DP Physics SL E.1 Structure of the Atom

DP Physics SL E.1 Structure of the Atom - Full notes

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DP Physics SL | E.1 Structure of the Atom

DP PHYSICS · THEME E: NUCLEAR AND QUANTUM PHYSICS

E.1

Structure of the Atom

Full notes for SL · IB-style

Contents

1	Models of the atom
2	Nuclei and isotopes
3	Photons and spectra
4	Atomic physics in context
5	Exam technique
6	Data-based questions and further practice
7	Common misconceptions and links
8	Exam-style question with markscheme
9	Check your understanding
10	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
describe the Geiger–Marsden experiment and the nuclear model
use nuclide notation and describe isotopes
use E = hf and the electronvolt
explain emission and absorption spectra using energy levels
calculate photon wavelengths for transitions, including in hydrogen
describe ionisation and excitation

This booklet covers the SL content of this subtopic. Additional Higher Level material is not included.

Guiding questions

How do the energy levels of electrons in atoms explain spectra? What evidence supports the nuclear model of the atom? How are the size and density of nuclei determined?

How this topic is examined

E.1 appears in Papers 1 and 2. Expect descriptions of Geiger–Marsden, nuclide notation, energy-level transitions and spectra; HL adds nuclear radius, density and closest approach.

Convert between eV and J carefully and use hc for wavelength.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Models of the atom

From indivisible spheres to a nucleus.

Think about it

How could scientists learn about something as small as an atom without ever seeing it?

Our picture of the atom changed dramatically between 1897 and 1913 as new experiments overturned old models. This section follows that story and the key experiment that revealed the nucleus.

1.1 Early models

Dalton (1803) proposed indivisible atoms. Thomson discovered the electron (1897) and proposed the 'plum pudding' model: negative electrons embedded in a sphere of spread-out positive charge. Rutherford (1911) proposed a tiny dense positive nucleus, and Bohr (1913) added quantised electron energy levels.

Figure 1. Three historical models of the atom.

Quick check

1. What discovery led to the plum pudding model?

Answers: 1. the electron

1.2 Discovering the electron

Thomson found that cathode rays were deflected by electric and magnetic fields towards a positive plate, so they were negatively charged particles. Their charge-to-mass ratio was about 1800 times that of a hydrogen ion, so they were far lighter than atoms: the first subatomic particle, the electron.

Figure 2. Cathode rays deflected by an electric field.

Quick check

1. What did the direction of deflection show about cathode rays?

Answers: 1. they are negatively charged

1.3 The Geiger–Marsden experiment

Alpha particles from a radioactive source were fired at very thin gold foil in a vacuum, and a detector was moved around to count scattered alphas at different angles. Most passed straight through; a few were deflected through large angles; about 1 in 8000 bounced back.

Observation	Conclusion
most alphas pass straight through	the atom is mostly empty space
some deflected through small angles	positive charge is concentrated
very few bounce back	the nucleus is tiny, dense and positive

Figure 3. Geiger–Marsden scattering apparatus (in a vacuum).

Quick check

1. Why was a very thin foil used?

Answers: 1. so that each alpha is scattered by at most one nucleus

1.4 Interpreting the scattering

An alpha particle aimed close to a nucleus feels a strong repulsive force and is deflected through a large angle; one passing far away is barely deflected. The number of alphas falls very steeply with scattering angle, consistent with a point-like positive nucleus.

Figure 4. Alpha particle paths near a nucleus: the closer the approach, the larger the deflection.

Figure 5. Relative number of alphas against scattering angle (logarithmic scale).

WORKED EXAMPLE 1 Plum pudding failure (exam level)

Explain why the plum pudding model could not explain the large-angle scattering.

Solution

Spread-out positive charge produces only weak, small deflecting forcesR1

Large deflections need positive charge and mass concentrated in a tiny regionR1

Common mistake

Saying that most alpha particles bounced back. Only about 1 in 8000 did.

1.5 Deflection and miss distance

The closer an alpha particle's line of approach is to the nucleus (the smaller its impact parameter), the larger its deflection. Since the nucleus is tiny, only a very small fraction of alphas pass close enough to be strongly deflected.

Figure 6. Deflection angle against impact parameter (schematic).

Quick check

1. Why do so few alpha particles get large deflections?

Answers: 1. very few pass close enough to the tiny nucleus

1.6 Sizes in the atom

Atoms have radii of about 10⁻¹⁰ m; nuclei about 10⁻¹⁵ m (1 fm). The nucleus contains almost all the mass. If the nucleus were the size of a marble, the atom would be the size of a football stadium.

WORKED EXAMPLE 2 Scale (standard)

If a nucleus were 1 cm across, estimate the size of the atom.

Solution

Ratio 10⁻¹⁰/10⁻¹⁵ = 10⁵M1

Atom ≈ 1 km acrossA1

WORKED EXAMPLE 3 Volume fraction (exam level)

Estimate the fraction of an atom's volume occupied by its nucleus.

Solution

(10⁻¹⁵/10⁻¹⁰)³M1

= 10⁻¹⁵A1

2 Nuclei and isotopes

What the nucleus is made of.

2.1 Nucleons and nuclide notation

The nucleus contains protons and neutrons (nucleons). A nuclide is written ᴬₓX: nucleon (mass) number A at the top, proton (atomic) number Z at the bottom. Neutron number N = A − Z. In a neutral atom the number of electrons equals Z.

WORKED EXAMPLE 4 Particles (basic)

State the numbers of protons, neutrons and electrons in a neutral atom of uranium-238 (Z = 92).

Solution

92 protons, 92 electronsA1

146 neutronsA1

Quick check

1. How many neutrons are in carbon-14 (Z = 6)?

Answers: 1. 8

2.2 Isotopes

Isotopes are atoms of the same element (same Z) with different numbers of neutrons (different A). They have the same chemistry but different masses and nuclear stability. Many elements are mixtures of isotopes, which is why relative atomic masses are not whole numbers.

Isotope	Protons	Neutrons	Stable?
hydrogen-1	1	0	yes
deuterium (H-2)	1	1	yes
tritium (H-3)	1	2	no (beta emitter)
carbon-12	6	6	yes
carbon-14	6	8	no (beta emitter)

WORKED EXAMPLE 5 Average mass (standard)

Chlorine is 75% Cl-35 and 25% Cl-37. Calculate its average relative atomic mass.

Solution

0.75 × 35 + 0.25 × 37M1

= 35.5A1

2.3 Atomic mass units

The unified atomic mass unit u = 1.661 × 10⁻²⁷ kg, one-twelfth of the mass of a carbon-12 atom. Proton mass ≈ 1.007 u; neutron ≈ 1.009 u; electron ≈ 0.000549 u.

WORKED EXAMPLE 6 Conversion (standard)

Convert 4.0026 u (helium-4 atom) to kilograms.

Solution

4.0026 × 1.661 × 10⁻²⁷M1

= 6.65 × 10⁻²⁷ kgA1

2.4 The isotopes of hydrogen

Hydrogen has three isotopes: hydrogen-1 (one proton), deuterium (one proton, one neutron) and tritium (one proton, two neutrons). Deuterium and tritium are the fuels of fusion reactors (E.5).

Figure 7. The three isotopes of hydrogen (red: proton, grey: neutron, blue: electron).

2.5 Measuring isotope abundance

A mass spectrometer (D.3) separates ions by mass, and the detector current at each position gives the relative abundance of each isotope. Averaging the masses weighted by abundance gives the relative atomic mass.

Figure 8. Isotope abundances of neon from a mass spectrometer.

WORKED EXAMPLE 7 Neon (exam level)

Use the abundances in the chart to find the relative atomic mass of neon.

Solution

0.905 × 20 + 0.003 × 21 + 0.092 × 22M1

= 20.2A1

2.6 Stable and unstable nuclei

Light stable nuclei have roughly equal numbers of protons and neutrons; heavier stable nuclei have more neutrons than protons. Nuclei away from the band of stability are radioactive (E.3).

Figure 9. Neutron number against proton number: stable nuclei lie in a band above N = Z for heavy elements.

Quick check

1. Why do heavy nuclei need extra neutrons?

Answers: 1. neutrons add strong-force attraction without electric repulsion

2.7 Forces in the nucleus

Protons in the nucleus repel each other electrically. The strong nuclear force, which acts between all nucleons at very short range (about 1–3 fm), is attractive and much stronger, holding the nucleus together. At even shorter distances it becomes repulsive, keeping nucleons apart.

Quick check

1. Why do nuclei not fly apart under electrostatic repulsion?

Answers: 1. the strong nuclear force overcomes it at short range

3 Photons and spectra

Light carries energy in packets.

Think about it

Why does a sodium street lamp glow orange-yellow while a neon sign glows red?

3.1 Photons

Light and other electromagnetic radiation is emitted and absorbed in packets (photons) of energy E = hf = hc/λ, where h = 6.63 × 10⁻³⁴ J s. Higher-frequency (shorter-wavelength) photons carry more energy.

Figure 10. Photon energy against wavelength (E in eV = 1240/λ in nm).

WORKED EXAMPLE 8 Photon energy (standard)

Calculate the energy of a 500 nm photon in J and eV.

Solution

E = 6.63 × 10⁻³⁴ × 3.0 × 10⁸/5.0 × 10⁻⁷ = 3.98 × 10⁻¹⁹ JM1 A1

= 2.5 eVA1

WORKED EXAMPLE 9 Photons per second (exam level)

A 1.0 mW laser emits 633 nm light. How many photons per second does it emit?

Solution

E = 3.14 × 10⁻¹⁹ J per photonM1

N = 1.0 × 10⁻³/3.14 × 10⁻¹⁹ = 3.2 × 10¹⁵ s⁻¹A1

3.2 Photon energies of visible light

Visible photons carry between about 1.8 eV (red) and 3.1 eV (violet). This range matches the energy gaps of many atoms and molecules, which is why we see colours from flames, lamps and LEDs.

Figure 11. Photon energies of visible colours (wavelengths in nm).

WORKED EXAMPLE 10 LED colour (standard)

An LED emits photons of energy 2.3 eV. Find the wavelength and colour.

Solution

λ = 1240/2.3M1

= 540 nm (green)A1

WORKED EXAMPLE 11 Flame test (exam level)

Lithium gives a red flame at 671 nm. Find the energy gap and explain why the colour identifies lithium.

Solution

ΔE = 1240/671 = 1.85 eVA1

This gap is characteristic of lithium's energy levelsR1

3.3 Emission and absorption spectra

A hot, dense source gives a continuous spectrum. A hot low-pressure gas emits light only at particular wavelengths (an emission line spectrum). Cool gas in front of a continuous source absorbs the same wavelengths, giving dark lines (an absorption spectrum). Each element has its own unique set of lines.

Figure 12. Continuous, emission and absorption spectra (hydrogen lines shown).

WORKED EXAMPLE 12 Stars (standard)

Explain how astronomers detect helium in the Sun.

Solution

Dark absorption lines in the Sun's spectrumR1

Their wavelengths match helium's characteristic linesR1

In the real world

Helium was discovered in 1868 from an unexplained yellow line in the Sun's spectrum, 27 years before it was found on Earth.

3.4 Energy levels

Electrons in atoms can only occupy discrete energy levels. When an electron drops from a higher level to a lower one, a photon is emitted with energy equal to the difference: hf = Eupper − Elower. Absorption is the reverse: only photons of exactly the right energy are absorbed.

Key point

Energy levels are negative: zero is the energy of a free electron at rest (ionised atom).

WORKED EXAMPLE 13 Transition (standard)

An electron falls from −1.51 eV to −3.40 eV in hydrogen. Find the wavelength of the emitted photon.

Solution

ΔE = 1.89 eV = 3.02 × 10⁻¹⁹ JM1

λ = hc/ΔE = 658 nm (red)A1

3.5 Quantisation: a staircase analogy

A ball on a ramp can rest at any height, but a ball on a staircase can only rest on particular steps. Electrons in atoms are like the staircase: only certain energies are allowed, so only certain energy changes (and photon energies) are possible. The steps get closer together near the top, just as hydrogen's levels crowd together near E = 0.

Figure 13. Quantised energy levels are like steps on a staircase.

Quick check

1. Why can an atom not absorb a photon whose energy lies between two allowed gaps?

Answers: 1. there is no allowed level for the electron to move to

3.6 The hydrogen spectrum

Hydrogen's levels are En = −13.6/n² eV. Transitions down to n = 1 emit ultraviolet (Lyman series); to n = 2 emit visible light (Balmer series); to n = 3 infrared (Paschen series).

Figure 14. Energy levels of hydrogen with Lyman and Balmer transitions.

Figure 15. Balmer series wavelengths against upper level: lines crowd together towards a limit.

WORKED EXAMPLE 14 Lyman line (exam level)

Calculate the wavelength of the photon emitted in the n = 2 to n = 1 transition.

Solution

ΔE = 10.2 eV = 1.63 × 10⁻¹⁸ JM1

λ = 122 nm (UV)A1

3.7 Ionisation

The ionisation energy is the energy needed to remove an electron from the ground state completely (to E = 0). For hydrogen it is 13.6 eV. Photons with more energy than this can ionise the atom, with the excess becoming the electron's kinetic energy.

WORKED EXAMPLE 15 Ionising photon (standard)

Find the maximum wavelength of a photon that can ionise hydrogen from the ground state.

Solution

λ = hc/(13.6 × 1.6 × 10⁻¹⁹)M1

= 91 nmA1

3.8 Counting spectral lines

An electron in level n can return to the ground state by various routes, producing up to n(n − 1)/2 different photon energies.

WORKED EXAMPLE 16 Lines (standard)

How many different photon energies are possible from atoms excited to n = 4?

Solution

4 × 3/2M1

= 6A1

3.9 Excitation by collision

Atoms can also be excited by collisions with electrons, as in fluorescent lamps and neon signs. An electron with enough kinetic energy raises an atomic electron to a higher level; light is emitted when it falls back.

Quick check

1. How does a fluorescent tube produce visible light?

Answers: 1. mercury atoms emit UV; the phosphor coating absorbs UV and re-emits visible light

3.10 Energy levels of other atoms

Every element has its own pattern of energy levels and hence its own spectrum. Mercury, used in fluorescent lamps, has a strong transition of 4.9 eV that emits ultraviolet at 253 nm.

Figure 16. Simplified energy levels of mercury.

WORKED EXAMPLE 17 Mercury line (standard)

Show that the 4.9 eV transition gives a wavelength of about 253 nm.

Solution

λ = 1240/4.9M1

= 253 nmA1

3.11 The solar spectrum

The Sun's spectrum is crossed by thousands of dark absorption lines (Fraunhofer lines) from elements in its cooler outer layers. Matching them to laboratory spectra reveals the Sun's composition.

Figure 17. Some of the main Fraunhofer absorption lines in sunlight.

Quick check

1. Which element produces the D lines at 589 nm?

Answers: 1. sodium

3.12 Fluorescence

Some materials absorb a high-energy (UV) photon and re-emit the energy as two or more lower-energy photons, some of them visible. Fluorescent lamps, highlighter pens and security markings on banknotes use this.

Figure 18. Fluorescence: one UV photon absorbed, lower-energy photons emitted.

Quick check

1. Why is the emitted light always of longer wavelength than the absorbed light?

Answers: 1. each emitted photon has less energy than the absorbed one

3.13 Continuous spectra and lasers

In hot dense matter, atoms interact strongly and the energy levels smear out, so all wavelengths are emitted. Lasers, by contrast, use stimulated emission from one particular transition to produce intense, monochromatic, coherent light.

WORKED EXAMPLE 18 Classifying (standard)

Classify the spectrum from (a) a sodium street lamp; (b) a car headlamp filament.

Solution

(a) line (emission) spectrumA1

(b) continuous spectrumA1

4 Atomic physics in context

Applications of energy levels.

4.1 How lasers work

An atom can absorb a photon, emit one spontaneously, or be stimulated to emit by a passing photon of the same energy. Stimulated emission produces a second photon identical in frequency, phase and direction. In a laser, a population inversion (more atoms excited than not) lets stimulated emission dominate, and mirrors build up an intense, coherent beam.

Figure 19. Absorption, spontaneous emission and stimulated emission.

Quick check

1. Why is laser light monochromatic?

Answers: 1. all photons come from the same energy-level transition

4.2 Measuring wavelengths of spectral lines

A diffraction grating (C.3) separates light from a discharge lamp into its spectral lines; measuring the angles gives the wavelengths using nλ = d sin θ, and hence the energy-level differences.

Figure 20. A grating spectrometer separates the spectral lines of a lamp.

WORKED EXAMPLE 19 Hydrogen line (exam level)

A grating with 600 lines per mm diffracts a hydrogen line to 23.2° in first order. Find its wavelength and the energy gap.

Solution

λ = d sin θ = 1.667 × 10⁻⁶ × sin 23.2° = 657 nmM1 A1

E = 1240/657 = 1.89 eVA1

4.3 Atomic clocks

Atomic clocks use the extremely stable frequency of radiation from a particular transition in caesium or strontium atoms. The second is defined as 9 192 631 770 periods of the caesium-133 transition. GPS depends on such clocks.

WORKED EXAMPLE 20 Caesium photon (standard)

Find the energy and wavelength of the caesium clock radiation (f = 9.19 × 10⁹ Hz).

Solution

E = hf = 6.1 × 10⁻²⁴ JA1

λ = c/f = 3.3 cm (microwave)A1

4.4 Spectra across the universe

The same spectral lines appear in light from distant stars and galaxies, shifted to longer wavelengths (redshift) by the expansion of the universe. Recognising the pattern of lines lets astronomers measure the shift and the speed of recession (C.3).

Quick check

1. How do astronomers know that distant galaxies are made of the same elements as Earth?

Answers: 1. their spectra show the same patterns of lines, only shifted

5 Exam technique

Getting atomic physics questions right.

5.1 Multi-step problem: discharge tube

WORKED EXAMPLE 21 Neon sign (exam level)

A neon sign emits a strong line at 640 nm. Electrons in the tube are accelerated through 20 V between collisions. Find (a) the photon energy; (b) the electron kinetic energy; (c) whether such electrons can excite this transition.

Solution

(a) E = 1240/640 = 1.94 eVA1

(b) Ek = 20 eVA1

5.2 Multi-step problem: transitions

WORKED EXAMPLE 22 Hydrogen photons (exam level)

Hydrogen atoms are excited to n = 4. Find (a) the number of possible lines; (b) the longest wavelength emitted; (c) the shortest wavelength emitted.

Solution

(a) 6 linesA1

(b) n = 4 → 3: ΔE = 0.66 eV, λ = 1.88 μm (infrared)M1 A1

5.3 Multi-step problem: star spectrum

WORKED EXAMPLE 23 Stellar hydrogen (exam level)

The absorption line of hydrogen at 656.3 nm in a laboratory appears at 662.9 nm in a star's spectrum. Find (a) the energy of the laboratory photon; (b) which hydrogen transition it corresponds to; (c) the star's recession speed (Δλ/λ ≈ v/c).

Solution

(a) E = 1240/656.3 = 1.89 eVA1

(b) n = 3 to n = 2 (absorption: n = 2 to n = 3)A1

6 Data-based questions and further practice

Applying atomic physics.

6.1 Data: hydrogen lines

The four visible lines of hydrogen have wavelengths 656, 486, 434 and 410 nm.

λ / nm	656	486	434	410
photon energy / eV	1.89	2.55	2.86	3.03
upper level n	3	4	5	6

WORKED EXAMPLE 24 Checking a line (exam level)

Show that the 486 nm line corresponds to the n = 4 to n = 2 transition.

Solution

E = 1240/486 = 2.55 eVM1

−0.85 − (−3.40) = 2.55 eV ✓A1

6.2 Method checklist

•Convert eV to J (× 1.6 × 10⁻¹⁹) before using λ = hc/E, or use λ(nm) = 1240/E(eV).
•Energy differences between levels give photon energies; levels themselves are negative.
•Describe Geiger–Marsden observations and conclusions separately.
•Distinguish emission (bright lines) from absorption (dark lines).

6.3 Graded practice

WORKED EXAMPLE 25 Frequency (standard)

Calculate the frequency of a photon of energy 4.0 eV.

Solution

f = E/h = 6.4 × 10⁻¹⁹/6.63 × 10⁻³⁴M1

= 9.7 × 10¹⁴ HzA1

WORKED EXAMPLE 26 Absorption (exam level)

Hydrogen in the ground state is illuminated with photons of 10.2, 11.0 and 12.1 eV. Which are absorbed?

Solution

Allowed jumps from −13.6 eV: 10.2 eV (to n = 2) and 12.1 eV (to n = 3)M1

10.2 and 12.1 eV absorbed; 11.0 eV is notA1

WORKED EXAMPLE 27 Excited atom (exam level)

A hydrogen atom absorbs a 12.1 eV photon from the ground state. List the photon energies it can emit on returning.

Solution

Reaches n = 3M1

12.1, 1.89 and 10.2 eVA1

6.4 Graded practice: atoms and nuclei

WORKED EXAMPLE 28 Nuclide (standard)

An ion of magnesium-24 (Z = 12) has charge +2e. State its numbers of protons, neutrons and electrons.

Solution

12 protons, 12 neutronsA1

10 electronsA1

WORKED EXAMPLE 29 Gold atoms in foil (exam level)

A gold foil is 0.40 μm thick. Gold atoms are about 2.6 × 10⁻¹⁰ m across. Roughly how many atomic layers does an alpha cross?

Solution

0.40 × 10⁻⁶/2.6 × 10⁻¹⁰M1

≈ 1500 layersA1

WORKED EXAMPLE 30 Density comparison (standard)

Compare the density of a nucleus (2.3 × 10¹⁷ kg m⁻³) with that of water.

Solution

2.3 × 10¹⁷/1000M1

≈ 10¹⁴ times denserA1

6.5 Graded practice: spectra

WORKED EXAMPLE 31 Balmer limit (exam level)

Find the shortest wavelength in the Balmer series (n = ∞ to n = 2).

Solution

ΔE = 3.40 eVM1

λ = 1240/3.40 = 365 nmA1

WORKED EXAMPLE 32 Sodium (standard)

Sodium emits a strong 589 nm line. Find the energy gap in eV.

Solution

E = 1240/589M1

= 2.1 eVA1

WORKED EXAMPLE 33 Absorbed energy (exam level)

An electron with 12.5 eV kinetic energy collides with a ground-state hydrogen atom and excites it to n = 3. Find the electron's kinetic energy afterwards.

Solution

Energy used = 12.09 eVM1

Remaining = 0.4 eVA1

6.6 Graded practice: photons

WORKED EXAMPLE 34 Photon from frequency (standard)

A photon has frequency 6.0 × 10¹⁴ Hz. Find its energy in eV and its wavelength.

Solution

E = hf = 3.98 × 10⁻¹⁹ J = 2.5 eVA1

λ = c/f = 500 nmA1

WORKED EXAMPLE 35 Bulb photons (exam level)

A 60 W lamp converts 5% of its power into visible light of average wavelength 560 nm. Estimate the number of visible photons emitted per second.

Solution

Visible power 3.0 W; E per photon = 3.55 × 10⁻¹⁹ JM1

N = 8.4 × 10¹⁸ s⁻¹A1

WORKED EXAMPLE 36 X-ray photon (standard)

Find the energy in keV of an X-ray photon of wavelength 0.10 nm.

Solution

E = 1240/0.10 eVM1

= 12.4 keVA1

6.7 Graded practice: Geiger–Marsden

WORKED EXAMPLE 37 Why gold? (standard)

Suggest why gold was chosen for the foil.

Solution

Gold can be beaten into extremely thin foilR1

Large Z gives strong, easily detected scatteringR1

WORKED EXAMPLE 38 Why a vacuum? (standard)

Explain why the apparatus was evacuated.

Solution

Alpha particles have a short range in air (a few cm)R1

Air molecules would scatter or absorb themR1

WORKED EXAMPLE 39 Changing the foil (exam level)

Predict the effect on the number of large-angle scatterings of using aluminium (Z = 13) instead of gold (Z = 79) of the same thickness.

Solution

Smaller nuclear charge gives weaker repulsionR1

Far fewer large-angle deflectionsA1

6.8 Graded practice: energy levels

WORKED EXAMPLE 40 Paschen line (exam level)

Find the wavelength of the n = 4 to n = 3 transition in hydrogen and state its region.

Solution

ΔE = 1.51 − 0.85 = 0.66 eVM1

λ = 1.9 μm (infrared)A1

WORKED EXAMPLE 41 Absorbed wavelength (standard)

What wavelength must a photon have to raise a hydrogen atom from n = 2 to n = 4?

Solution

ΔE = 3.40 − 0.85 = 2.55 eVM1

λ = 486 nmA1

WORKED EXAMPLE 42 Ionisation from n = 3 (standard)

Find the minimum energy and maximum wavelength to ionise hydrogen from n = 3.

Solution

E = 1.51 eVA1

λ = 1240/1.51 = 821 nmA1

6.9 Graded practice: explanations

WORKED EXAMPLE 43 Line spectrum (exam level)

Explain why hot hydrogen gas emits only certain wavelengths.

Solution

Electrons occupy discrete energy levelsR1

Photons are emitted only with energies equal to differences between levels: hf = ΔER1

WORKED EXAMPLE 44 Absorption lines (exam level)

Explain why absorption lines are dark even though the absorbing atoms later re-emit the energy.

Solution

Re-emission occurs in all directionsR1

So much less light at those wavelengths continues towards the observerR1

WORKED EXAMPLE 45 Strong force (standard)

State two properties of the strong nuclear force.

Solution

Acts between nucleons; very short range (about 1–3 fm)A1

Attractive at typical separations, repulsive at very short range; much stronger than electric forceA1

6.10 Key terms

Term	Meaning
nucleon	proton or neutron
nucleon number A	number of protons + neutrons
proton number Z	number of protons
isotope	same Z, different A
photon	packet of electromagnetic energy E = hf
energy level	allowed energy of an electron in an atom
ionisation energy	energy to remove an electron from the ground state

6.11 Self-test

Try these without notes, then check the answers.

Quick check

1. What fraction of alpha particles bounced back in Geiger–Marsden?

2. How many neutrons in ²³₈₉₂U?

3. Find the energy in eV of a 620 nm photon.

4. Why are energy levels negative?

5. What series of hydrogen lines is visible?

6. Why does each element have a unique spectrum?

Answers: 1. about 1 in 8000 2. 146 3. 2.0 eV 4. zero is defined as a free electron; bound electrons have less energy 5. the Balmer series (transitions to n = 2) 6. each has a unique set of energy levels

6.12 Data: ionisation energies

The first ionisation energies of elements vary periodically, peaking at the noble gases, whose electrons are most tightly bound.

Figure 21. First ionisation energy against proton number.

WORKED EXAMPLE 46 Ionising photon (exam level)

Find the maximum wavelength of light that can ionise sodium (5.1 eV) and state the region of the spectrum.

Solution

λ = 1240/5.1M1

= 243 nm (ultraviolet)A1

6.13 Comparing atomic models

Model	Key idea	Evidence or problem
Thomson	electrons in spread-out positive charge	failed to explain large-angle alpha scattering
Rutherford	tiny dense positive nucleus	explained scattering; could not explain line spectra or stability
Bohr	quantised electron energy levels	explained hydrogen spectrum; failed for larger atoms
quantum (modern)	electrons described by wave functions	explains all atomic spectra and chemistry

Quick check

1. Which observation did the Rutherford model fail to explain?

Answers: 1. line spectra (and why electrons do not spiral into the nucleus)

6.14 Self-test 2

More quick questions on spectra and atoms.

Quick check

1. What is the energy of the n = 3 level of hydrogen?

2. State the region of the spectrum of the Lyman series.

3. Which spectrum shows dark lines on a bright background?

4. Find the wavelength of a 3.0 eV photon.

5. What is an isotope?

6. Name the force that holds nucleons together.

Answers: 1. −1.51 eV 2. ultraviolet 3. an absorption spectrum 4. 413 nm 5. a nucleus with the same Z but different A 6. the strong nuclear force

6.15 Common exam errors

•Forgetting to convert eV to joules.
•Using a level's energy instead of the difference between levels.
•Saying most alpha particles bounced back.
•Confusing A and Z in nuclide notation.
•Mixing up emission and absorption spectra.

7 Common misconceptions and links

Students often think…	Actually…
electrons orbit like planets	they occupy discrete energy levels
absorption and emission lines differ in wavelength	they occur at the same wavelengths
most alpha particles bounced back	very few did
isotopes have different chemistry	they have the same chemistry
energy levels are positive	bound levels are negative; zero at ionisation

How E.1 connects to the rest of the course

Topic	Connection
E.2 Quantum physics (HL)	photons and quantisation
E.3 Radioactive decay	nuclear stability
C.3 Wave phenomena	diffraction gratings for spectra
E.5 Fusion and stars	stellar spectra

8 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

The ground-state energy of hydrogen is −13.6 eV and the first two excited states are −3.40 eV and −1.51 eV.

(a) Describe what is meant by the ground state. [1]

(b) Calculate the wavelength of the photon emitted in a transition from −1.51 eV to −3.40 eV. [3]

(d) Explain why hydrogen gas produces a line spectrum rather than a continuous spectrum. [2]

(e) Outline one conclusion from the Geiger–Marsden experiment. [2]

Markscheme

Part	Answer	Marks
(a)	lowest energy state of the atom	A1
(b)	ΔE = 1.89 eV = 3.02 × 10⁻¹⁹ J	M1
	λ = hc/ΔE	M1
	658 nm	A1
(c)	visible (red)	A1
(d)	electrons have discrete energy levels	R1
	only specific photon energies emitted	R1
(e)	atom mostly empty space / nucleus small and dense	R1
	supported by most alphas passing through / few large deflections	R1

Examiner's comments. In (b), the commonest error was failing to convert eV to joules before using λ = hc/E.

9 Check your understanding

Section A: multiple choice

1. Large-angle alpha scattering shows that

A. the nucleus is small, dense and positive B. electrons are massive C. atoms are solid D. the nucleus is negative

2. A nuclide with A = 23 and Z = 11 has

A. 11 neutrons B. 23 neutrons C. 12 neutrons D. 34 neutrons

3. Isotopes have the same

A. number of neutrons B. number of protons C. mass number D. nuclear mass

4. 1 eV equals

A. 1 J B. 6.63 × 10⁻³⁴ J C. 9.11 × 10⁻³¹ J D. 1.6 × 10⁻¹⁹ J

5. Photon energy is proportional to

A. wavelength B. frequency C. amplitude D. speed

6. Absorption lines are produced when

A. cool gas absorbs specific photon energies B. hot gas emits light C. light is reflected D. photons gain mass

7. The ionisation energy of hydrogen is

A. 3.4 eV B. 1.0 eV C. 0.85 eV D. 13.6 eV

8. Electrons excited to n = 3 can produce how many lines?

A. 2 B. 1 C. 3 D. 6

Section B: structured

9. Calculate the frequency of a photon of energy 4.0 eV. [2]

10. Distinguish between emission and absorption spectra. [3]

11. Explain how the Geiger–Marsden results led to the nuclear model. [3]

12. Calculate the energy needed to ionise hydrogen from n = 2. [1]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. f = E/h = 9.7 × 10¹⁴ Hz.

10. Emission: bright lines on dark background from hot gas; absorption: dark lines on continuous spectrum from cool gas; same wavelengths.

11. Most through (empty space); few large deflections (small, massive, positive nucleus).

12. 3.40 eV.

10 Summary and topic map

Idea	Remember
Nuclear model	tiny dense positive nucleus
Notation	A = protons + neutrons; Z = protons
Photons	E = hf = hc/λ
Spectra	emission and absorption lines
Energy levels	hf = E₂ − E₁; hydrogen −13.6/n² eV
Ionisation	remove electron to E = 0

Before the exam, make sure you…

•convert eV to J
•use hc/ΔE for wavelengths
•explain spectra using discrete levels
•describe Geiger–Marsden observations and conclusions

Structure of the Atom on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 22. Topic map for Structure of the Atom.

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