Atomic structure and spectra — CSCA (China)

In this lesson you will learn

Describe Thomson's and Rutherford's models and explain what the α-particle scattering experiment showed.
State Bohr's postulates and use E_n = E₁/n² and r_n = n²r₁ for hydrogen.
Find the energy, frequency and wavelength of the photon in a transition and count spectral lines.
Decide what can be absorbed, calculate ionisation energies and recognise line, continuous and absorption spectra.

Walk past a neon sign at night: a neon tube glows red-orange, a hydrogen tube glows pink-violet. Send that light through a spectroscope and you see not a rainbow but a few sharp coloured lines. Every element has its own set of lines, like a barcode. Why does an atom emit only certain colours? The answer leads from the first models of the atom to Bohr's energy levels, the idea behind every CSCA question on atomic structure. The photon itself (E = hν) is explained in the lesson «The photoelectric effect»; what happens inside the nucleus is the topic of «Fundamentals of nuclear physics».

From the plum pudding to the nucleus: Thomson and Rutherford

In 1897 J. J. Thomson discovered the electron, a tiny negative particle found in every atom. An atom as a whole is neutral, so it must also contain positive charge. But how is that positive charge arranged? Thomson's answer (1904) became the first model of the atom.

Definition

Thomson's model (“plum pudding”)

The atom is a sphere of evenly spread positive charge with the electrons embedded in it like raisins in a pudding. The mass and the positive charge fill the whole volume of the atom. Chinese textbooks call it the “date cake model”, 枣糕模型.

To test the model, Rutherford's co-workers Geiger and Marsden (1909) fired fast, positively charged α particles (helium nuclei) at a very thin gold foil. Each hit made a tiny flash on a fluorescent screen that could be moved around the foil, and the flashes were counted through a microscope. The whole apparatus was in a vacuum so that air would not scatter the α particles. Rutherford explained the results in 1911.

Observation	Conclusion
Most α particles pass straight through the foil or are deflected only slightly.	Almost the whole volume of the atom is empty space.
A few are deflected through large angles, and a very small number by more than 90° — some bounce almost straight back.	All the positive charge and almost all the mass are packed into a tiny region, the nucleus; only such a concentrated charge makes a field strong enough to turn an α particle back.
Thomson's model cannot produce such large deflections.	A spread-out charge gives only a weak field, and electrons are about 7300 times lighter than an α particle, so they cannot turn it round.

The α-particle scattering experiment: what was seen and what it means

Definition

Rutherford's nuclear (planetary) model

At the centre of the atom there is a tiny nucleus that carries all the positive charge and more than 99.9 % of the mass; the electrons move around it through empty space, like planets around the Sun. The nucleus has a radius of the order of 10⁻¹⁵ m, the atom about 10⁻¹⁰ m.

r(atom) : r(nucleus) ≈ 10⁻¹⁰ m : 10⁻¹⁵ m = 10⁵ : 1

where:

r(atom)radius of the atom (the size of the electron cloud), m
r(nucleus)radius of the nucleus, m

Lengths differ by 10⁵, so areas differ by 10¹⁰ and volumes by 10¹⁵. That is why most α particles pass without ever coming close to a nucleus.

The scale of the atom

1) If a nucleus were enlarged to a bead 1 cm across, how wide would the atom be?
2) What fraction of the atom's volume does the nucleus fill?

Show solution

1) Everything scales by the same factor: 1 cm · 10⁵ = 10⁵ cm = 1 km. A bead in the middle of a stadium, with a kilometre of emptiness around it!
2) Volume goes as the radius cubed: (10⁻⁵)³ = 10⁻¹⁵, one part in a million billion. Yet almost all the mass sits in that volume, so nuclear matter is enormously dense.

CSCA-style item: the scattering experiment

Which statement about the α-particle scattering experiment is correct?
A) Most α particles bounced back, so the atom is a solid ball
B) The large-angle deflections are caused by collisions with electrons
C) The experiment showed that the nucleus consists of protons and neutrons
D) Most α particles went straight through, so most of the atom is empty space

Show solution

Answer D: the majority going straight through shows the emptiness; the rare large deflections show the small heavy nucleus.
Traps: A turns the observation upside down — only a tiny number bounced back. B: an electron is ~7300 times lighter than an α particle and cannot turn it round. C: the experiment revealed only a small positive nucleus; the neutron was discovered in 1932.

Rutherford's model had a serious problem. According to classical electromagnetism, an electron moving in a circle is accelerating and must radiate electromagnetic waves all the time. It would lose energy, spiral into the nucleus within a tiny fraction of a second and give a continuous spectrum on the way. Real atoms, however, are stable and emit line spectra. Bohr removed this contradiction.

Line spectra: the fingerprints of elements

Definition

Emission spectrum: continuous and line

A hot solid, a liquid or a dense gas (a lamp filament, the deep layers of the Sun) gives all wavelengths: a continuous spectrum. A hot rarefied gas, whose atoms are far apart, emits only separate wavelengths: a line spectrum. The positions of the lines depend only on the element.

Definition

Absorption spectrum

When white light passes through a cooler gas, dark lines appear on the continuous spectrum. They sit exactly where the gas's bright emission lines are: an atom absorbs precisely the wavelengths it can emit. The dark (Fraunhofer) lines in sunlight reveal the elements of the Sun's atmosphere; helium was first found in the solar spectrum in 1868 and on Earth only in 1895. Finding what a substance is made of from its spectrum is called spectral analysis.

Hydrogen has four lines in the visible range: 656 nm (red), 486 nm (blue-green), 434 nm (blue-violet) and 410 nm (violet). In 1885 the Swiss schoolteacher Balmer described them all with one formula, long before anyone knew why it worked.

1/λ = R(1/2² − 1/n²), n = 3, 4, 5, …1/λ = R(1/2² − 1/n²), n = 3, 4, 5, …

where:

λwavelength of the line, m
Rthe Rydberg constant, R ≈ 1.10 × 10⁷ m⁻¹
na whole number from 3 up; each n gives one line

Balmer's formula (巴耳末公式 in Chinese textbooks). As n grows, the lines crowd together and approach the series limit at n → ∞.

Working with Balmer's formula

Take R = 1.10 × 10⁷ m⁻¹. Find 1) the wavelength of the n = 3 line; 2) the wavelength of the n = 4 line; 3) the shortest wavelength of the series.

Show solution

1) 1/λ = R(1/4 − 1/9) = 5R/36 ⇒ λ = 36/(5R) = 36/(5.5 × 10⁷) ≈ 6.55 × 10⁻⁷ m ≈ 655 nm, the red line.
2) 1/λ = R(1/4 − 1/16) = 3R/16 ⇒ λ = 16/(3.3 × 10⁷) ≈ 4.85 × 10⁻⁷ m ≈ 485 nm, the blue-green line.
3) n → ∞: 1/n² → 0, 1/λ = R/4 ⇒ λ = 4/R ≈ 3.64 × 10⁻⁷ m ≈ 364 nm, already ultraviolet. The small differences from the measured 656 and 486 nm come from rounding R.

Bohr's model: stationary states and energy levels

In 1913 Niels Bohr kept Rutherford's nucleus but added quantum rules. His three postulates contradict classical physics, yet they explain the hydrogen spectrum exactly:

Stationary states (定态): the electron can move only in certain allowed orbits, and in them it does not radiate even though it is accelerating. Each state has a definite energy of the atom: E₁, E₂, E₃, …
Transition rule (跃迁): an atom emits or absorbs a photon only when the electron jumps from one stationary state to another, and the photon energy equals the difference of the two energies: hν = E_m − E_n.
Quantised orbits: only orbits with radius r_n = n²r₁ are allowed (from the condition on angular momentum, mvr = nh/2π).

Definition

Energy level, ground state, excited state

The energies of the stationary states are the atom's energy levels (能级). The lowest level (n = 1) is the ground state (基态), the most stable one. The states with n ≥ 2 are excited states (激发态); the atom stays in them only briefly and returns to lower levels by emitting photons.

E_n = E₁/n², r_n = n²r₁ (E₁ = −13.6 eV, r₁ = 0.53 × 10⁻¹⁰ m)E_n = E₁/n², r_n = n²r₁ (E₁ = −13.6 eV, r₁ = 0.53 × 10⁻¹⁰ m)

where:

nthe principal quantum number: 1, 2, 3, …
E_nenergy of the atom in level n, eV; zero means the electron at rest infinitely far from the nucleus, so all bound levels are negative
E₁ground-state energy, −13.6 eV (1 eV = 1.6 × 10⁻¹⁹ J)
r_n, r₁radius of orbit n and the first (Bohr) radius

Energy goes as 1/n², radius as n². As n grows, the levels approach zero and crowd together.

n	E_n, eV	r_n
1	−13.6	r₁
2	−3.4	4r₁
3	−1.51	9r₁
4	−0.85	16r₁
5	−0.54	25r₁
∞	0	—

The first levels of the hydrogen atom

Levels and radii

1) Find E₃ and E₆.
2) Which level has an energy of −0.544 eV?
3) Find r₄. What is the ratio r₃ : r₂?

Show solution

1) E₃ = −13.6/9 ≈ −1.51 eV; E₆ = −13.6/36 ≈ −0.38 eV.
2) n² = 13.6/0.544 = 25 ⇒ n = 5.
3) r₄ = 16r₁ = 16 × 0.53 × 10⁻¹⁰ ≈ 8.5 × 10⁻¹⁰ m; r₃ : r₂ = 9 : 4 (the squares, not 3 : 2!).

The atom's energy is the sum of the electron's kinetic energy and the potential energy of its attraction to the nucleus. On a circular orbit the Coulomb force supplies the centripetal force, ke²/r² = mv²/r, which gives two simple relations:

E_k = ke²/(2r_n) = −E_n, E_p = −ke²/r_n = 2E_nE_k = ke²/(2r_n) = −E_n, E_p = −ke²/r_n = 2E_n

where:

E_kkinetic energy of the electron
E_ppotential energy of the electron–nucleus system (zero at infinity)
kCoulomb's constant, 9 × 10⁹ N · m²/C²

As n increases: r grows, v falls, E_k falls, E_p rises and the total energy E_n rises (towards zero).

When the electron moves up

A hydrogen atom goes from n = 1 to n = 2. 1) Find E_k and E_p in both states. 2) How do they change? 3) Find v₁ : v₂.

Show solution

1) n = 1: E_k = 13.6 eV, E_p = −27.2 eV (check: 13.6 − 27.2 = −13.6). n = 2: E_k = 3.4 eV, E_p = −6.8 eV.
2) E_k falls by 10.2 eV, E_p rises by 20.4 eV, and the total goes from −13.6 to −3.4 eV, rising by 10.2 eV, the energy of the absorbed photon.
3) E_k ∝ v² and E_k ∝ 1/n² ⇒ v ∝ 1/n, so v₁ : v₂ = 2 : 1.

Transitions, spectral lines and ionisation

hν = E_m − E_n, λ = hc/(E_m − E_n) (m > n)hν = E_m − E_n, λ = hc/(E_m − E_n) (m > n)

where:

hPlanck's constant, 6.63 × 10⁻³⁴ J · s
ν, λfrequency and wavelength of the emitted or absorbed photon
E_m, E_nenergies of the upper and the lower level
cspeed of light, 3 × 10⁸ m/s

A jump down emits a photon; a jump up absorbs one. Because the photon energy is fixed exactly, the spectrum consists of lines.

The photon from a jump

Find the photon energy and wavelength and say which part of the spectrum it is in: 1) 4 → 2; 2) 2 → 1 (also find the frequency); 3) 5 → 3.

Show solution

1) ΔE = −0.85 − (−3.4) = 2.55 eV; λ ≈ 1240/2.55 ≈ 486 nm, the visible blue-green Balmer line (the same as from Balmer's formula!).
2) ΔE = −3.4 − (−13.6) = 10.2 eV = 10.2 × 1.6 × 10⁻¹⁹ ≈ 1.63 × 10⁻¹⁸ J; ν = ΔE/h ≈ 2.5 × 10¹⁵ Hz; λ ≈ 1240/10.2 ≈ 122 nm, ultraviolet (Lyman series).
3) ΔE = 13.6 × (1/9 − 1/25) = 13.6 × 16/225 ≈ 0.97 eV; λ ≈ 1280 nm, infrared (Paschen series).

The length of an arrow is the photon energy: a long arrow means a high frequency and a short wavelength. The Balmer lines (jumps to n = 2) are exactly the lines given by Balmer's formula above.

Series	Jumps end on	Region	Longest-wavelength line
Lyman	n = 1	ultraviolet	2 → 1: 10.2 eV (122 nm)
Balmer	n = 2	visible (and near ultraviolet)	3 → 2: 1.89 eV (656 nm)
Paschen	n = 3	infrared	4 → 3: 0.66 eV (1875 nm)

The main spectral series of hydrogen (in Chinese a series is 线系, e.g. 巴耳末系)

N = C(n, 2) = n(n − 1)/2N = C(n, 2) = n(n − 1)/2

where:

nthe highest level the atoms are excited to
Nnumber of different lines (frequencies) that many atoms emit while returning to the ground state

Every pair of levels gives one line. A single atom, however, can emit at most n − 1 photons (going down one step at a time).

Counting the lines

1) Many hydrogen atoms are excited to n = 3. How many lines do they emit, and what are the photon energies?
2) The atoms are in n = 6. How many lines are there, and how many of them belong to the Balmer series?
3) One atom is in n = 5. What are the largest and the smallest number of photons it can emit?

Show solution

1) C(3, 2) = 3: 3 → 2 (1.89 eV), 2 → 1 (10.2 eV), 3 → 1 (12.09 eV). Check: 1.89 + 10.2 = 12.09.
2) C(6, 2) = 6 · 5/2 = 15; Balmer lines are the jumps to n = 2: from 6, 5, 4 and 3, so 4 lines.
3) At most 4 (5 → 4 → 3 → 2 → 1), at least 1 (straight 5 → 1).

CSCA-style item: coming down from n = 4

Many hydrogen atoms return from the n = 4 level to the ground state. Which statement is correct?
A) The atoms emit light of only 3 different frequencies
B) The photon from the 4 → 1 jump has the longest wavelength
C) The photon from the 4 → 3 jump has the lowest frequency
D) All the emitted photons are visible light

Show solution

4 → 3 is the smallest gap: 1.51 − 0.85 = 0.66 eV, so it has the lowest frequency; answer C.
Traps: A is the single-atom count n − 1 = 3; many atoms give C(4, 2) = 6 lines. B is reversed: 4 → 1 (12.75 eV) has the largest energy, hence the shortest wavelength. D: jumps to n = 1 are ultraviolet and 4 → 3 is infrared.

Definition

Ionisation and ionisation energy

Ionisation (电离) is removing the electron from the atom completely (n → ∞, E = 0). The least energy needed for it is the ionisation energy: 13.6 eV for hydrogen in the ground state.

A photon is all or nothing: the atom absorbs it only if hν exactly equals the gap between two levels.
Exception, ionisation: any photon with energy equal to or above the ionisation energy can be absorbed; the extra energy becomes the kinetic energy of the freed electron.
A colliding electron can hand over just part of its energy: the atom is excited if the electron's kinetic energy is at least ΔE.

E_ion = 0 − E_n = 13.6/n² eV, E_k = hν − E_ionE_ion = 0 − E_n = 13.6/n² eV, E_k = hν − E_ion

where:

E_ionionisation energy from level n
E_kkinetic energy of the electron after ionisation

An excited atom is easier to ionise: its electron already sits on a level closer to zero.

Absorption and ionisation

1) Find the ionisation energy of hydrogen from the ground state and from n = 3.
2) A 15 eV photon ionises a ground-state atom. What is the kinetic energy of the electron?
3) Photons of 10.2, 11, 12.75 and 14 eV fall on ground-state atoms. What happens with each? And with an electron of kinetic energy 11 eV?

Show solution

1) From the ground state 13.6 eV, from n = 3: 13.6/9 ≈ 1.51 eV.
2) E_k = 15 − 13.6 = 1.4 eV.
3) 10.2 eV = E₂ − E₁ ⇒ absorbed, the atom goes to n = 2. 11 eV matches no gap (there is no level at −13.6 + 11 = −2.6 eV) ⇒ not absorbed. 12.75 eV = E₄ − E₁ ⇒ the atom goes to n = 4. 14 eV > 13.6 eV ⇒ ionisation, E_k = 0.4 eV. The 11 eV electron, however, can give 10.2 eV and raise the atom to n = 2, keeping 0.8 eV itself.

CSCA-style item: ionising an excited atom

Hydrogen atoms are in the n = 3 state (Eₙ = −13.6/n² eV). Which of these photons can ionise them?
A) 0.66 eV B) 0.85 eV C) 1.20 eV D) 1.89 eV

Show solution

Ionisation from n = 3 needs at least |E₃| = 1.51 eV, and for ionisation no exact match is required: any larger energy works. Only D is ≥ 1.51 eV; the electron leaves with 1.89 − 1.51 = 0.38 eV of kinetic energy.
Traps: A matches the 3 → 4 jump, so it is absorbed but does not ionise; B is the size of the n = 4 level, not of level 3; C is below 1.51 eV. Do not throw out D because “1.89 eV is the 2 → 3 gap”: from n = 3 it is enough to ionise.

How CSCA asks about this

The test is taken in English or in Chinese, so learn to recognise the key terms in both languages:

Term	中文	Pinyin
atomic nucleus	原子核	yuánzǐhé
electron	电子	diànzǐ
α-particle scattering experiment	α粒子散射实验	α lìzǐ sǎnshè shíyàn
nuclear model of the atom	核式结构模型	héshì jiégòu móxíng
plum pudding model	枣糕模型	zǎogāo móxíng
stationary state	定态	dìngtài
energy level	能级	néngjí
ground state	基态	jītài
excited state	激发态	jīfātài
transition	跃迁	yuèqiān
ionisation	电离	diànlí
line spectrum	线状谱	xiànzhuàngpǔ
continuous spectrum	连续谱	liánxùpǔ
absorption spectrum	吸收光谱	xīshōu guāngpǔ
photon	光子	guāngzǐ

Scattering experiment: observation → conclusion (“most go straight → empty space; a few at large angles → tiny heavy positive nucleus”), and why electrons cannot turn an α particle. These are 15–20-second items.
Energy-level diagram: read E_n, find ΔE, compare frequencies and wavelengths, name the series and the region of a line.
Counting lines: “a group of atoms in n = 4” → 6; “one atom” → at most 3; “how many are visible (Balmer)”.
What can be absorbed: photons that match a gap, photons that ionise, photons that do neither; colliding electrons.
What changes when n changes: r, v, E_k, E_p and the total energy, with their signs.

Key points

α scattering: most go straight through → the atom is mostly empty; a few are deflected strongly → all the positive charge and almost all the mass are in a tiny nucleus (10⁻¹⁵ m vs 10⁻¹⁰ m for the atom).
Bohr: no radiation in stationary states; hν = E_m − E_n; for hydrogen E_n = −13.6/n² eV and r_n = n²r₁.
As n increases, E_k falls while E_p and the total energy rise: E_k = −E_n, E_p = 2E_n.
Many atoms give C(n, 2) lines, one atom at most n − 1 photons; λ (nm) ≈ 1240 / ΔE (eV); a bigger ΔE means a shorter wavelength.
A photon is absorbed only if it matches a gap exactly or ionises the atom; a colliding electron can give part of its energy. Ionisation from level n takes 13.6/n² eV.
A line spectrum is an element's fingerprint; absorption lines sit where the emission lines are. Lyman is ultraviolet, Balmer visible, Paschen infrared.

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Who proposed the nuclear model of the atom from the results of the α-particle scattering experiment?

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