IB Physics SL · first assessment 2025 · Theme E

E.1 Structure of the atom: IB Physics SL exam-style questions

E.1 follows the evidence for the modern model of the atom. At SL the Geiger–Marsden–Rutherford experiment is treated qualitatively: what the scattering of alpha particles showed about the size, mass and charge of the nucleus. You also need nuclear notation with nucleon number and proton number.

Emission and absorption spectra are evidence for discrete energy levels. Photons are emitted or absorbed in transitions between levels, with energy E = hf, and spectra reveal the chemical composition of a gas or a star. The Bohr model formula and distance of closest approach are Higher Level.

  • 22 questions
  • 128 marks
  • Paper 1A: 12
  • Paper 1B: 3
  • Paper 2: 7
  • Full mark schemes

Showing 506 of 506 questions · 3452 marks

Tick questions to build a test

22 practice questions on E.1 Structure of the atom

1E-1A-01
Nuclear notation·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksState

A nucleus of iodine is represented by 12753I.

Which row gives the number of protons and the number of neutrons in this nucleus?

Number of protonsNumber of neutrons
Show mark scheme
Marking pointMarkNotes
Step 1The lower number is the proton number Z = 53; the neutron number is A − Z = 127 − 53 = 74.✓ 1Answer B

Answer: B  ·  1 stage of work, one mark

Every option, and why

  • AThe two numbers have been swapped: 74 is the neutron number, and the proton number is the lower figure, Z = 53.
  • BCorrect: Z = 53 protons and N = A − Z = 127 − 53 = 74 neutrons.
  • C127 is the nucleon number A, which counts the protons and the neutrons together.
  • D180 = 127 + 53 adds Z to A; the neutron number is A − Z.

Syllabus understandingE.1 — nuclear notation AZX where A is the nucleon number, Z is the proton number and X is the chemical symbol Command term: State

2E-1A-02
Rutherford scattering·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksIdentify

In a Geiger–Marsden–Rutherford experiment, alpha particles are directed at a thin gold foil. The diagrams show possible paths of one alpha particle that passes close to a single gold nucleus N.

Which diagram shows a possible path?

Four possible paths A to D of an alpha particle passing close to a gold nucleus N: A curves towards and round the nucleus, B is a straight line, C curves smoothly away from the nucleus, D is two straight lines meeting at a sharp corner near the nucleus.A.NB.NC.ND.N
Diagram NOT accurately drawn
Show mark scheme
Marking pointMarkNotes
Step 1The alpha particle and the nucleus are both positively charged, so the alpha particle is repelled by a force that grows smoothly as it approaches N and dies away as it leaves: the path curves gradually away from the nucleus.✓ 1Answer C

Answer: C  ·  1 stage of work, one mark

Every option, and why

  • AThis path curves towards N and round it, as if the nucleus attracted the alpha particle; both are positive, so the force is repulsive.
  • BAn alpha particle passing this close to a nucleus feels a large repulsive force, so it cannot continue undeflected.
  • CCorrect: the electric repulsion increases as the particle approaches N and decreases as it moves away, so the path bends smoothly away from the nucleus.
  • DThe direction changes suddenly at one point, as if the particle bounced off a hard surface; the electric force acts at a distance, so the path is a smooth curve.

Syllabus understandingE.1 — the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus Command term: Identify

3E-1A-03
Energy levels & photons·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine

An electron in an atom moves from an energy level of −3.40 eV to an energy level of −13.6 eV.

Which row describes the photon involved in this transition?

PhotonFrequency / Hz
Show mark scheme
Marking pointMarkNotes
Step 1The electron moves to a lower level, so the atom loses energy and a photon of energy ΔE = −3.40 − (−13.6) = 10.2 eV = 1.63 × 10−18 J is emitted.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2f = ΔE/h = 1.63 × 10−18/6.63 × 10−34 = 2.5 × 1015 Hz.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThe frequency is right, but the electron falls to a lower level, so energy is given out: the photon is emitted, not absorbed.
  • BCorrect: ΔE = 10.2 eV = 1.63 × 10−18 J is given out as a photon of frequency ΔE/h = 2.5 × 1015 Hz.
  • CThis uses 3.40 + 13.6 = 17.0 eV, adding the sizes of the two levels instead of finding their difference.
  • DThis divides 10.2 by h without converting the energy from electronvolts to joules.

Syllabus understandingE.1 — that photons are emitted and absorbed during atomic transitions; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf Command term: Determine

4E-1A-04
Energy levels & photons·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDetermine

An electron in an atom falls from an energy level E2 to a lower energy level E1.

What is the wavelength of the photon emitted? (h is the Planck constant and c is the speed of light.)

Show mark scheme
Marking pointMarkNotes
Step 1The photon carries the energy lost by the atom: hf = E2 − E1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2With c = fλ, λ = c/f = hc/(E2 − E1).✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: hf = E2 − E1 and λ = c/f, so λ = hc/(E2 − E1).
  • Bh/(E2 − E1) is 1/f, the period of the wave; the speed of light from c = fλ has been left out.
  • CThe photon energy is the difference between the two levels, not their sum.
  • DThis is 1/λ: the relation has been inverted.

Syllabus understandingE.1 — that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf Command term: Determine

5E-1A-05
Emission & absorption spectra·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce

The emission spectrum of a hot gas consists of bright lines. When white light passes through the same gas when it is cool, dark lines appear on a continuous spectrum at the same wavelengths as the bright lines.

What do these observations show?

Show mark scheme
Marking pointMarkNotes
Step 1A sharp line means photons of one definite energy hf, so an atom can gain or lose only certain amounts of energy: its energy levels are discrete.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The bright and dark lines coincide because the same pairs of levels are involved whether a photon is emitted or absorbed.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AAbsorption also happens only at particular energies — that is why the absorption spectrum shows dark lines rather than a general dimming.
  • BA continuous range of energies would give continuous spectra, not sharp lines.
  • CCorrect: only certain photon energies are emitted or absorbed, and the lines occur at the same wavelengths, so the same set of level differences is involved.
  • DThe bright and dark lines occur at exactly the same wavelengths, so they come from the same atoms.

Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels Command term: Deduce

6E-1A-06
Emission & absorption spectra·E.1 Structure of the atom
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDeduce

The diagram shows the four lowest energy levels of an atom. A cool gas of these atoms, all in the ground state, is illuminated with light containing a wide range of photon energies.

Which photon energy can be absorbed by the gas?

Four energy levels of an atom at −1.6 eV, −3.7 eV, −5.5 eV and −10.4 eV (ground state), drawn to scale.Energy / eV−10.4ground state−5.5−3.7−1.6
Diagram drawn to scale
Show mark scheme
Marking pointMarkNotes
Step 1In a cool gas the atoms are in the ground state, so a photon is absorbed only if its energy equals the difference between the ground state and a higher level: 4.9 eV, 6.7 eV or 8.8 eV.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Only 4.9 eV = −5.5 − (−10.4) is among the options; 1.8 eV and 2.1 eV are gaps between excited levels, which are empty.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • A1.8 eV = −3.7 − (−5.5) would need atoms already in the −5.5 eV level; in a cool gas that level is empty.
  • B2.1 eV = −1.6 − (−3.7) would need atoms already in the −3.7 eV level; in a cool gas that level is empty.
  • CCorrect: 4.9 eV = −5.5 − (−10.4) exactly matches the gap from the ground state to the first excited level.
  • D5.5 eV is the size of the −5.5 eV level read as a photon energy; it matches no gap from the ground state (the photon energy must be a difference between two levels).

Syllabus understandingE.1 — that photons are emitted and absorbed during atomic transitions; that emission and absorption spectra provide evidence for discrete atomic energy levels Command term: Deduce

7E-1A-07
Emission & absorption spectra·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce

Dark lines are seen in the spectrum of sunlight at wavelengths that match the bright lines of helium measured in a laboratory.

What can be concluded?

Show mark scheme
Marking pointMarkNotes
Step 1The continuous spectrum from the hot interior passes through the cooler gas of the Sun's outer layers, whose atoms absorb photons with energies equal to their own level differences.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Each element has its own set of level differences, so dark lines at helium's wavelengths show that helium is present in that gas.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe matching wavelengths are the fingerprint of helium wherever it is found, including the Sun.
  • BOther elements have their own lines in the solar spectrum; a match shows that helium is present, not that it is the only element.
  • CThe positions of the lines depend on the energy levels of the atom, not on the temperature of the gas.
  • DCorrect: helium atoms in the Sun's outer layers absorb exactly the photon energies they would emit, removing those wavelengths from the continuous spectrum.

Syllabus understandingE.1 — that emission and absorption spectra provide information on the chemical composition Command term: Deduce

8E-1B-05
Emission & absorption spectra·E.1 Structure of the atom
Paper 1BMedium10 marks
Data-based question10 steps to full marksDeduce

A student uses a calibrated spectrometer to measure the wavelengths of the seven brightest visible lines emitted by a gas-discharge tube whose label is missing. Each measured wavelength has an uncertainty of ±2 nm.

Measured wavelengths / nm: 446, 472, 493, 503, 589, 666, 707.

The table lists the brightest visible lines of four gases from a spectral database.

GasWavelengths of brightest visible lines / nm
Hydrogen410.2, 434.0, 486.1, 656.3
Helium447.1, 471.3, 492.2, 501.6, 587.6, 667.8, 706.5
Mercury404.7, 435.8, 546.1, 577.0, 579.1
Sodium498.3, 568.8, 589.0, 589.6, 615.4
(a)

Explain why the light from the tube consists of a set of separate wavelengths.

(2)
(b)

Identify the gas in the tube. Justify your answer with reference to the uncertainty in the measurements.

(2)
(c)

Determine, in eV, the photon energy for the 707 nm line and its absolute uncertainty.

(3)
(d)

The database gives four energy levels of this gas at 20.96 eV, 21.22 eV, 22.72 eV and 23.07 eV above the ground state. Deduce which transition emits the 707 nm line, and whether your answer to (c) supports this.

(2)
(e)

Suggest how the student could check that the spectrometer does not have a systematic error.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Electrons in the atoms can occupy only discrete energy levels✓ 1
A photon is emitted when an electron moves to a lower level; its energy E = hf equals the difference between the levels, so only certain frequencies/wavelengths occur✓ 1OWTTE
Part (b)
Helium✓ 1
Every measured line matches a helium line within ±2 nm; the 589 nm line alone would also match sodium, but sodium (and hydrogen, mercury) do not account for the other lines✓ 1Award MP2 for a justification that uses the ±2 nm tolerance for all the lines.
Part (c)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/707 × 10−9 = 2.81 × 10−19 J✓ 1
= 2.813 × 10−19/1.60 × 10−19 = 1.758 eV✓ 1Accept 1.76 eV.
Fractional uncertainty in E = 2/707 = 0.28 %, so E = (1.758 ± 0.005) eV✓ 1MP3 is for the uncertainty with precision matching the value. Allow ECF.
Part (d)
22.72 − 20.96 = 1.76 eV (the only pair of levels giving close to 1.76 eV)✓ 1
1.76 eV lies within 1.753–1.763 eV, so the measurement supports this transition✓ 1Allow ECF from (c).
Part (e)
Measure the lines of a lamp whose wavelengths are known (e.g. a mercury or sodium lamp) and compare the readings with the accepted values✓ 1Accept: calibrate against a known source. OWTTE

Answers: (b) helium  ·  (c) (1.758 ± 0.005) eV  ·  (d) 22.72 eV → 20.96 eV (1.76 eV) (the remaining parts are explanations — see the table above)

Syllabus understandingE.1 — that emission and absorption spectra provide information on the chemical composition; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf. Command term: Deduce

9E-2-01
Rutherford scattering·E.1 Structure of the atom
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksExplain

In an experiment of the Geiger–Marsden–Rutherford type, a narrow beam of alpha particles, all with the same kinetic energy, is directed at a very thin gold foil, 19779Au, inside an evacuated chamber. A detector that can be moved around the foil counts the alpha particles scattered through different angles.

Figure 1 shows three alpha particles, P, Q and R, approaching one gold nucleus. R is moving directly towards the nucleus.

Figure 1: three alpha particles P, Q and R approach a gold nucleus along parallel lines; R is aimed directly at the nucleus, Q passes close to it and P further away.gold nucleusPQRalpha particles approaching the nucleus
Figure 1 — diagram not to scale
(a)

The gold atom.

(i)

State the number of protons and the number of neutrons in a nucleus of 19779Au.

(1)
(ii)

Outline why the electrons in the gold atoms have a negligible effect on the paths of the alpha particles.

(1)
(b)

Draw, on Figure 1, the paths of alpha particles P, Q and R as they pass the nucleus and move away from it.

(3)
(c)

The detector shows that most alpha particles pass through the foil with very small deflections, and that a very small fraction are scattered through more than 90°. Explain what each of these observations shows about the structure of the atom.

(3)
(d)

The gold foil is replaced by an aluminium foil, 2713Al, of the same thickness. The number of nuclei per unit volume is almost the same in aluminium as in gold. Predict, with a reason, how the fraction of alpha particles scattered through more than 90° changes.

(2)
(e)

Suggest why the foil must be very thin and why the chamber is evacuated.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
79 protons and 197 − 79 = 118 neutrons✓ 1Both values needed.
Part (a)(ii)
The mass of an electron is tiny compared with that of an alpha particle «about 1/7300», so collisions with electrons hardly change the alpha particle's velocity✓ 1Accept: the electrons' negative charge is spread thinly through the atom, so the force it exerts is small. OWTTE
Part (b)
R slows down, stops short of the nucleus and returns back along its line of approach✓ 1Do not accept a path that touches or passes through the nucleus.
P and Q both curve away from the nucleus, the bending occurring close to it, and then continue in straight lines✓ 1Do not accept paths that bend towards the nucleus.
Q is deflected through a clearly larger angle than P✓ 1MP3 only scores if MP2 scores.
Part (c)
Most pass with very small deflections: the atom is mostly empty space / the nucleus occupies a tiny fraction of the volume of the atom✓ 1
A few are scattered through large angles: the positive charge of the atom is concentrated in a very small region, where the repulsive «electric» force is very large✓ 1
Turning back a fast, massive alpha particle shows that this small region «the nucleus» also contains most of the mass of the atom✓ 1Award [1 max] for 'the nucleus is small, dense and positive' with no link to the observations.
Part (d)
The fraction decreases✓ 1
The aluminium nucleus has a much smaller charge «13e compared with 79e», so the repulsive force at a given distance is smaller and fewer alpha particles are turned through large angles✓ 1MP2 only scores if MP1 scores. Do not accept answers based only on the mass of the foil.
Part (e)
Thin foil: so that each alpha particle is deflected by at most one nucleus «and the observed angle is due to a single encounter»✓ 1Accept: so that the alpha particles are not absorbed or slowed significantly in the foil.
Vacuum: alpha particles have a range of only a few centimetres in air; collisions with air molecules would absorb or deflect them✓ 1OWTTE

Answers: (a)(i) 79 protons, 118 neutrons (the remaining parts are explanations — see the table above)

Syllabus understandingE.1 — the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus; nuclear notation AZX where A is the nucleon number, Z is the proton number and X is the chemical symbol Command term: Explain

10E-2-02
Energy levels & photons·E.1 Structure of the atom
Paper 2Hard15 marks
Short answer & extended response15 steps to full marksDeduce

In a gas-discharge lamp, electrons are accelerated by a potential difference and then collide with atoms of a gas. A collision can transfer energy from an electron to an atom and excite the atom.

The figure shows the four lowest energy levels, E1 to E4, of an atom of the gas. Before a collision every atom is in the ground state E1. Ignore the kinetic energy given to the atom as a whole in a collision.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

Energy-level diagram of the gas atom: E1 = −9.00 eV (ground state), E2 = −6.60 eV, E3 = −4.80 eV, E4 = −2.60 eV, with the zero of energy shown dashed.0E₁−9.00 eVE₂−6.60 eVE₃−4.80 eVE₄−2.60 eVenergy levels of the gas atom
Energy levels drawn to scale
(a)

Exciting the atoms.

(i)

State the minimum potential difference through which an electron, initially at rest, must be accelerated to be able to excite an atom from E1 to E2.

(1)
(ii)

In the lamp, each electron strikes an atom with a kinetic energy of 5.00 eV. Deduce which of the levels E2, E3 and E4 can be reached.

(2)
(iii)

A photon of energy 5.00 eV passes through the gas. Explain why it is not absorbed, although an electron of kinetic energy 5.00 eV can excite an atom.

(2)
(b)

The light emitted by the lamp.

(i)

Draw, on the figure, arrows to show all the transitions that produce photons in this lamp.

(2)
(ii)

Determine which of these transitions emit visible light (wavelengths from 400 nm to 700 nm).

(3)
(c)

The electrical power input to the lamp is 12 W. 3.0 % of this power is emitted as photons from the transition E2 → E1. Calculate the number of these photons emitted per second.

(2)
(d)

The accelerating potential difference is increased so that each electron strikes an atom with a kinetic energy of 7.00 eV. Predict, with a reason, the change in the visible spectrum of the lamp.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
2.40 V✓ 1Accept 2.4 V. Do not accept 2.40 eV.
Part (a)(ii)
Energies needed from the ground state: E2 2.40 eV, E3 4.20 eV, E4 6.40 eV✓ 1
E2 and E3 can be reached; E4 cannot, because 6.40 eV > 5.00 eV✓ 1MP2 only scores if MP1 scores.
Part (a)(iii)
A photon is absorbed only if its whole energy equals the difference between two levels; 5.00 eV matches no difference from E1 «2.40, 4.20 or 6.40 eV»✓ 1
An electron can transfer only part of its kinetic energy to the atom and keep the rest «e.g. it keeps 0.80 eV after exciting E3»✓ 1OWTTE
Part (b)(i)
Downward arrows from E3 to E2 and from E3 to E1✓ 1
Downward arrow from E2 to E1, and no arrow starting at E4✓ 1Award [1 max] if the arrows point upwards. Allow ECF from (a)(ii).
Part (b)(ii)
λ = hc/ΔE used, e.g. λ = 6.63 × 10−34 × 3.00 × 108/(2.40 × 1.60 × 10−19)✓ 1
Wavelengths 518 nm «E2 → E1», 691 nm «E3 → E2» and 296 nm «E3 → E1»✓ 1Accept 517–518 nm, 690–691 nm and 296 nm.
E2 → E1 and E3 → E2 are visible; E3 → E1 is ultraviolet✓ 1Allow ECF from (b)(i).
Part (c)
Energy of one photon = 2.40 × 1.60 × 10−19 = 3.84 × 10−19 J✓ 1
N = 0.030 × 12/3.84 × 10−19 = 9.4 × 1017 s−1✓ 1Award [2] for CNA. Award [1 max] for 3.1 × 1019 s−1 «3.0 % not applied».
Part (d)
7.00 eV > 6.40 eV, so E4 can now be reached✓ 1
New transitions E4 → E3 «2.20 eV», E4 → E2 «4.00 eV» and E4 → E1 «6.40 eV» give 565 nm, 311 nm and 194 nm✓ 1Accept 564–565 nm, 310–311 nm, 194 nm.
Only the 565 nm line is visible, so one new «yellow-green» line appears in the visible spectrum; the other two are ultraviolet✓ 1Do not award MP3 for 'three new visible lines'.

Answers: (a)(i) 2.40 V  ·  (b)(ii) 518 nm and 691 nm visible; 296 nm ultraviolet  ·  (c) 9.4 × 1017 s−1  ·  (d) new visible line at 565 nm (the remaining parts are explanations — see the table above)

Syllabus understandingE.1 — that photons are emitted and absorbed during atomic transitions; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf; B.5 — that the electric potential difference V is the work done per unit charge on moving a positive charge between two points as given by V = W/q Command term: Deduce

11E-2-03
Emission & absorption spectra·E.1 Structure of the atom
Paper 2Easy9 marks
Short answer & extended response9 steps to full marksExplain

A student looks through a spectroscope at two light sources: a filament lamp and a discharge tube containing helium gas. Figure 1 shows the two spectra between 400 nm and 700 nm.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

Figure 1: the visible spectrum of a filament lamp is continuous from 400 nm to 700 nm; the helium discharge tube gives bright lines at 447, 471, 492, 502, 588 and 668 nm on a dark background.filament lamphelium discharge tube400450500550600650700wavelength / nm
Figure 1 — wavelength scale drawn to scale
(a)

The two spectra.

(i)

Describe how the spectrum of the helium tube differs from that of the filament lamp.

(2)
(ii)

Explain how the helium spectrum provides evidence that atoms have discrete energy levels.

(3)
(b)

The yellow line in the helium spectrum has a wavelength of 587.6 nm.

(i)

Calculate the energy, in eV, of a photon of this wavelength.

(2)
(ii)

This photon is emitted when an electron in a helium atom moves from a level of energy −1.51 eV to a lower level. Calculate the energy of the lower level.

(1)
(c)

A second discharge tube contains an unknown gas. Outline how its spectrum could be used to identify the gas.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The filament lamp gives a continuous spectrum, containing all visible wavelengths «colours»✓ 1
The helium tube gives a line spectrum: bright lines at a few particular wavelengths, with dark regions between them✓ 1
Part (a)(ii)
Each line consists of photons of one wavelength/frequency and therefore of one energy, E = hf✓ 1
A photon is emitted when an electron in an atom moves from a higher to a lower energy level; its energy equals the difference between the two levels✓ 1
Only certain photon energies are emitted, so only certain energy differences exist: the energy levels are discrete✓ 1OWTTE
Part (b)(i)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/587.6 × 10−9 = 3.38 × 10−19 J✓ 1
E = 3.38 × 10−19/1.60 × 10−19 = 2.12 eV✓ 1Award [2] for CNA. Accept 2.11–2.12 eV.
Part (b)(ii)
−1.51 − 2.12 = −3.63 eV✓ 1Allow ECF from (b)(i). Accept −3.62 eV. The answer must be negative.
Part (c)
Measure the wavelengths of its lines and compare them with the known line spectra of the elements; each element has its own unique set of lines✓ 1OWTTE

Answers: (b)(i) 2.12 eV  ·  (b)(ii) −3.63 eV (the remaining parts are explanations — see the table above)

Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf Command term: Explain

12E-1A-26
Energy levels & photons·E.1 Structure of the atom
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine

The diagram shows energy levels of an atom. An atom in the ground state absorbs a photon of energy 12.09 eV.

Which row gives the level to which the electron is raised and the number of different photon energies that can then be emitted as it returns to the ground state?

Energy levels of an atom at −13.6 eV (ground state), −3.40 eV, −1.51 eV and −0.85 eV, with the ionization level at 0.Energy / eV0ionization−0.85−1.51−3.40−13.6ground state
Diagram NOT accurately drawn
Level reachedNumber of different photon energies
Show mark scheme
Marking pointMarkNotes
Step 1New energy = −13.6 + 12.09 = −1.51 eV, which is exactly a level, so the photon is absorbed.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2From −1.51 eV the electron can fall straight to −13.6 eV, or to −3.40 eV and then to −13.6 eV: three different photon energies.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe electron need not return in a single jump: it can also stop at −3.40 eV on the way down, giving two further photon energies.
  • BThis counts only the two-step route (−1.51 → −3.40 → −13.6 eV) and forgets the direct fall from −1.51 eV to −13.6 eV.
  • CCorrect: −13.6 + 12.09 = −1.51 eV; the routes −1.51 → −13.6, −1.51 → −3.40 and −3.40 → −13.6 eV give three different photon energies.
  • DThis assumes the photon lifts the electron to the highest level shown; reaching −0.85 eV would need 12.75 eV, not 12.09 eV.

Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels; that photons are emitted and absorbed during atomic transitions Command term: Determine

13E-1A-27
Ionization·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine

The ground state of an atom is at −13.6 eV. A photon of energy 15.0 eV is incident on an atom in its ground state.

Which row describes what happens?

What happens to the photonKinetic energy of the freed electron
Show mark scheme
Marking pointMarkNotes
Step 1Freeing the electron from the ground state needs 0 − (−13.6) = 13.6 eV; the photon carries more, so it is absorbed and the atom is ionized (a free electron can have any energy, so no exact match is needed).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Kinetic energy of the freed electron = 15.0 − 13.6 = 1.4 eV.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AAn exact match is needed only between two bound levels; above 13.6 eV the electron is freed and can take any kinetic energy.
  • BCorrect: 13.6 eV is used to free the electron and the remaining 15.0 − 13.6 = 1.4 eV becomes its kinetic energy.
  • CThis gives the electron all of the photon's energy, forgetting the 13.6 eV needed to free it.
  • DThis calculates 15.0 − (−13.6) = 28.6 eV, adding the ionization energy instead of subtracting it.

Syllabus understandingE.1 — that photons are emitted and absorbed during atomic transitions Command term: Determine

14E-1A-28
Emission & absorption spectra·E.1 Structure of the atom
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksIdentify

The diagram shows the emission spectrum of a gas. All four bright lines are produced by transitions that end on the ground state.

White light passes through a cool sample of the same gas. Which spectrum shows the light that emerges?

Top: emission spectrum of the gas, four bright lines on a dark background. Below: four spectra A to D. A: continuous spectrum with dark lines at the same four wavelengths. B: continuous spectrum with dark lines at four different wavelengths. C: bright lines on a dark background at the same four wavelengths. D: continuous spectrum with no lines.Emission spectrum of the gas400500600700wavelength / nmA.B.C.D.400500600700wavelength / nm
Diagram NOT accurately drawn
Show mark scheme
Marking pointMarkNotes
Step 1Atoms in the cool gas absorb photons whose energies equal the gaps between their levels — the same gaps that produce the emission lines — so the continuous spectrum emerges with dark lines at exactly the emission-line wavelengths.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: the cool gas removes only the photons whose energies match its level differences, leaving dark lines at the emission wavelengths on a continuous spectrum.
  • BThe dark lines are at different wavelengths from the bright lines; absorption and emission involve the same level differences, so they occur at the same wavelengths.
  • CThis is an emission spectrum again; light passing through a cool gas gives a continuous spectrum with dark lines missing from it.
  • DThe gas does absorb light, but only at particular wavelengths, so the spectrum cannot be continuous and unbroken.

Syllabus understandingE.1 — that emission and absorption spectra provide evidence for discrete atomic energy levels Command term: Identify

15E-1A-29
Energy levels & photons·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine

The diagram shows three energy levels of an atom. Photon P is emitted in the transition from −0.85 eV to −3.40 eV and photon Q in the transition from −1.51 eV to −3.40 eV.

What is (wavelength of Q)/(wavelength of P)?

Three energy levels at −0.85 eV, −1.51 eV and −3.40 eV; transition P from −0.85 eV to −3.40 eV and transition Q from −1.51 eV to −3.40 eV.Energy / eV−0.85−1.51−3.40PQ
Diagram NOT accurately drawn
Show mark scheme
Marking pointMarkNotes
Step 1Photon energies: EP = −0.85 − (−3.40) = 2.55 eV and EQ = −1.51 − (−3.40) = 1.89 eV.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2λ = hc/E, so λQ/λP = EP/EQ = 2.55/1.89 = 1.35.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis is 0.85/1.51, the ratio of the starting levels; the photon energy is the difference between two levels.
  • BThis is EQ/EP = 1.89/2.55, which is the ratio of P's wavelength to Q's — wavelength is inversely proportional to photon energy.
  • CCorrect: λ ∝ 1/E, so λQ/λP = 2.55/1.89 = 1.35.
  • DThis is (2.55/1.89)2, as if the wavelength were inversely proportional to the square of the energy.

Syllabus understandingE.1 — that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf Command term: Determine

16E-1A-30
Rutherford scattering·E.1 Structure of the atom
Paper 1AMedium1 mark
Multiple choice · 1 mark1 step to full marksIdentify

In the Geiger–Marsden–Rutherford experiment, alpha particles were directed at a thin gold foil.

Which row correctly links an observation to the conclusion drawn from it?

ObservationConclusion
Show mark scheme
Marking pointMarkNotes
Step 1Turning a fast alpha particle back needs a very large repulsive force, which is possible only if the positive charge and most of the mass are concentrated in a very small volume — the nucleus.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AUndeflected paths show that the atom is mostly empty space; on their own they say nothing about the sign of the charge.
  • BElectrons are far too light to deflect alpha particles noticeably; small deflections come from passing a nucleus at a larger distance.
  • CThe scattering depends on the charge and mass of the nucleus; it gives no information about neutrons, which were discovered later.
  • DCorrect: only a small, massive, positively charged nucleus can exert a force large enough to turn an alpha particle back.

Syllabus understandingE.1 — the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus Command term: Identify

17E-1B-13
Energy levels & photons·E.1 Structure of the atom
Paper 1BMedium11 marks
Data-based question11 steps to full marksDetermine

A student estimates the Planck constant using six light-emitting diodes (LEDs). For each LED the potential difference is increased slowly until the LED just begins to glow; this threshold p.d. V is read from a digital voltmeter (uncertainty ±0.05 V). The peak wavelength λ of each LED (uncertainty ±5 nm) is taken from the manufacturer's data sheet. The student assumes that the energy of each photon emitted equals eV.

(e = 1.60 × 10−19 C, c = 3.00 × 108 m s−1)

LEDλ / nm1/λ / 106 m−1V / V
red6351.571.85
orange6101.641.90
amber5921.692.00
green5252.24
blue4682.142.57
violet4052.98
Threshold p.d. V against 1/λ for six LEDs, points with error bars1.41.61.82.02.22.42.61/λ / 10⁶ m⁻¹1.61.82.02.22.42.62.83.03.2V / V
Graph drawn to scale
(a)

Complete the table for the green and the violet LEDs.

(1)
(b)

Show that the model predicts a straight-line graph of V against 1/λ with gradient hc/e.

(1)
(c)

Draw the line of best fit and determine its gradient.

(3)
(d)

Draw lines of maximum and minimum gradient and determine the absolute uncertainty in the gradient.

(2)
(e)

Determine the Planck constant and its absolute uncertainty.

(2)
(f)

The accepted value of the Planck constant is 6.63 × 10−34 J s. Comment on your result.

(1)
(g)

When extended, the line of best fit does not pass through the origin. Suggest one reason for this systematic error.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
green: 1.90; violet: 2.47 (× 106 m−1)✓ 1Both needed. Accept 3 s.f.
Part (b)
eV = hf = hc/λ, so V = (hc/e) × (1/λ), which is y = mx with m = hc/e✓ 1
Part (c)
Straight line through all the error bars✓ 1
Gradient from a large triangle✓ 1
Gradient = 1.28 × 10−6 V m✓ 1Accept 1.22–1.36 × 10−6 V m. Unit and power of ten needed.
Part (d)
Steepest and shallowest lines drawn through all the error bars✓ 1
Uncertainty = (max − min)/2, e.g. (1.44 − 1.12)/2 = ±0.16 × 10−6 V m✓ 1Accept ±0.08 to ±0.19 × 10−6 V m from the candidate's lines.
Part (e)
h = e × gradient/c = 1.60 × 10−19 × 1.28 × 10−6/3.00 × 108 = 6.8 × 10−34 J s✓ 1Allow ECF from (c).
Same percentage uncertainty as the gradient: h = (6.8 ± 0.9) × 10−34 J s✓ 1MP2 is for matching precision. Allow ECF from (d).
Part (f)
6.63 × 10−34 J s lies within the range of the result, so the result is consistent with the accepted value✓ 1Allow ECF.
Part (g)
Every V differs from hc/eλ by a similar amount: e.g. the photons are emitted over a range of wavelengths, not only at the peak λ, and electrons can gain extra energy from thermal motion, so the LED starts to emit at a p.d. slightly below hc/eλ✓ 1Accept any systematic effect that shifts every V by a similar amount, e.g. the onset of the glow is judged by eye; p.d. across the LED's internal resistance. OWTTE

Answers: (a) 1.90 and 2.47 × 106 m−1  ·  (c) 1.28 × 10−6 V m  ·  (d) ±0.16 × 10−6 V m  ·  (e) (6.8 ± 0.9) × 10−34 J s (the remaining parts are explanations — see the table above)

Syllabus understandingE.1 — that photons are emitted and absorbed during atomic transitions; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf. Command term: Determine

18E-1B-18
Rutherford scattering·E.1 Structure of the atom
Paper 1BEasy8 marks
Data-based question8 steps to full marksExplain

Students use a computer simulation of the Geiger–Marsden–Rutherford experiment. Alpha particles, all with the same kinetic energy, are directed at a thin gold foil in a vacuum, and a detector counts the alpha particles scattered through an angle θ in 10.0 minutes. Like a real experiment, the simulation includes random variation; the uncertainty in a count N is ±√N.

θ / °Counts in 10.0 minutes
1589664
305756
451233
60408
90106
12042
15033
(a)

State what the very large counts at small angles show about the atoms of the foil.

(1)
(b)

Calculate the percentage uncertainty in the count at θ = 150°.

(1)
(c)

Suggest why, in a real experiment, the counting time at large angles would be made much longer than at small angles.

(1)
(d)

Explain what the counts at angles greater than 90° show about the nucleus.

(2)
(e)

The simulation is repeated at θ = 60° with a foil of twice the thickness, giving 835 counts. A student predicts that, for a thin foil, the count is proportional to the thickness. Deduce whether the data support this prediction.

(2)
(f)

Suggest why a very thin foil is used.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Most alpha particles pass through with little or no deflection, so the atom is mostly empty space✓ 1OWTTE
Part (b)
√33/33 × 100 = 17 %✓ 1Accept 17.4 %.
Part (c)
The counts at large angles are small, so their percentage uncertainty is large; a longer time gives more counts and a smaller percentage uncertainty✓ 1OWTTE
Part (d)
A few alpha particles are turned back, which needs a very large repulsive force✓ 1
so the positive charge and most of the mass of the atom are concentrated in a very small nucleus✓ 1Award MP2 for "small, dense, positively charged nucleus". OWTTE
Part (e)
Prediction: 2 × 408 = 816 ± 40; measured 835 ± 29✓ 1
The difference (19) is smaller than the combined uncertainty (≈ 69), so the data support the prediction✓ 1MP2 requires a comparison with the uncertainties. Accept a ratio of 2.05 ≈ 2 within uncertainty.
Part (f)
So that each alpha particle is deflected by at most one nucleus / is not stopped in the foil✓ 1OWTTE

Answers: (b) 17 % (the remaining parts are explanations — see the table above)

Syllabus understandingE.1 — the Geiger–Marsden–Rutherford experiment and the discovery of the nucleus (guidance: only a qualitative approach is required for standard level). Command term: Explain

19E-2-16
Emission & absorption spectra·E.1 Structure of the atom
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksExplain

Figure 1 shows four energy levels, E1 to E4, of the atoms of an element Q. E1 is the ground state.

When Q is very hot it emits a line spectrum. In the visible region, 400 nm to 700 nm, there are three lines, at 478 nm, 497 nm and 654 nm (Figure 2, upper strip).

In a second experiment, white light is passed through a tube of Q vapour at room temperature and the transmitted light is examined with a spectroscope. The lower strip of Figure 2 shows the spectrum of the white light before it enters the tube.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

Figure 1: energy levels of element Q: E1 = −6.00 eV (ground state), E2 = −3.50 eV, E3 = −1.60 eV, E4 = −0.90 eV.0E₁−6.00 eVE₂−3.50 eVE₃−1.60 eVE₄−0.90 eV
Figure 1 — energy levels drawn to scale
Figure 2: upper strip, emission spectrum of hot Q with bright lines at 478, 497 and 654 nm; lower strip, a continuous visible spectrum on which the candidate draws the absorption lines.emission spectrum of hot Qlight transmitted through cool Q (draw the dark lines here)400450500550600650700wavelength / nm
Figure 2 — wavelength scale drawn to scale
(a)

Calculate the wavelength of the photon emitted when an atom of Q moves from E2 to E1.

(2)
(b)

Identify the transitions that produce the emission lines at 478 nm and 654 nm.

(2)
(c)

The absorption spectrum of cool Q.

(i)

Draw, on the lower strip of Figure 2, the dark lines that are seen in the visible spectrum of the light transmitted through the cool vapour.

(2)
(ii)

Explain your answer to (c)(i).

(3)
(iii)

Atoms that absorb a photon re-emit a photon of the same wavelength. Explain why the absorption lines nevertheless appear dark.

(2)
(d)

Suggest how the visible absorption spectrum would change if the vapour were heated to a very high temperature, with the white light still passing through it.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
ΔE = −3.50 − (−6.00) = 2.50 eV = 4.00 × 10−19 J✓ 1
λ = hc/ΔE = 6.63 × 10−34 × 3.00 × 108/4.00 × 10−19 = 4.97 × 10−7 m «497 nm»✓ 1Award [2] for CNA.
Part (b)
478 nm: photon energy hc/λ = 2.60 eV, the transition E4 → E2✓ 1
654 nm: photon energy 1.90 eV, the transition E3 → E2✓ 1Award [1 max] for both transitions identified with no supporting energies.
Part (c)(i)
One dark line, at 497 nm «directly below the 497 nm emission line»✓ 1
No dark lines at 478 nm or at 654 nm✓ 1MP2 only scores if MP1 scores.
Part (c)(ii)
At room temperature almost all the atoms of Q are in the ground state E1✓ 1
So only photons of energy En − E1 can be absorbed; the 478 nm and 654 nm lines need atoms already in E2, and there are almost none✓ 1
The other absorptions from E1 «to E3 and E4, about 283 nm and 244 nm» are in the ultraviolet, so they are not seen✓ 1Allow ECF from (c)(i).
Part (c)(iii)
The re-emitted photons travel in all «random» directions✓ 1
So far fewer photons of that wavelength continue towards the observer, and the line looks dark compared with the neighbouring wavelengths✓ 1OWTTE
Part (d)
Collisions would raise some atoms to E2, so dark lines could also appear at 478 nm and 654 nm✓ 1Accept: extra absorption lines due to transitions from excited states.

Answers: (a) 497 nm (the remaining parts are explanations — see the table above)

Syllabus understandingE.1 — that photons are emitted and absorbed during atomic transitions; that emission and absorption spectra provide evidence for discrete atomic energy levels Command term: Explain

20E-2-17
Emission & absorption spectra·E.1 Structure of the atom
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDeduce

The spectrum of a star S shows dark absorption lines on a continuous background. Table 1 gives the wavelengths of the absorption lines of S. Table 2 gives laboratory wavelengths of some lines of four elements.

(c = 3.00 × 108 m s−1, Wien's displacement constant = 2.9 × 10−3 m K)

Table 1: wavelengths of the absorption lines of star S / nm
393.49 · 396.97 · 434.18 · 486.28 · 589.17 · 589.77 · 656.48
Table 2: elementLaboratory wavelengths / nm
Hydrogen434.05 · 486.13 · 656.28
Helium447.15 · 587.56 · 667.82
Sodium588.99 · 589.59
Calcium393.37 · 396.85
(a)

Outline why the dark lines of an element in the spectrum of a star occur at wavelengths that also appear as bright lines in the laboratory emission spectrum of that element.

(2)
(b)

Deduce, with reasons, which of the four elements in Table 2 are present in the outer layers of S.

(3)
(c)

All the lines of S are shifted from their laboratory wavelengths.

(i)

Show that the fractional shift Δλ/λ of the hydrogen line at 656.28 nm is about 3.0 × 10−4.

(1)
(ii)

Determine the speed of S along the line of sight, and state whether S is moving towards or away from the Earth.

(2)
(iii)

State why every line in the spectrum of S is shifted by the same fraction of its wavelength.

(1)
(d)

The continuous spectrum of S has its maximum intensity at a wavelength of 480 nm. Estimate the surface temperature of S.

(1)
(e)

The spectrometer measures wavelengths with an uncertainty of ±0.05 nm. A second star moves directly away from the Earth at 15 km s−1. Determine whether the spectrometer can detect this motion using the calcium line at 393.37 nm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
An atom absorbs or emits only photons whose energy equals the difference between two of its energy levels, E = hc/λ✓ 1
Absorption between two levels involves the same energy difference, and so the same wavelength, as emission between them; each element has its own set of levels, so its lines identify it✓ 1OWTTE
Part (b)
Hydrogen and calcium are present: each of their laboratory lines has a matching line of S, slightly «0.1–0.2 nm» longer✓ 1
Sodium is present: both of its lines are matched, by 589.17 nm and 589.77 nm✓ 1
Helium is not present: S has no lines near 447.15, 587.56 or 667.82 nm «589.17 nm is the sodium line at 588.99 nm, not the helium line at 587.56 nm»✓ 1Award [1 max] for the three correct elements with no reasons.
Part (c)(i)
Δλ/λ = (656.48 − 656.28)/656.28 = 3.05 × 10−4✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)(ii)
v = cΔλ/λ = 3.00 × 108 × 3.05 × 10−4 = 9.1 × 104 m s−1✓ 1Accept 8.9 × 104 to 9.3 × 104 m s−1 from other lines; allow use of 3.0 × 10−4 «9.0 × 104 m s−1».
Away from the Earth, because the observed wavelengths are longer «red-shifted»✓ 1
Part (c)(iii)
Δλ/λ ≈ v/c depends only on the speed of the source, not on the wavelength, and all the light from S comes from a source with the same speed✓ 1OWTTE
Part (d)
T = 2.9 × 10−3/480 × 10−9 = 6.0 × 103 K✓ 1Accept 6040 K.
Part (e)
Δλ = λv/c = 393.37 × 1.5 × 104/3.00 × 108 = 0.020 nm✓ 1
0.020 nm is smaller than the uncertainty of 0.05 nm, so the motion cannot be detected «reliably»✓ 1Award MP2 for a conclusion consistent with the candidate's Δλ.

Answers: (c)(i) 3.05 × 10−4  ·  (c)(ii) 9.1 × 104 m s−1, away from the Earth  ·  (d) 6.0 × 103 K  ·  (e) Δλ = 0.020 nm — not detectable (the remaining parts are explanations — see the table above)

Syllabus understandingE.1 — that emission and absorption spectra provide information on the chemical composition; C.5 — the relative change in wavelength of light due to the Doppler effect, Δλ/λ ≈ v/c, and the use of spectral line shifts to determine the motion of stars; B.1 — Wien's displacement law Command term: Deduce

21E-2-18
Nuclear notation·E.1 Structure of the atom
Paper 2Medium14 marks
Short answer & extended response14 steps to full marksDetermine

Copper has two stable isotopes, copper-63 and copper-65. In a mass spectrometer, copper atoms are ionized so that every ion carries a charge of +e. The ions pass through a velocity selector and then through a narrow slit S into a region of uniform magnetic field of flux density 0.400 T directed into the page, where they move in semicircles and strike a detector plate (Figure 1).

In the velocity selector the ions move between two parallel plates 2.00 cm apart with a potential difference of 640 V between them. A uniform magnetic field of flux density 0.160 T, perpendicular to both the electric field and the velocity of the ions, also acts in the selector.

Masses of the ions: 63Cu+ 62.93 u, 65Cu+ 64.93 u. (1 u = 1.661 × 10−27 kg, e = 1.60 × 10−19 C)

Figure 1: copper ions leave a velocity selector, travel up the page through slit S in a detector plate and enter a uniform magnetic field directed into the page; the detector plate extends on both sides of S.×××××××××××××××××××××××××××××××××××××××××××××××××××××××Svelocity selectordetector plateuniform magnetic field, 0.400 T, directed into the pageion beam
Figure 1 — diagram not to scale
(a)

The ions.

(i)

State the number of protons, neutrons and electrons in one 6529Cu+ ion.

(2)
(ii)

Outline why the two isotopes of copper cannot be separated by chemical reactions.

(1)
(b)

The velocity selector.

(i)

Show that the ions leave the velocity selector with a speed of about 2.0 × 105 m s−1.

(2)
(ii)

Explain why ions of both isotopes leave the velocity selector with the same speed.

(2)
(c)

The region of uniform magnetic field.

(i)

Draw, on Figure 1, the paths of the 63Cu+ ions and of the 65Cu+ ions after they pass through S. Label each path.

(2)
(ii)

Determine the distance between the points at which the two types of ion strike the detector plate.

(3)
(d)

The current at the detector due to 63Cu+ ions is 6.4 × 10−10 A. Calculate the number of 63Cu+ ions arriving at the detector each second.

(1)
(e)

Suggest why the whole spectrometer must be evacuated.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
29 protons and 65 − 29 = 36 neutrons✓ 1
28 electrons «one electron has been removed from the neutral atom»✓ 1
Part (a)(ii)
Isotopes have the same number of protons and so the same number and arrangement of electrons in the atom; their chemical behaviour is identical✓ 1OWTTE
Part (b)(i)
E = V/d = 640/0.0200 = 3.20 × 104 V m−1✓ 1
For undeflected ions qE = qvB, so v = E/B = 3.20 × 104/0.160 = 2.00 × 105 m s−1✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
Only ions for which the electric force qE is balanced by the magnetic force qvB travel straight through to the exit slit; others are deflected✓ 1
The condition v = E/B does not depend on the mass «or charge» of the ion, so both isotopes emerge with the same speed✓ 1
Part (c)(i)
Both paths are semicircles that start at S and curve to the LEFT, ending on the detector plate✓ 1F = qvB on a positive ion moving up the page in a field into the page is to the left «Fleming's left-hand rule». Do not accept paths curving to the right.
The 65Cu+ path has the larger radius and both paths are labelled✓ 1Any visibly larger radius is acceptable.
Part (c)(ii)
qvB = mv2/r gives r = mv/qB; for 63Cu+ r = 62.93 × 1.661 × 10−27 × 2.00 × 105/(1.60 × 10−19 × 0.400) = 0.327 m✓ 1
For 65Cu+ r = 0.337 m✓ 1
Distance = 2 × (0.3370 − 0.3266) = 2.1 × 10−2 m✓ 1Accept 0.020–0.021 m. Award [3] for CNA. Allow ECF from (b)(i). Award [2 max] for 1.0 × 10−2 m «difference in radii, not diameters».
Part (d)
N = I/e = 6.4 × 10−10/1.60 × 10−19 = 4.0 × 109 s−1✓ 1
Part (e)
Collisions with gas molecules would deflect the ions or change their speed «or neutralise them», so they would not follow the predicted paths✓ 1OWTTE

Answers: (a)(i) 29 p, 36 n, 28 e  ·  (b)(i) 2.00 × 105 m s−1  ·  (c)(ii) 2.1 × 10−2 m  ·  (d) 4.0 × 109 s−1 (the remaining parts are explanations — see the table above)

Syllabus understandingE.1 — nuclear notation AZX where A is the nucleon number, Z is the proton number and X is the chemical symbol; E.3 — isotopes; D.3 — the motion of a charged particle in perpendicularly orientated uniform electric and magnetic fields Command term: Determine

22E-2-33
Energy levels & photons·E.1 Structure of the atom
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksDetermine whether

In a fluorescent lamp, mercury atoms emit ultraviolet photons of wavelength 254 nm. The inside of the glass tube is coated with a phosphor, which absorbs these photons and emits visible light.

The diagram shows four energy levels of an ion in the phosphor. The energies are measured from the ground state.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, Wien's displacement constant = 2.90 × 10−3 m K)

Four energy levels of a phosphor ion, measured from the ground state: 0, 2.10 eV, 2.65 eV and 4.89 eV, drawn to scale.energy / eV0ground state2.10 eV2.65 eV4.89 eV
Diagram drawn to scale
(a)

Show that the energy of a photon of wavelength 254 nm is about 4.9 eV.

(1)
(b)

The filament of a filament lamp is at a temperature of 2900 K.

(i)

Calculate the wavelength at which the spectrum of the filament has its maximum intensity.

(1)
(ii)

Hence suggest why a fluorescent lamp gives out much more visible light than a filament lamp with the same input power.

(2)
(c)

An ion in the ground state absorbs a 254 nm photon.

(i)

Draw arrows on the diagram to show this absorption and one way in which the ion can return to the ground state by emitting exactly two photons.

(2)
(ii)

Calculate the wavelength of the photon emitted in the transition from the 4.89 eV level to the 2.10 eV level.

(2)
(iii)

Determine whether a transition from the 2.65 eV level to the 2.10 eV level produces visible light.

(2)
(d)

Suggest why a lamp without the phosphor coating would give out very little visible light.

(1)
(e)

Some ions return to the ground state through the 2.65 eV and 2.10 eV levels in turn, emitting three photons. Determine the percentage of the energy of the absorbed ultraviolet photon that is emitted as visible light by these ions.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = 6.63 × 10−34 × 3.00 × 108/254 × 10−9 = 7.83 × 10−19 J «= 4.89 eV»✓ 1Must see full substitution OR answer to 3 s.f. (4.89 eV).
Part (b)(i)
λmax = 2.90 × 10−3/2900 = 1.0 × 10−6 m «1000 nm»✓ 1
Part (b)(ii)
The filament gives a continuous spectrum whose peak is in the infrared, so most of its energy is emitted as infrared and only a small fraction between 400 nm and 700 nm✓ 1Allow ECF from (b)(i).
In the fluorescent lamp the energy is emitted at a few particular wavelengths: the ultraviolet line of mercury is converted by the phosphor into visible lines, so a large fraction of the energy is visible✓ 1OWTTE
Part (c)(i)
upward arrow from the ground state to the 4.89 eV level✓ 1
two downward arrows: from 4.89 eV to 2.65 eV OR to 2.10 eV, and then to the ground state✓ 1Arrows must start and end on levels. Do not award MP2 for a route with three emissions.
Part (c)(ii)
ΔE = 2.79 eV = 2.79 × 1.60 × 10−19 = 4.46 × 10−19 J✓ 1
λ = hc/ΔE = 4.46 × 10−7 m «446 nm»✓ 1Award [2] for CNA.
Part (c)(iii)
ΔE = 0.55 eV, so λ = hc/ΔE = 2.3 × 10−6 m✓ 1ALTERNATIVE: compare 0.55 eV with the photon energies of visible light, about 1.8–3.1 eV.
this is longer than 700 nm «infrared», so the light is not visible✓ 1Conclusion must be consistent with the calculation.
Part (d)
most of the energy is emitted by mercury as ultraviolet radiation, which is invisible «and is absorbed by the glass»✓ 1OWTTE
Part (e)
visible photons: 4.89 − 2.65 = 2.24 eV «555 nm» and 2.10 eV «592 nm»; the 0.55 eV photon is infrared✓ 1Allow ECF from (c)(iii).
(2.24 + 2.10)/4.89 × 100 = 89 %✓ 1Award [2] for CNA.

Answers: (a) 4.89 eV  ·  (b)(i) 1.0 × 10−6 m  ·  (c)(ii) 446 nm  ·  (c)(iii) 2.3 × 10−6 m — infrared, not visible  ·  (e) 89 % (the remaining parts are explanations — see the table above)

Syllabus understandingE.1 — that photons are emitted and absorbed during atomic transitions; that the frequency of the photon released during an atomic transition depends on the difference in energy level as given by E = hf; that emission and absorption spectra provide evidence for discrete atomic energy levels; B.1 — the emission spectrum of a black body and Wien's displacement law Command term: Determine whether

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