The Nucleus & Atomic Spectra (SL)

This workbook tells the story of a single, tiny experiment that changed how physicists pictured the atom forever, and then turns to the light that atoms give out and take in. In 1909, a beam of alpha particles fired at gold foil revealed that atoms have a small, dense, positively charged nucleus — not a uniform blob of charge, as had been assumed. Decades of studying the coloured lines in emission and absorption spectra then showed that electrons in an atom can only exist at certain fixed energies, and that every time one drops between two of these levels, it emits a single photon of light. Work through it in order; each section builds on the one before it.

By the end of this workbook you should be able to:

Explain how the Geiger–Marsden–Rutherford scattering experiment provided evidence for a small, dense, positively charged nucleus at the centre of the atom.
Use nuclear notation (Z, A, X) to describe a nuclide, and identify isotopes of the same element.
Explain how emission and absorption spectra provide evidence for discrete atomic energy levels, and how a spectrum reveals the chemical composition of a sample.
Describe how photons are emitted and absorbed during atomic transitions, and apply E = hf to relate a photon's frequency to the energy-level difference that produced it.

Name

1. The nucleus: the Geiger–Marsden–Rutherford experiment

By 1904, physicists knew that atoms contained tiny, negatively charged electrons — J. J. Thomson had found them using cathode rays. But since atoms are electrically neutral overall, there had to be positive charge somewhere too. Thomson's best guess was the "plum pudding" model: a blob of spread-out positive charge with electrons dotted through it, like fruit suspended in a pudding.

Fig. 1.1 Two competing models of the atom, before and after 1909.

In 1909, at Rutherford's laboratory in Manchester, two of his students, Hans Geiger and Ernest Marsden, fired a narrow beam of fast, positively charged alpha particles at an extremely thin sheet of gold foil (only a few hundred atoms thick). A screen coated in zinc sulfide, which produces a tiny flash of light whenever an alpha particle hits it, was used to detect the particles after they had passed through — or bounced off — the foil.

Fig. 1.2 The alpha-particle scattering apparatus (plan view).

In their own words

Rutherford later described the large-angle results as "quite the most incredible event that has ever happened to me... It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

The results were:

The overwhelming majority of alpha particles passed straight through the foil with little or no deflection — exactly as expected if the atom is mostly empty space.
A small fraction were deflected through moderate angles (more than about 10°).
An extremely small fraction — roughly 1 in 8000 — were deflected through more than 90°, and a few came almost straight back the way they came.

Rutherford's interpretation. Because most alpha particles passed straight through, most of the atom must be empty space. Because a few bounced back almost the way they came, there had to be something extremely small, extremely dense, and positively charged at the centre of the atom, able to repel a fast-moving positive alpha particle head-on. Rutherford called this the nucleus.

This was a genuine paradigm shift: the plum-pudding model was replaced by the nuclear model, in which almost all the mass and all the positive charge of an atom is concentrated in a nucleus roughly 10⁴–10⁵ times smaller than the atom itself, with electrons occupying the mostly-empty space around it.

PhET simulation — Rutherford Scattering

Fire alpha particles at a plum-pudding atom and then at a nuclear atom, and see the difference in scattering pattern for yourself.

Open the PhET simulation ↗

Simulation: Rutherford Scattering, PhET Interactive Simulations, University of Colorado Boulder — phet.colorado.edu.

Check your understanding

1Using the PhET simulation above, switch between the "Plum Pudding Atom" and "Rutherford Atom" screens. Describe one difference you observe in how the alpha particles scatter.

This is a descriptive, simulation-based question (compare the scattering patterns you observe) — check your wording against the description of the experiment in this section and with your teacher.

2In your own words (fewer than 100 words), explain why Rutherford concluded that atoms contain a small, positively charged, dense nucleus. Do not use a diagram.

This is a descriptive question (explain in your own words) — check your wording against the description of the experiment in this section and with your teacher.

3Suggest what would have happened in the Geiger–Marsden experiment if neutrons had been used instead of alpha particles. Explain your answer.

Neutrons carry no charge, so they would not be repelled electrically by the nucleus at all — almost none would be deflected by more than a small angle from occasional direct nuclear ("strong-force") collisions, so the large-angle scattering pattern that revealed the nucleus's charge and size would not appear in the same way.

2. Nuclear notation and isotopes

After Rutherford's discovery, it became clear that the nucleus itself is made of two kinds of particle: positively charged protons and uncharged neutrons, together called nucleons. Every atom of a given element has the same number of protons, but the number of neutrons can vary.

Particle	Relative mass	Relative charge	Location
proton	1	+1	nucleus
neutron	1	0	nucleus
electron	1/1840	−1	surrounding the nucleus

Key definitions.

Proton number, Z — the number of protons in the nucleus. This determines which element the atom is.
Nucleon number, A — the total number of protons and neutrons (also called the mass number).
Neutron number, N — the number of neutrons, where N = A − Z.

Fig. 2.1 Standard nuclide notation: ^A_ZX.

Two or more atoms of the same element (same Z) with different nucleon numbers A are called isotopes. They have identical chemical properties but different masses. For example, the three isotopes of hydrogen are:

¹₁H (hydrogen, no neutrons) · ²₁H (deuterium, 1 neutron) · ³₁H (tritium, 2 neutrons)

Worked example 2.1

A particular element has proton number 19.

Identify the element.
Its most common isotope has a nucleon number of 39. State the number of protons, neutrons and electrons in a neutral atom of this isotope.
Write the full nuclide symbol for this isotope.

Answer:
a) Potassium (K).
b) 19 protons, 20 neutrons (39 − 19), 19 electrons (the atom is neutral).
c) ³⁹₁₉K

Check your understanding

4Consider a neutral atom.

What do you get when you change the number of protons in an atom?
What do you get when you change the number of neutrons in an atom?
What do you get when you change the number of electrons in an atom?

a) a different element — changing the number of protons (Z) changes which element it is; b) a different isotope of the same element — changing the number of neutrons keeps Z (and so the element) the same, but changes the nucleon number A and the mass; c) an ion — changing the number of electrons means the atom is no longer electrically neutral, so it becomes a charged ion (positive if electrons are removed, negative if electrons are added).

5Chlorine, Cl, has proton number 17. Its two most common isotopes are chlorine-35 and chlorine-37. Write the full nuclide symbol for each, and state the number of neutrons in each.

³⁵₁₇Cl (18 neutrons); ³⁷₁₇Cl (20 neutrons).

6An ion is formed from a nitrogen atom (Z = 7, A = 14) by removing three electrons. State the number of protons, neutrons and electrons in this ion, and its overall charge.

7 protons, 7 neutrons, 4 electrons; overall charge = +3.

7Carbon-14 is an isotope of carbon (Z = 6) used in radiocarbon dating. Write its full nuclide symbol, and state the number of neutrons it contains.

¹⁴₆C (8 neutrons, since N = A − Z = 14 − 6).

3. Evidence for energy levels: emission and absorption spectra

The simple picture of electrons orbiting a nucleus (rather like planets orbiting a star) turns out to be seriously incomplete. Orbiting satellites can have any orbital energy — a continuous range. Electrons in atoms cannot: they can only exist with certain very precise, separated (discrete) energies, called atomic energy levels. The lowest of these is the ground state.

The evidence for this comes from studying the light that atoms give out or take in.

When a gas is excited (heated, or given energy by an electric current), it emits light only at certain specific frequencies. Viewed through a prism or diffraction grating, this appears as a series of bright, separate lines on a dark background: an emission spectrum.
When white light (a continuous spectrum) is passed through a cool gas, the gas absorbs light at exactly those same frequencies. This produces dark lines on an otherwise continuous, bright spectrum: an absorption spectrum.

Key idea. Every line in an emission or absorption spectrum corresponds to an electron transition between two discrete energy levels. Since different elements have different sets of energy levels, every element produces its own unique "fingerprint" of spectral lines. This is why spectra can be used to identify the chemical composition of a distant star, a nebula, or an unknown sample in a lab.

Live simulation: build your own line spectrum

This tool (Foothill College AstroSims) lets you add individual coloured emission lines onto a continuous spectrum and see the resulting pattern, just as astronomers use spectrometers to read the chemical fingerprint of starlight.

Interactive simulation — open the online version of this workbook to launch it.

Simulation: Foothill College AstroSims — Spectroscopy Demonstrator (opens the full version in a new tab).

Explore real spectra — atomic-spectra.net

Browse the real emission-line spectrum of any element in the periodic table.

Open atomic-spectra.net ↗

External resource: atomic-spectra.net. Opens in a new tab.

Check your understanding

8Using the AstroSims tool above (or atomic-spectra.net), compare the emission spectra of hydrogen and helium. Describe two differences.

This is a descriptive, simulation-based question — see the key idea box in this section for the underlying physics; check your specific comparison with your teacher.

9Explain how an astronomer could use the absorption spectrum of a star to determine which elements are present in its outer atmosphere.

This is a descriptive question — see the key idea box in this section, which explains why each element's absorption lines act as a unique fingerprint; check your wording with your teacher.

10Explain why the existence of line spectra (rather than continuous, "rainbow" spectra) is evidence that atomic energy levels are discrete rather than continuous.

Discrete spectral lines correspond to discrete photon frequencies (E = hf); since only specific frequencies appear, only specific energy differences — and therefore only discrete energy levels — can exist within the atom.

4. Photons and atomic transitions

When an electron in an atom drops from a higher energy level to a lower one, the atom emits a single "packet" of electromagnetic energy called a photon. Conversely, an atom can absorb a photon and jump from a lower level to a higher one — but only if the photon carries exactly the right amount of energy to bridge the gap between the two levels.

Photon: a discrete packet ("quantum") of electromagnetic energy, carrying energy E = hf, where h is the Planck constant.

energy of one photon: E = hf (h = Planck's constant = 6.63 × 10⁻³⁴ J s)

Because atomic energy levels are discrete, the differences between them are also discrete — so only certain photon energies (and therefore only certain frequencies, since E = hf) can be emitted or absorbed by a given atom. This is exactly why line spectra exist.

Physically, each energy level corresponds to an electron occupying one of the atom's "shells" around the nucleus. But throughout this topic, energy levels are drawn as flat horizontal lines rather than as circles, since that makes it far easier to see the transitions between them clearly — and the line spectra you explored in Section 3 are themselves just another picture of the very same spacing. Fig. 4.1 shows how all three pictures (shells → flat lines → spectral lines) are connected.

Fig. 4.1 The circular electron shells are then shown as a horizontal "sliver" to save space and make the model as clear as possible.

Fig. 4.2 Emission transitions (downward, coloured) release photons of different energies; absorption (upward, dashed) requires a photon of exactly the right energy. Notice the gaps between levels shrink higher up, as in a real atom.

Worked example 4.1

An electron in an atom drops from an energy level of −1.20 × 10⁻¹⁸ J to a level of −3.06 × 10⁻¹⁸ J. Calculate the frequency of the photon emitted, and state the part of the electromagnetic spectrum it belongs to.

Answer:
ΔE = (−1.20 × 10⁻¹⁸) − (−3.06 × 10⁻¹⁸) = 1.86 × 10⁻¹⁸ J
E = hf ⟹ f = E/h = (1.86 × 10⁻¹⁸)/(6.63 × 10⁻³⁴) ... (complete the division to find f, in Hz, then identify the part of the electromagnetic spectrum this frequency belongs to)

Check your understanding

11A photon of frequency 5.0 × 10¹⁴ Hz is absorbed by an atom. Calculate the energy gained by the atom, in both joules and electronvolts.

E = hf = (6.63 × 10⁻³⁴) × (5.0 × 10¹⁴) ... (complete the calculation, in J, then divide by 1.60 × 10⁻¹⁹ J to convert to eV).

12An atom has just four energy levels. State the maximum possible number of different spectral lines (transitions) it could produce.

6 possible transitions — every pair of the 4 levels gives one possible line, and there are 4 × 3 / 2 = 6 distinct pairs.

13Explain why a photon can only be absorbed by a particular atom if its energy exactly matches the gap between two of that atom's energy levels — rather than being absorbed partially.

Because atomic energy levels are discrete, an electron can only ever be in one exact level or another — there is no "in-between" state to absorb partial energy into, so the photon energy must exactly bridge two allowed levels.

14An electron in an atom drops from an energy level of −0.85 × 10⁻¹⁸ J to a level of −2.42 × 10⁻¹⁸ J. Calculate the frequency of the photon emitted.

ΔE = (−0.85 × 10⁻¹⁸) − (−2.42 × 10⁻¹⁸) = 1.57 × 10⁻¹⁸ J
f = ΔE/h = (1.57 × 10⁻¹⁸)/(6.63 × 10⁻³⁴) ... (complete the division to find f, in Hz).

Glossary

Nucleon: A proton or a neutron; the particles that make up an atomic nucleus.
Nuclide: A specific type of nucleus, defined by its proton number Z and nucleon number A, written as ^A_ZX.
Isotope: One of two or more atoms of the same element (same Z) with different numbers of neutrons (different A).
Energy level: One of the discrete, allowed values of energy that an electron bound in an atom may have.
Ground state: The lowest-energy, most stable energy level available to an electron in an atom.
Photon: A discrete packet ("quantum") of electromagnetic energy, carrying energy E = hf.
Emission spectrum: A series of bright, discrete lines produced when an excited gas emits light only at frequencies corresponding to transitions between its atoms' energy levels.
Absorption spectrum: A continuous spectrum with dark lines produced when a cool gas absorbs light at the same frequencies it would otherwise emit, as electrons are excited to higher energy levels.