error carried forward.

3. (a) E = V/d = 1200/0.040 = 3.0 × 10⁴ V m⁻¹ [1]; eE = Bev, so v = E/B = 2.0 × 10⁷ m s⁻¹ [1] (b) Bev = mv²/r, so e/m = v/(Br) [1] = 2.0 × 10⁷ / (1.5 × 10⁻³ × 0.076) = 1.75 × 10¹¹ C kg⁻¹ [1] (c) It is about 1800 times the hydrogen ion’s value (9.6 × 10⁷ C kg⁻¹) [1]; for equal charge the electron is far less massive than any atom, so it is a sub-atomic particle [1]. Examiner insight: A comparison alone earns one mark in (c); the second needs the conclusion about mass.

4. (a) mg = 6πηrv with m = (4/3)πr³ρ [1]; r = √(9ηv/(2ρg)) [1] = √(9 × 1.8 × 10⁻⁵ × 8.5 × 10⁻⁵ / (2 × 870 × 9.81)) = 8.98 × 10⁻⁷ m [1] (b) m = (4/3)π(8.98 × 10⁻⁷)³ × 870 [1] = 2.64 × 10⁻¹⁵ kg [1] (c) Q = mgd/V [1] = 2.64 × 10⁻¹⁵ × 9.81 × 8.0 × 10⁻³ / 324 = 6.39 × 10⁻¹⁹ C [1]; ÷ 1.60 × 10⁻¹⁹ = 4.0, so Q = 4e [1] (d) Charge is quantised: always a whole-number multiple of e [1]. Examiner insight: On a “show that”, give the answer to more significant figures than the stated value (8.98, not 9.0) or the final mark is lost.

5. (a) Two from: Newton’s reputation [1]; it explained reflection and refraction [1]; the speed of light in water could not yet be measured. [2] (b) Waves from the two slits overlap; where they arrive in phase they reinforce to give bright fringes [1]; in antiphase they cancel to give dark fringes [1]. Examiner insight: No calculation is expected in (b); a formula without the phase explanation earns nothing.

6. (a) ε₀ = 1/(μ₀c²) [1] = 1/(4π × 10⁻⁷ × (3.00 × 10⁸)²) = 8.84 × 10⁻¹² F m⁻¹ [1] (b) λ = 2 × 0.75 = 1.50 m [1]; c = fλ = 200 × 10⁶ × 1.50 = 3.0 × 10⁸ m s⁻¹ [1] (c) The electric field strength due to a charged object in free space [1]. Examiner insight: Node spacing is λ/2; taking 0.75 m as λ loses the first mark, and the second may be allowed as error carried forward.

7. (a) t = 1/(2Nf) = 1/(2 × 400 × 25) = 5.0 × 10⁻⁵ s [1]; light travels 2d = 15 km [1]; c = 15 000 / 5.0 × 10⁻⁵ = 3.0 × 10⁸ m s⁻¹ [1] (b) The values agree, so light is an electromagnetic wave [1]. Examiner insight: Using d instead of 2d halves the answer and loses the last two marks.

8. (a) Classical theory predicted black-body intensity rising without limit as wavelength decreases [1], but the observed intensity peaks and falls to zero at short wavelengths [1]. (b) Energy is emitted in quanta of E = hf [1]. (c) Below a threshold frequency no electrons are emitted, however intense the light [1]; emission is instant, with no build-up time at low intensity [1]; maximum kinetic energy depends on frequency, not intensity [1]. (d) EM radiation has a particle (photon) nature [1]. Examiner insight: Each mark in (c) needs the observation, not just “wave theory is wrong”.

9. (a) p = h/λ = 6.63 × 10⁻³⁴ / 1.2 × 10⁻¹⁰ = 5.53 × 10⁻²⁴ kg m s⁻¹ [1]; eV = p²/(2m) [1]; V = (5.53 × 10⁻²⁴)² / (2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹) = 105 V [1] (b) Electrons go faster, so momentum rises and wavelength falls [1]; less diffraction, so the rings get smaller [1]. (c) Electrons must pass through it [1]; a thick sample slows them, changing their wavelength and blurring the image [1]. Examiner insight: “Estimate” still needs the working; a bare 105 V gets only the final mark.

10. A fine conducting tip is held about 1 nm above a conducting surface with a small pd [1]. Electrons cross the gap by quantum tunnelling [1]. The current depends very strongly on the gap width [1]. The tip height is adjusted to keep the current constant, so the tip follows the surface contours [1]. [4] Examiner insight: Saying “electrons jump the gap” without the word tunnelling usually loses that mark.

11. (a) A semi-silvered plate splits light into two perpendicular beams, reflected back by mirrors [1]; the beams recombine to give fringes [1]; motion through the ether should change the travel times, so rotating the apparatus by 90° should shift the fringes [1]. (b) No shift was detected, so absolute motion cannot be detected [1]; the speed of light is invariant [1]. (c) A frame that is not accelerating, in which Newton’s first law holds [1]. (d) Earth-frame time = 1800/(0.98 × 3.00 × 10⁸) = 6.12 μs [1]; γ = 5.03, dilated half-life = 7.54 μs [1]; with dilation 1000 × 0.5^(6.12/7.54) = 570 [1]; without, 1000 × 0.5^(6.12/1.5) = 59 [1] (e) Contracted distance = 1800/5.03 = 358 m, time = 358/(0.98c) = 1.22 μs [1]; 1.22/1.5 = 0.81 half-lives, giving the same 570 [1]. Examiner insight: Pair the proper half-life with muon-frame time, or the dilated half-life with Earth-frame time; mixing frames loses the dilation and answer marks.

12. (a) m₀c² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 8.20 × 10⁻¹⁴ J [1]; E = 4.80 × 10⁻¹³ + 8.20 × 10⁻¹⁴ = 5.62 × 10⁻¹³ J, γ = 6.85 [1]; v = c√(1 − 1/γ²) = 0.989c = 2.97 × 10⁸ m s⁻¹ [1] (b) m = γm₀ = 6.24 × 10⁻³⁰ kg [1] (c) v = √(2Ek/m₀) = √(2 × 4.80 × 10⁻¹³ / 9.11 × 10⁻³¹) = 1.03 × 10⁹ m s⁻¹ [1], which exceeds c [1]. (d) Electrons of increasing known energy were timed over a known distance [1]; speed approached but never exceeded c, matching the relativistic graph [1]. Examiner insight: Convert MeV to joules (× 1.60 × 10⁻¹³) before combining with m₀c²; mixing units loses the method mark.

Where marks are usually lost

• Giving a “show that” Millikan radius only to 2 s.f.
• Taking node spacing as the wavelength.
• Using d instead of 2d in Fizeau’s method.
• Writing “light is particles” instead of naming the photoelectric observations.
• Describing the STM without tunnelling, or the TEM without a thin sample.
• Mixing reference frames in muon calculations.

Next steps

• Turning points revision notes
• Turning points study guide
• AQA A-level Physics course hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA AS and A-level Physics specification (7407/7408), version 1.4, July 2026, AS and A-level exams June 2016 onwards: section 3.12 Turning points in physics (A-level only). Published by AQA.