(b) mass = $14 \times 1.67 \times 10^{-27} = 2.34 \times 10^{-26}$ kg (allow $6m_p + 8m_n$ from the data booklet = $2.34\times10^{-26}$ kg)(1)
(b) specific charge = $\dfrac{9.60\times10^{-19}}{2.34\times10^{-26}} = 4.1 \times 10^{7}$ C kg$^{-1}$ unit required(1)
Answer: (a) 8 (b) $4.1 \times 10^{7}$ C kg$^{-1}$
Examiner tip: Specific charge is charge ÷ mass with unit C kg⁻¹; use only the protons for the charge but all 14 nucleons for the mass.
2Standard4 marks
Strontium-90 ($^{90}_{38}\mathrm{Sr}$) decays by $\beta^-$ emission to an isotope of yttrium (Y).
(a) Write the complete equation for this decay. [2]
(b) Explain why the existence of the antineutrino was proposed. [1]
(c) State the change in quark character that occurs in $\beta^-$ decay. [1]
(a) $^{90}_{38}\mathrm{Sr} \rightarrow {}^{90}_{39}\mathrm{Y} + {}^{\;\,0}_{-1}\mathrm{e}$ with correct nucleon and proton numbers on Y and the electron (accept $^{\;\,0}_{-1}\beta$, β$^-$ or e$^-$ for the electron)(1)
(a) $+\ \bar{\nu}_e$ included (accept $\bar{\nu}$ without subscript; reject $\nu_e$ / neutrino)(1)
(b) the beta particles are emitted with a range of (kinetic) energies / a continuous energy spectrum, so another (undetected) particle must carry away the missing energy (and momentum) (allow 'to conserve energy (in beta decay)' / 'otherwise energy would not be conserved'; ignore reference to momentum; reject 'to conserve charge' alone)(1)
(c) a down quark changes to an up quark (d → u) (accept udd → uud; reject 'neutron changes to proton' with no reference to quarks)(1)
Answer: (a) $^{90}_{38}\mathrm{Sr} \rightarrow {}^{90}_{39}\mathrm{Y} + \beta^- + \bar{\nu}_e$ (b) beta energies vary so energy not conserved without it (c) d → u
Examiner tip: An electron-antineutrino, not a neutrino, accompanies β⁻ decay — check lepton number: 0 = +1 + (−1).
3Standard5 marks
The lambda particle $\Lambda^0$ has quark composition uds. It decays as follows:
$\Lambda^0 \rightarrow p + \pi^-$
(a) Show that charge and baryon number are conserved in this decay. [2]
(b) Explain why this decay must be a weak interaction. [2]
(c) State the quark composition of the $\pi^-$. [1]
(a) charge: 0 → (+1) + (−1) = 0(1)
(a) baryon number: 1 → 1 + 0 = 1(1)
(b) strangeness of $\Lambda^0$ is −1 and strangeness of products is 0 (strangeness not conserved)(1)
(b) strangeness can change (by 1) only in weak interactions(1)
(c) $\mathrm{d}\bar{\mathrm{u}}$(1)
Answer: (a) Q: 0 → 0; B: 1 → 1 (b) S changes from −1 to 0, only the weak interaction allows this (c) $\mathrm{d}\bar{\mathrm{u}}$
Examiner tip: A strange quark has strangeness −1 (not +1) — the sign trips many students up.
4Standard6 marks
Ultraviolet radiation of wavelength 200 nm is incident on a clean zinc plate. The work function of zinc is 4.3 eV.
(a) Calculate the maximum kinetic energy of the photoelectrons, in J. [4]
(b) Visible light of wavelength 500 nm does not release electrons from zinc, however intense it is. Explain why. [2]
(a) $E_{k(max)} = 9.95\times10^{-19} - 6.88\times10^{-19} = 3.1 \times 10^{-19}$ J allow 3.0 × 10-19 J if intermediate values are rounded to 2 s.f.(1)
(b) photon energy at 500 nm (2.5 eV / $4.0\times10^{-19}$ J) is less than the work function(1)
(b) one electron absorbs one photon, so increasing intensity only increases the number of photons, not the energy of each(1)
Answer: (a) $3.1\times10^{-19}$ J (b) each photon has energy below the work function; one photon–one electron
Examiner tip: Convert the work function to joules before subtracting — mixing eV and J is the classic error.
5Standard5 marks
The diagram shows some energy levels of a hydrogen atom.
(a) An electron moves from the $n = 3$ level to the $n = 2$ level. Calculate the wavelength of the photon emitted. [3]
(b) State the number of different photon energies that could be emitted by atoms excited to the $n = 4$ level. [1]
(c) A free electron with kinetic energy 12.5 eV collides with a hydrogen atom in its ground state. State the highest level to which the atom can be excited. [1]
(a) $\Delta E = 3.40 - 1.51 = 1.89$ eV $= 1.89\times1.60\times10^{-19} = 3.02\times10^{-19}$ J(1)
(c) $n = 3$ (needs 12.09 eV; $n = 4$ would need 12.75 eV)(1)
Answer: (a) $6.6\times10^{-7}$ m (b) 6 (c) $n = 3$
Examiner tip: A colliding electron can give part of its energy, but a photon must be absorbed whole — read whether it is an electron or a photon.
6Hard6 marks
A fluorescent tube contains mercury vapour at low pressure. The inside of the glass is coated with a phosphor.
Explain how the tube produces visible light when a high potential difference is applied across it. [6]
the p.d. accelerates free electrons through the tube (from the cathode)(1)
electrons collide (inelastically) with mercury atoms(1)
atomic electrons in mercury are excited to higher energy levels (some atoms are ionised)(1)
the excited electrons return to lower levels / de-excite, emitting photons(1)
the energy gaps in mercury are large, so these photons are mainly ultraviolet(1)
the UV photons are absorbed by the phosphor coating, exciting its atoms(1)
the phosphor's electrons de-excite in several smaller steps, emitting lower-energy (visible) photons(1)
Level
Marks
What it looks like
Level 3
5–6
A full, coherent account covering most of the indicative points in a logical order; physics terminology used correctly throughout.
Level 2
3–4
A partial account covering several indicative points with some linking; may have gaps or minor errors in the sequence or terminology.
Level 1
1–2
Isolated, relevant points with little or no linking between them.
Level 0
0
No relevant content.
Answer: Accelerated electrons excite mercury atoms → UV emitted on de-excitation → phosphor absorbs UV and re-emits visible light in smaller energy steps.
Examiner tip: Top-band answers explain why the phosphor is needed: mercury emits UV, and the phosphor converts it to visible light in smaller energy drops.
7Standard4 marks
An electron and a positron, both effectively at rest, annihilate each other.
(a) Explain why two photons are produced, rather than one. [1]
(b) Calculate the wavelength of each photon. [3]
(a) total momentum before is zero, so two photons moving in opposite directions are needed to conserve momentum(1)
(b) energy of each photon = rest energy of an electron = 0.511 MeV $= 0.511\times10^{6}\times1.60\times10^{-19} = 8.2\times10^{-14}$ J (allow $m_ec^2 = 9.11\times10^{-31}\times(3.00\times10^{8})^2$)(1)
(a) $eV = \frac{1}{2}mv^2$ so $V = \dfrac{mv^2}{2e}$(1)
(a) $V = \dfrac{9.11\times10^{-31}\times(6.06\times10^{6})^2}{2\times1.60\times10^{-19}} = 105$ V allow 100 V (2 s.f.); allow 104–106 V if $v$ is rounded to 6.1 × 106 m s-1; allow full marks for the route $p = h/\lambda$, $E_k = p^2/2m$, $V = E_k/e$(1)
(b) electrons have greater speed / momentum, so de Broglie wavelength decreases(1)
(b) less diffraction, so the rings become smaller / closer together(1)
Answer: (a) 105 V (100 V to 2 s.f.) (b) shorter wavelength → less diffraction → rings get smaller
Examiner tip: Link each step: higher p.d. → higher momentum → shorter λ → less diffraction. Skipping the wavelength step loses the mark.
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