A-LEVEL EXAM PREP
AQA AS LEVEL PHYSICS
7407/1 PAPER 1
JUNE 2026/2027
ACTUAL EXAM TEST BANK A+
WITH REAL EXAM QUESTIONS AND
100% CORRECT VERIFIED ANSWERS
(GUARANTEED PASS!!)
Aligned to AQA AS Physics 7407 Specification • AS Level
A+ QUESTIONS 4 SECTIONS 100%
VERIFIED COMPLETE RATIONALES
CATEGORIES
SECTION 1: Particles and Radiation
SECTION 2: Waves
SECTION 3: Mechanics and Materials
SECTION 4: Electricity and Measurements
STUVIAACTUALEXAM
, SECTION 1: PARTICLES AND RADIATION
Q1. A student investigates the photoelectric effect using a metal surface with a work function of 2.3 eV.
Monochromatic light of wavelength 450 nm is incident on the surface. Calculate the maximum kinetic
energy of the emitted photoelectrons and determine whether the light is capable of causing emission.
A. 2.76 eV; emission occurs because photon energy exceeds work function
B. 0.46 eV; emission occurs because photon energy exceeds work function
C. 0.46 eV; emission does not occur because photon energy is less than work function
D. 2.76 eV; emission does not occur because photon energy is less than work function
Correct Answer: B
Rationale: Photon energy E = hc/λ ≈ 2.76 eV. Maximum KE = E − φ = 2.76 − 2.3 = 0.46 eV. Since E > φ,
photoemission occurs. Option B incorrectly reports the photon energy as the KE; C and D reverse the emission
conclusion.
Q2. In a particle physics experiment, a proton collides with an antiproton and annihilates. The collision
products include several pions. Explain the possible quark composition of a π■ meson produced in this
process and identify the conservation laws that must be satisfied.
A. u u-bar; baryon number and strangeness conserved
B. d u-bar; charge and lepton number conserved but baryon number not
C. s s-bar; all quantum numbers conserved
D. u d-bar; baryon number, charge and lepton number conserved
Correct Answer: D
Rationale: A π■ has quark content u d-bar. Annihilation of p + p-bar (both baryon number ±1) produces mesons
(baryon number 0), conserving baryon number, charge and lepton number. Options B and D describe neutral mesons;
C has incorrect quark assignment.
Q3. An electron is accelerated through a potential difference of 120 V in an electron diffraction tube. The
de Broglie wavelength of the electron is measured to investigate wave-particle duality. Calculate the
wavelength and state how increasing the accelerating voltage affects the diffraction pattern.
A. 0.112 nm; fringe spacing increases as wavelength decreases
B. 1.12 nm; fringe spacing decreases as wavelength decreases
C. 0.112 nm; fringe spacing decreases as wavelength decreases
D. 0.0112 nm; fringe spacing remains unchanged
Correct Answer: C
Rationale: λ = h / √(2m e V) ≈ 0.112 nm. Higher voltage increases momentum, decreases λ, and therefore reduces
the diffraction angle and fringe spacing. B reverses the spacing change; C and D have incorrect wavelength
magnitudes.
STUVIAACTUALEXAM • Page 2
AQA AS LEVEL PHYSICS
7407/1 PAPER 1
JUNE 2026/2027
ACTUAL EXAM TEST BANK A+
WITH REAL EXAM QUESTIONS AND
100% CORRECT VERIFIED ANSWERS
(GUARANTEED PASS!!)
Aligned to AQA AS Physics 7407 Specification • AS Level
A+ QUESTIONS 4 SECTIONS 100%
VERIFIED COMPLETE RATIONALES
CATEGORIES
SECTION 1: Particles and Radiation
SECTION 2: Waves
SECTION 3: Mechanics and Materials
SECTION 4: Electricity and Measurements
STUVIAACTUALEXAM
, SECTION 1: PARTICLES AND RADIATION
Q1. A student investigates the photoelectric effect using a metal surface with a work function of 2.3 eV.
Monochromatic light of wavelength 450 nm is incident on the surface. Calculate the maximum kinetic
energy of the emitted photoelectrons and determine whether the light is capable of causing emission.
A. 2.76 eV; emission occurs because photon energy exceeds work function
B. 0.46 eV; emission occurs because photon energy exceeds work function
C. 0.46 eV; emission does not occur because photon energy is less than work function
D. 2.76 eV; emission does not occur because photon energy is less than work function
Correct Answer: B
Rationale: Photon energy E = hc/λ ≈ 2.76 eV. Maximum KE = E − φ = 2.76 − 2.3 = 0.46 eV. Since E > φ,
photoemission occurs. Option B incorrectly reports the photon energy as the KE; C and D reverse the emission
conclusion.
Q2. In a particle physics experiment, a proton collides with an antiproton and annihilates. The collision
products include several pions. Explain the possible quark composition of a π■ meson produced in this
process and identify the conservation laws that must be satisfied.
A. u u-bar; baryon number and strangeness conserved
B. d u-bar; charge and lepton number conserved but baryon number not
C. s s-bar; all quantum numbers conserved
D. u d-bar; baryon number, charge and lepton number conserved
Correct Answer: D
Rationale: A π■ has quark content u d-bar. Annihilation of p + p-bar (both baryon number ±1) produces mesons
(baryon number 0), conserving baryon number, charge and lepton number. Options B and D describe neutral mesons;
C has incorrect quark assignment.
Q3. An electron is accelerated through a potential difference of 120 V in an electron diffraction tube. The
de Broglie wavelength of the electron is measured to investigate wave-particle duality. Calculate the
wavelength and state how increasing the accelerating voltage affects the diffraction pattern.
A. 0.112 nm; fringe spacing increases as wavelength decreases
B. 1.12 nm; fringe spacing decreases as wavelength decreases
C. 0.112 nm; fringe spacing decreases as wavelength decreases
D. 0.0112 nm; fringe spacing remains unchanged
Correct Answer: C
Rationale: λ = h / √(2m e V) ≈ 0.112 nm. Higher voltage increases momentum, decreases λ, and therefore reduces
the diffraction angle and fringe spacing. B reverses the spacing change; C and D have incorrect wavelength
magnitudes.
STUVIAACTUALEXAM • Page 2
