Question bank › A-level Physics › Particles and radiation
Collisions of electrons, energy levels and wave–particle duality
Answer on paper first, then reveal the mark scheme one mark at a time.
1Standard5 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)
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.
(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.