Skip to main content
GIOPHYSICS
Menu

A-Level Physics · guided topic map

Atomic, nuclear and quantum physics for Cambridge International AS & A Level Physics

Atomic, nuclear and quantum physics for A-Level Physics, organized into 3 syllabus topics and 8 mapped concept guides.

Syllabus topics
3
Mapped concept guides
8
Educational level
Cambridge International AS & A Level

Syllabus to lesson

Choose the exact concept

Work in order or jump to the concept named in your specification, course outline, or assignment.

11

Particle physics

AS Level foundations

2 guides
  1. 01Atoms, nuclei, quarks, and leptonsMapped lesson
  2. 02Nuclear radiation and decay equationsMapped lesson
22

Quantum physics

A Level extension

3 guides
  1. 01Photon energy, momentum, and the photoelectric effectMapped lesson
  2. 02Matter waves and electron diffractionMapped lesson
  3. 03Discrete energy levels and line spectraMapped lesson
23

Nuclear physics

A Level extension

3 guides
  1. 01Mass defect, binding energy, fission, and fusionMapped lesson
  2. 02Radioactive decay, activity, and half-lifeMapped lesson
  3. 03Exponential decay and the decay constantMapped lesson

Diagrams

Atomic, nuclear and quantum physics as A-Level Physics draws it

The figures from the A-Level Physics practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Fig. 11.1Quantum physics · Nuclear physicsA Level
An electron diffraction tube: cathode, anode, thin crystal and screencathodeanodethin crystalfluorescent screenelectron beamvacuum250 V

Figure comment

Fig. 11.1An evacuated tube is drawn as a long horizontal rectangle, closed at its right-hand end by a bar labelled fluorescent screen. Near the left-hand end is a short vertical cathode, and beyond it a vertical anode plate with a gap at its centre; leads from both pass out of the tube to a cell labelled 250 V, whose positive terminal is joined to the anode. An arrow along the axis shows a beam of electrons passing through the gap in the anode and travelling to a thin crystal mounted upright across the middle of the tube. The screen is drawn blank.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

22 Quantum physics23 Nuclear physicsOpen A Level question · A Level extension Q11
Guided questions 5 parts

Reading cue. The 250 V acts only between cathode and anode; beyond the anode gap there is no field, so the electrons meet the crystal at the speed they had on leaving it.

  1. aExplain Explain why the cell in Fig. 11.1 is connected with its positive terminal to the anode.

    recall2 marks

    Check answer 2 marks
    1. electrons carry negative charge
    2. they are repelled from the negative cathode and attracted to the positive anode, so this connection accelerates them along the tube towards the screen

  2. bDescribe Describe what is seen on the blank fluorescent screen once the electron beam has passed through the thin crystal.

    routine3 marks

    Check answer 3 marks
    1. a set of concentric bright rings
    2. centred on the point where the undeviated beam meets the screen
    3. the pattern is brightest at the centre, with the rings becoming fainter further out

  3. cDetermine The accelerating p.d. is increased from 250 V to 1000 V. Determine the factor by which the de Broglie wavelength of the electrons changes, and describe the effect on the pattern on the screen.

    demanding3 marks

    Check answer 3 marks
    1. eV = ½mv² and λ = h/mv, so λ ∝ 1/√V
    2. V is 4 times greater, so λ is halved: factor 0.50 (7.8 × 10⁻¹¹ m falls to 3.9 × 10⁻¹¹ m)
    3. the diffraction angles are smaller, so the rings close in towards the centre of the screen

  4. dSuggest Protons are accelerated through the same 250 V and directed at the same crystal. Suggest why no ring pattern appears on the screen.

    top of the paper4 marks

    Check answer 4 marks
    1. λ = h/√(2meV), so the proton wavelength is smaller than the electron wavelength by √(mp/me) ≈ 43
    2. λ ≈ 1.8 × 10⁻¹² m for the protons
    3. this is far smaller than the spacing of the atoms in the crystal, so the diffraction angles are too small to be seen
    4. (the much heavier protons would also be absorbed within the crystal)

Transfer challenge

Neutrons in a reactor are slowed until their kinetic energy is 0.025 eV. Calculate their de Broglie wavelength, and suggest why such neutrons are useful for investigating crystal structure. (mass of a neutron = 1.67 × 10⁻²⁷ kg)

Check answer 3 marks
  1. E = 0.025 × 1.60 × 10⁻¹⁹ = 4.0 × 10⁻²¹ J, and p = √(2mE) = 3.7 × 10⁻²⁴ N s
  2. λ = h/p = 6.63 × 10⁻³⁴ / 3.7 × 10⁻²⁴ = 1.8 × 10⁻¹⁰ m
  3. this is comparable with the spacing of atoms in a crystal, so the neutrons are strongly diffracted and the pattern reveals that spacing
02Fig. 12.1Capacitance · Nuclear physicsA Level
A charged capacitor discharging through a resistor, with a voltmeter across itCVSR = 100 kΩ

Figure comment

Fig. 12.1A circuit with three branches between an upper and a lower horizontal wire. On the left is a capacitor labelled C; in the middle is a voltmeter connected permanently across it, with junction dots where its branch joins each of the two wires; on the right is a resistor labelled R = 100 kΩ. An open switch S lies in the upper wire between the voltmeter branch and the resistor branch, so that closing it connects the resistor across the capacitor.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

19 Capacitance23 Nuclear physicsOpen A Level question · A Level extension Q12
Guided questions 5 parts

Reading cue. The voltmeter is permanently across C, so it is part of the discharge path even before S closes; R joins the circuit only when S is closed.

  1. aCalculate The capacitor in Fig. 12.1 has capacitance 470 μF. Calculate the time constant of the circuit after S is closed.

    recall2 marks

    Check answer 2 marks
    1. τ = RC = 100 × 10³ × 470 × 10⁻⁶
    2. τ = 47 s

  2. bDetermine The voltmeter reads 9.0 V at the instant S is closed. Determine its reading 60 s later.

    routine3 marks

    Check answer 3 marks
    1. V = V₀e^(−t/RC), with t/RC = 60/47 = 1.28
    2. V = 9.0 × e^(−1.28)
    3. V = 2.5 V

  3. cDetermine Determine the time taken for the voltmeter reading to halve, and state whether this time would differ if the reading at the instant of closing S were 4.5 V instead of 9.0 V.

    demanding3 marks

    Check answer 3 marks
    1. ½ = e^(−t/RC), so t = RC ln 2
    2. t = 47 × 0.693 = 33 s
    3. the time to halve is independent of the starting p.d., so it would still be 33 s

  4. dDeduce Deduce the energy transferred to R while the voltmeter reading falls from 9.0 V to 4.5 V, and state the assumption you make about the voltmeter.

    top of the paper4 marks

    Check answer 4 marks
    1. energy stored at 9.0 V = ½CV² = ½ × 470 × 10⁻⁶ × 9.0² = 1.9 × 10⁻² J
    2. energy stored at 4.5 V = ½ × 470 × 10⁻⁶ × 4.5² = 4.8 × 10⁻³ J
    3. energy transferred to R = 1.9 × 10⁻² − 4.8 × 10⁻³ = 1.4 × 10⁻² J
    4. assumes no charge flows through the voltmeter branch, i.e. its resistance is effectively infinite

Transfer challenge

The activity of a radioactive source falls from 4.0 × 10³ Bq to 5.0 × 10² Bq in 24 hours. Determine the half-life of the source and its decay constant.

Check answer 3 marks
  1. the activity falls by a factor of 8 = 2³, so 24 hours is three half-lives
  2. half-life = 8.0 hours
  3. λ = ln2 / t½ = 0.693 / (8.0 × 3600) = 2.4 × 10⁻⁵ s⁻¹