IB DP Physics · Theme E · Nuclear and quantum physics

Structure of the atom.

E.1 · The nucleus, energy levels, spectra
Standard level and higher level

SL + HL · 6 hours Additional HL · 3 hours Atomic Models sim ↗

The journey

Three parts.

Part 01 · SL + HL

How was the nucleus discovered?

Geiger–Marsden–Rutherford, nuclear notation

Part 02 · SL + HL

What do spectra tell us?

Energy levels, photons, \(E = hf\)

Part 03 · HL only

How big is a nucleus? Why these levels?

Nuclear radius, closest approach, Bohr’s hydrogen

Slides with a crimson edge and an HL only flag are additional higher level. Every other slide is for SL and HL.

E.1 STRUCTURE OF THE ATOM · SL + HL

Part 01 · SL + HL

Discovering the nucleus.

New evidence → new model

E.1 STRUCTURE OF THE ATOM · SL + HL

Changing models

A model that kept changing.

Before 1897

Solid spheres

Tiny spheres that cannot be divided

1904

Plum pudding

After the electron was discovered

1911

Nuclear model

From alpha scattering

1913

Bohr model

Electrons in energy levels

The Atomic Models sim ↗ adds the next step: the 1926 electron cloud, the model used today.

E.1 STRUCTURE OF THE ATOM · SL + HL

The story · Cambridge, 1897

Background story · not assessed

Thomson finds the electron.

Portrait of J. J. Thomson, an older man with glasses and a mustache — **J. J. Thomson**

A long glass tube with a bulb at each end and wire coils around the middle — The **cathode ray tube** he used in 1897, now in the Science Museum, London (photo mirrored)

Glowing rays crossed the tube. Nobody knew what they were
Magnets and electric fields bent the rays: they were negative particles
Every metal he tried gave the same particles
So they must be inside every atom: the electron

Atoms can be divided. The solid sphere model had to go.

Portrait: Bain News Service, Library of Congress, public domain. Tube: Science Museum London, CC BY-SA 2.0.

E.1 STRUCTURE OF THE ATOM · SL + HL

Thomson (1904)

The plum pudding model.

Electron discovered → atoms can be divided
A ball of positive charge
Negative electrons embedded in it
No nucleus

Try it: Atomic Models sim ↗

PLUM PUDDING MODELNOT TO SCALE

− Electron+ Positive charge (spread)

E.1 STRUCTURE OF THE ATOM · SL + HL

The story · Manchester, 1909

Background story · not assessed

A professor, his assistant and a student.

Two men in dark suits sitting either side of a bench of laboratory apparatus — **Hans Geiger** (left) and **Ernest Rutherford** in the laboratory

Portrait of Ernest Marsden, a man with glasses in a suit — **Ernest Marsden**, pictured in 1921

Line drawing of a round metal chamber in cross-section, with parts labeled by letters — 1913 apparatus: source R, foil F, screen S, microscope M

Each alpha particle made a tiny flash on a screen. Geiger counted flashes through a microscope, for hours in a darkened lab
Rutherford asked Marsden, a 20-year-old student, to look for alpha particles bouncing back. Nobody expected any
In an early test with a platinum plate, about 1 in 8000 bounced back. The plum pudding model could not explain it

Photos: unknown photographer and S P Andrew Ltd, public domain. Diagram: Geiger and Marsden, 1913, public domain.

E.1 STRUCTURE OF THE ATOM · SL + HL

Geiger, Marsden & Rutherford (1909–1911)

The Geiger–Marsden–Rutherford experiment.

Fire alpha particles (+) at thin gold foil
Count them at all angles
Plum pudding predicts: all pass through
Only tiny deflections expected

ALPHA SCATTERINGNOT TO SCALE

E.1 STRUCTURE OF THE ATOM · SL + HL

Evidence → conclusion

What they saw. What it means.

OBSERVATIONCONCLUSION

Most went straight through

→

Atom is mostly empty space

Some were deflected

→

Center is positively charged

A very few bounced back

→

Mass concentrated in a tiny center

CLOSE-UPNOT TO SCALE

E.1 STRUCTURE OF THE ATOM · SL + HL

Rutherford (1911)

The nuclear model.

A tiny, dense nucleus at the center
The nucleus is positive
Almost all the mass is in it
Electrons outside the nucleus
Replaced the plum pudding model

NUCLEAR MODELNOT TO SCALE

− Electron+ Nucleus (positive)

E.1 STRUCTURE OF THE ATOM · SL + HL

Compare

Plum pudding vs nuclear.

	Plum pudding	Nuclear model
Positive charge	Spread through the atom	In a tiny nucleus
Mass	Spread through the atom	Concentrated in the nucleus
Empty space?	No	Mostly empty space
Electrons	Embedded in the positive ball	Outside the nucleus

E.1 STRUCTURE OF THE ATOM · SL + HL

Inside the atom

Three particles.

Particle	Charge	Mass	Where?
Proton	\(+e\)	≈ 1 u	Nucleus
Neutron	0	≈ 1 u	Nucleus
Electron	\(-e\)	≈ 0.0005 u	Around the nucleus

Nucleon = a proton or a neutron. A neutral atom has as many electrons as protons.

E.1 STRUCTURE OF THE ATOM · SL + HL

Inside the atom

Nuclear notation.

Nucleon number \(A\) = protons + neutrons
Proton number \(Z\) = protons
Neutrons: \(N = A - Z\)
X is the chemical symbol, fixed by \(Z\)

Try: \({}^{197}_{79}\mathrm{Au}\) → 79 p, 118 n

NUCLEAR NOTATION

E.1 STRUCTURE OF THE ATOM · SL + HL

Work through another one

Uranium with 235 on top and 92 below. Protons: 92. Neutrons: \(235 - 92 = 143\). Electrons in the neutral atom: 92.

Why do the protons decide the element?

\(Z\) sets the charge of the nucleus, so it sets how many electrons the neutral atom holds and how it behaves chemically. Change \(Z\) and it is a different element. Change \(N\) alone and it is another isotope of the same element: carbon-12 and carbon-14 both have \(Z = 6\).

You are not expected to recall chemical symbols. Questions give the symbol or the proton number.

The same notation is used for particles: an alpha particle is helium-4 with \(A = 4, Z = 2\), and an electron is written with \(A = 0, Z = -1\).

Inside the atom

Mostly empty space.

Atom radius ≈ \(\class{res}{10^{-10} \,\mathrm{m}}\)
Nucleus radius ≈ \(\class{res}{10^{-15} \,\mathrm{m}}\)
Almost all the mass is in the nucleus
Electrons have only certain energies: energy levels

AN ATOMNOT TO SCALE

E.1 STRUCTURE OF THE ATOM · SL + HL

Nature of science

Why models change.

A model must explain the evidence
It makes predictions we can test
New evidence can change or replace it
Results are published and peer reviewed

THE SCIENCE CYCLE

E.1 STRUCTURE OF THE ATOM · SL + HL

Quick check

The nucleus.

01Why did most alpha particles pass straight through?

The atom is mostly empty space

02Why did a very few come straight back?

A tiny, dense, positive nucleus repelled them

03How many neutrons in \({}^{56}_{26}\mathrm{Fe}\)?

\(56 - 26 = 30\)

04What is a nucleon?

A proton or a neutron

05Roughly what is the ratio of an atom’s diameter to its nucleus’s diameter?

About \(10^{5}\) (\(10^{-10}\,\mathrm{m}\) against \(10^{-15}\,\mathrm{m}\))

06What could the plum pudding model not explain?

Alpha particles scattered through large angles

PRESS → TO REVEAL EACH ANSWER

Part 02 · SL + HL

Energy levels and spectra.

Light as a message from inside the atom

E.1 STRUCTURE OF THE ATOM · SL + HL

The story · Copenhagen, 1913

Background story · not assessed

Bohr fixes the nuclear model.

Portrait of Niels Bohr in a suit and tie — **Niels Bohr**, pictured about 1922, the year he won the Nobel Prize

1912: Bohr, a 26-year-old from Denmark, works in Rutherford’s laboratory
He sees a problem. An orbiting electron should lose energy and spiral into the nucleus. Atoms would collapse
1913: his answer. Electrons can only orbit at certain distances
The test: his calculations matched the colors of light given out by hydrogen

He kept Rutherford’s nucleus and changed only what the evidence demanded.

Photo: AB Lagrelius & Westphal, public domain.

E.1 STRUCTURE OF THE ATOM · SL + HL

Energy levels

Only certain energies allowed.

An electron in an atom can have only certain energies
These are discrete energy levels
Lowest level: the ground state
Energies are negative: the electron is bound

Zero energy = electron just free of the atom.

HYDROGEN: ENERGY LEVELSNOT TO SCALE

E.1 STRUCTURE OF THE ATOM · SL + HL

Think of a ladder

You can stand on a rung, never between rungs. An electron in an atom is the same: it has one of the allowed energies and nothing in between. “Discrete” means separate, countable values.

Why negative?

Zero is chosen as the energy of an electron that has just escaped and is at rest. An electron held by the nucleus has less energy than that, so its energy is below zero. To free a ground-state hydrogen electron you must supply 13.6 eV: the ionization energy.

The electronvolt is the energy gained by an electron moving through 1 V: \(1 \,\mathrm{eV} = 1.60 \times 10^{-19} \,\mathrm{J}\). Convert to joules before using \(E = hf\).

Reading the levels as distances. The diagram shows energy, not where the electron is.

Photons

Light comes in packets.

Light is emitted and absorbed as photons
Each photon carries one fixed amount of energy
Higher frequency → more energy per photon

\(E = hf\)

\(h = 6.63 \times 10^{-34} \,\mathrm{J} \,\mathrm{s}\), the Planck constant. With \(c = f\lambda\): \(E = \frac{hc}{\lambda}\).

Worked example · green laser, \(\lambda = 532\,\mathrm{nm}\) \(f = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^{8}}{532 \times 10^{-9}}\) \(= \class{res}{5.64 \times 10^{14} \,\mathrm{Hz}}\) \(E = hf = 6.63 \times 10^{-34} \times 5.64 \times 10^{14}\) \(= \class{res}{3.74 \times 10^{-19} \,\mathrm{J}}\) \(= \dfrac{3.74 \times 10^{-19}}{1.60 \times 10^{-19}} = \class{res}{2.34 \,\mathrm{eV}}\)

E.1 STRUCTURE OF THE ATOM · SL + HL

What is a photon?

A packet (quantum) of electromagnetic energy. Light of one frequency is a stream of identical photons. A brighter beam of the same color has more photons each second, not bigger ones.

How small is one photon?

A 1 mW laser pointer giving out \(3.74 \times 10^{-19} \,\mathrm{J}\) photons emits \(\frac{0.001}{3.74 \times 10^{-19}} \approx 2.7 \times 10^{15}\) of them every second. That is why light normally looks continuous.

Visible photons carry roughly 1.8 eV (red) to 3.1 eV (violet). An answer far outside that for visible light means a slip with powers of ten or with nm.

Leaving the wavelength in nm. \(532 \,\mathrm{nm} = 532 \times 10^{-9} \,\mathrm{m}\).

Atomic transitions

One transition, one photon.

Absorb a photon → jump up a level
Drop down a level → emit a photon

HYDROGEN ATOMNOT TO SCALE

− Electron+ Proton~ Photon, \(hf = E_{2} - E_{1}\)

E.1 STRUCTURE OF THE ATOM · SL + HL

Worked example

From levels to a spectral line.

HYDROGEN: n = 3 → n = 2NOT TO SCALE

Find the wavelength of the photon emitted \(\Delta E_\mathrm{atom} = E_{2} - E_{3}\) \(= (-3.40) - (-1.51) = \class{res}{-1.89 \,\mathrm{eV}}\) \(E_\gamma = -\Delta E_\mathrm{atom} = \class{res}{3.02 \times 10^{-19} \,\mathrm{J}}\) \(f = \dfrac{E_\gamma}{h} = \dfrac{3.02 \times 10^{-19}}{6.63 \times 10^{-34}}\) \(= \class{res}{4.56 \times 10^{14} \,\mathrm{Hz}}\) \(\lambda = \dfrac{c}{f} = \class{res}{6.58 \times 10^{-7} \,\mathrm{m}}\) (red)

E.1 STRUCTURE OF THE ATOM · SL + HL

The method never changes

Use \(\Delta E_\mathrm{atom} = E_\mathrm{f} - E_\mathrm{i}\). For emission this is negative. Conservation gives \(\Delta E_\mathrm{total} = \Delta E_\mathrm{atom} + E_\gamma = 0\), so the photon has positive energy \(E_\gamma = -\Delta E_\mathrm{atom} = hf\). Convert eV to J before using \(h\), then use \(c = f\lambda\).

Does it match?

Hydrogen’s red line is measured at 656 nm. We got 658 nm because the level energies were rounded to three significant figures. A subtraction of two rounded numbers loses precision.

A quicker route: the data booklet gives \(hc = 1.24 \times 10^{-6} \,\mathrm{eV} \,\mathrm{m}\), so \(\lambda = \frac{hc}{E_\gamma} = \frac{1.24 \times 10^{-6}}{1.89} = 6.56 \times 10^{-7} \,\mathrm{m}\) with the energy left in eV.

Using one level energy instead of the difference between two.

Spectra

Emission spectra.

Hot, low-pressure gas glows
Split the light: a few bright lines
Each line: one frequency of photon
Not a continuous band of color

VISIBLE SPECTRAWAVELENGTH IN nm

E.1 STRUCTURE OF THE ATOM · SL + HL

Spectra

Absorption spectra.

White light through a cool gas
Continuous spectrum with dark lines
Dark lines: photons the gas absorbed
Same wavelengths as its emission lines

HYDROGENWAVELENGTH IN nm

E.1 STRUCTURE OF THE ATOM · SL + HL

Evidence → conclusion

Why lines mean levels.

OBSERVATIONCONCLUSION

Spectra are lines, not continuous bands

→

Photons have only certain energies

Each photon comes from one transition

→

Only certain energy differences exist

The same lines appear every time

→

The atom has fixed, discrete energy levels

If an electron could have any energy, the spectrum would be continuous.

E.1 STRUCTURE OF THE ATOM · SL + HL

Spectra

A fingerprint for each element.

Every element has its own set of levels
So its own pattern of lines
Match the lines → identify the element

The star’s light has crossed hydrogen and sodium, but not helium.

WHAT IS THE STAR MADE OF?

E.1 STRUCTURE OF THE ATOM · SL + HL

Why is the pattern unique?

The levels depend on the charge of the nucleus and on how the electrons arrange themselves around it. No two elements share both, so no two share a set of energy gaps.

How a star is analyzed

Its hot, dense interior gives a continuous spectrum. The cooler gas of its outer layers absorbs at its own wavelengths. Comparing the dark lines with spectra measured in the laboratory shows which elements are there.

One shared line is not enough. Helium’s yellow line (588 nm) sits almost on top of sodium’s (589 nm), but helium’s other lines are missing from this star.

Helium was found this way in the Sun’s spectrum in 1868, a yellow line that fitted no known element, before it was found on Earth. If all of a star’s lines are shifted by the same fraction, the shift gives its speed towards or away from us (Doppler effect, C.5).

Quick check

Levels and spectra.

01What is a photon?

A packet of electromagnetic energy, \(E = hf\)

02Electron drops from −1.51 eV to −3.40 eV. Photon energy?

1.89 eV

03Which photon has more energy: red or violet?

Violet: higher frequency

04Why is an absorption line dark?

The gas absorbed photons of that energy

05How do spectra show energy levels are discrete?

Only certain photon energies appear

06How can we tell what a star is made of?

Match its absorption lines to known elements

PRESS → TO REVEAL EACH ANSWER

Part 03 · Additional higher level

The nucleus and Bohr’s hydrogen.

Nuclear radius and density · closest approach
Bohr’s energy levels · quantized angular momentum

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

The story · Manchester, 1911

Background story · not assessed

Rutherford puts a number on the nucleus.

Portrait of Ernest Rutherford, a man with a mustache in a suit — **Ernest Rutherford**

About twenty men in suits posed in three rows outside a building — Rutherford with the staff of the **Manchester physics laboratories**, 1910

Geiger and Marsden had the counts. Rutherford wanted to know how small the center of the atom was
His argument: an alpha particle aimed straight at it stops when all its kinetic energy has become electric potential energy
His answer for gold: about \(3 \times 10^{-14}\,\mathrm{m}\), thousands of times smaller than the atom

Portrait: Bain Collection, Library of Congress, public domain. Group: unknown photographer, National Library of New Zealand, public domain.

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

Scattering, with numbers

Distance of closest approach.

Head-on alpha: slows, stops, returns
At the turning point: all kinetic energy → electric potential energy

\(E_{\mathrm{k}} = \dfrac{k(2e)(Ze)}{r}\)

An upper limit on the radius of the nucleus.

HEAD-ON COLLISIONNOT TO SCALE

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

Why energy conservation is enough

The alpha particle starts far away, where the electric potential energy is zero, with kinetic energy \(E_{\mathrm{k}}\). The repulsion does negative work on it all the way in. It stops when the potential energy \(\frac{kq_{1}q_{2}}{r}\) has taken all of \(E_{\mathrm{k}}\). Rearranged: \(r = \frac{2kZe^{2}}{E_{\mathrm{k}}}\).

What is assumed?

A head-on path; a nucleus so heavy that it does not recoil; and only the electric force acting, which is true while the alpha stays outside the nucleus.

More energy → smaller \(r\). Larger \(Z\) → larger \(r\). Both follow directly from the formula.

Forgetting that the alpha particle’s charge is \(2e\), or leaving \(E_{\mathrm{k}}\) in MeV.

Worked example

A 5.0 MeV alpha meets gold.

Gold: \(Z = 79\) · \(k = 8.99 \times 10^{9} \,\mathrm{N} \,\mathrm{m}^{2} \,\mathrm{C}^{-2}\) \(E_{\mathrm{k}} = 5.0 \,\mathrm{MeV} = 5.0 \times 10^{6} \times 1.60 \times 10^{-19} \,\mathrm{J}\) \(= \class{res}{8.0 \times 10^{-13} \,\mathrm{J}}\) \(r = \dfrac{2kZe^{2}}{E_{\mathrm{k}}}\) \(= \dfrac{2 \times 8.99 \times 10^{9} \times 79 \times (1.60 \times 10^{-19})^{2}}{8.0 \times 10^{-13}}\) \(= \class{res}{4.5 \times 10^{-14} \,\mathrm{m}}\)

The alpha turned back 45 fm from the center
So the gold nucleus is smaller than that
The atom is ≈ \(10^{-10} \,\mathrm{m}\): over 2000 times bigger

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

Scattering at high energies

Where Rutherford’s formula fails.

Rutherford assumed only electric repulsion
Low energy: prediction and count agree
High energy: the alpha gets close enough to touch
The strong nuclear force now acts too
Fewer are scattered than predicted

The energy where it fails gives the size of the nucleus.

ALPHAS SCATTERED AT ONE ANGLESKETCH

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

What changes at high energy?

The strong nuclear force attracts nucleons but reaches only a few femtometers. A low-energy alpha is turned back long before that, so it feels only electric repulsion and Rutherford’s formula works. A fast one reaches the surface of the nucleus, where the strong force pulls on it and may absorb it. The simple electric calculation no longer describes what happens.

Turning the failure into a measurement

At the energy where the counts first depart from the prediction, the distance of closest approach roughly equals the radius of the nucleus plus that of the alpha. For gold that is about \(7.0 + 1.9 = 8.9 \,\mathrm{fm}\), which needs an alpha of roughly 26 MeV.

In scattering questions the energies are kept low enough that you only consider the electric force, unless the question is about the deviation itself.

The story · Stanford, 1953

Background story · not assessed

Hofstadter measures the nucleus.

Portrait of Robert Hofstadter, a smiling man in a suit — **Robert Hofstadter**, pictured in 1961, the year of his Nobel Prize

A long straight pipe with equipment along it, running the length of a large hall — Stanford’s **Mark III** linear electron accelerator, 1950s

Alpha particles feel the strong force, so they blur the edge of a nucleus
Electrons do not. Hofstadter fired them at nuclei from Stanford’s new linear accelerator
The way they scattered gave the size of each nucleus: radius \(\propto A^{1/3}\)

The same method later showed that the proton has a size too.

Portrait: Nobel Foundation, public domain. Accelerator: US Office of Naval Research, public domain.

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

Size of the nucleus

Nuclear radius.

\(R = R_{0}A^{1/3}\)

\(R_{0} = 1.20 \times 10^{-15} \,\mathrm{m}\), the Fermi radius. \(A\) is the nucleon number.

FOUR NUCLEI, SAME SCALE\(1 \,\mathrm{fm} = 10^{-15} \,\mathrm{m}\)

+ Proton0 Neutron

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

Where does the cube root come from?

Nucleons pack like marbles in a bag: each takes up about the same space. So the volume of a nucleus is proportional to the number of nucleons, \(V \propto A\). Volume also goes as \(R^{3}\), so \(R^{3} \propto A\) and \(R \propto A^{1/3}\).

Worked example, gold-197

\(197^{1/3} = 5.82\), so \(R = 1.20 \times 10^{-15} \times 5.82 = 6.98 \times 10^{-15} \,\mathrm{m}\).

The radius grows slowly

Gold has 49 times the nucleons of helium-4 but less than 4 times the radius. To double the radius you need \(2^{3} = 8\) times as many nucleons.

On a calculator the power \(\frac{1}{3}\) needs brackets: \(A\) ^ (1/3).

Using the proton number \(Z\) in place of \(A\).

Size of the nucleus

Every nucleus has the same density.

Mass \(\approx A\,\mathrm{u}\) · volume \(= \tfrac{4}{3}\pi R^{3}\) \(\rho = \dfrac{A\,\mathrm{u}}{\tfrac{4}{3}\pi (R_{0}A^{1/3})^{3}}\) \(= \dfrac{3\,\mathrm{u}}{4\pi R_{0}^{3}}\) \(= \dfrac{3 \times 1.661 \times 10^{-27}}{4\pi \times (1.20 \times 10^{-15})^{3}}\) \(\approx \class{res}{2.3 \times 10^{17} \,\mathrm{kg} \,\mathrm{m}^{-3}}\)

\(A\) cancels: the same for all nuclei
About \(10^{14}\) times the density of water
Ordinary matter is mostly empty space

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

Why does \(A\) cancel?

Mass is proportional to \(A\) (each nucleon is about 1 u). Volume is proportional to \(R^{3}\), and \((A^{1/3})^{3} = A\), so volume is proportional to \(A\) as well. The ratio does not depend on \(A\). A nucleus behaves like a drop of liquid: add more and it gets bigger, not denser.

Making sense of \(10^{17}\)

A teaspoon (\(5\,\mathrm{cm}^{3}\)) of nuclear matter would have a mass of about \(10^{12} \,\mathrm{kg}\), a billion tonnes.

A neutron star is a star crushed until it is nuclear matter throughout: more than the mass of the Sun in a ball about 20 km across.

“Show that nuclear density is independent of \(A\)” is answered by the algebra on this slide. No numbers are needed.

The story · Basel, 1885 · Copenhagen, 1913

Background story · not assessed

A schoolteacher’s formula.

Johann Balmer

A thin vertical glass tube glowing pink in a dark room — Hydrogen glowing in a **discharge tube**

A dark strip crossed by a red line, a blue-green line and violet lines — Its light spread into a spectrum: the **visible lines of hydrogen**

1885: Balmer, a 60-year-old teacher, finds one formula that fits all four wavelengths
For 28 years nobody knows why it works
1913: Bohr sees the formula. His levels, \(E \propto -\frac{1}{n^{2}}\), produce it exactly

Bohr: “As soon as I saw Balmer’s formula, the whole thing was immediately clear to me.”

Portrait: unknown artist, public domain. Tube: Mohit Jhanjhotia, CC BY 4.0. Spectrum: Umop503, CC0.

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

The Bohr model for hydrogen

Bohr’s energy levels.

\(E = -\dfrac{13.6}{n^{2}}\,\mathrm{eV}\)

\(n = 1, 2, 3 \dots\) labels the level
\(n = 1\): −13.6 eV, the ground state
Levels crowd together towards 0
Ionization energy: 13.6 eV

HYDROGENNOT TO SCALE

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

Using the formula

\(n = 2\): \(\frac{-13.6}{4} = -3.40 \,\mathrm{eV}\). \(n = 3\): \(\frac{-13.6}{9} = -1.51 \,\mathrm{eV}\). \(n = 4\): \(\frac{-13.6}{16} = -0.85 \,\mathrm{eV}\). As \(n\) → ∞ the energy → 0 and the electron is free.

A transition, the quick way

For \(4 \to 2\): \(\Delta E_\mathrm{atom} = -3.40 - (-0.85) = -2.55 \,\mathrm{eV}\), so \(E_\gamma = -\Delta E_\mathrm{atom} = 2.55 \,\mathrm{eV}\). With \(hc = 1.24 \times 10^{-6} \,\mathrm{eV} \,\mathrm{m}, \lambda = \frac{1.24 \times 10^{-6}}{2.55} = 4.86 \times 10^{-7} \,\mathrm{m}\): hydrogen’s blue-green line at 486 nm.

Why it mattered

One formula with one constant reproduced every line of hydrogen then measured, and predicted further infrared lines that were found afterwards.

Dropping the minus sign, or squaring the 13.6 instead of the \(n\).

At SL the level energies are given. At HL you may have to calculate them from this formula.

The Bohr model for hydrogen

Why only those levels?

Bohr’s rule: angular momentum is quantized

\(mvr = \dfrac{nh}{2\pi}\)

Only certain orbits fit: \(r \propto n^{2}\)
Each orbit has one energy: \(E \propto -\frac{1}{n^{2}}\)

Quantized angular momentum → quantized orbits and energy.

BOHR ORBITS OF HYDROGENTO SCALE

− Electron+ Proton

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

How the rule fixes the orbits

The electric force keeps the electron in a circle: \(\frac{ke^{2}}{r^{2}} = \frac{mv^{2}}{r}\). Bohr’s rule gives \(v = \frac{nh}{2\pi mr}\). Put one into the other and \(r = \frac{n^{2}h^{2}}{4\pi^{2}kme^{2}}\). Everything except \(n\) is a constant, so \(r \propto n^{2}\). For \(n = 1\) it gives \(5.3 \times 10^{-11} \,\mathrm{m}\), the right size for an atom.

And the energies

The total energy of the orbiting electron is \(-\frac{ke^{2}}{2r}\). With \(r \propto n^{2}\) this is proportional to \(-\frac{1}{n^{2}}\), and the constants multiply out to 13.6 eV.

Why should angular momentum be quantized? Bohr could not say. E.2 gives a reason: an electron has a wavelength \(\lambda = \frac{h}{mv}\), and a whole number of wavelengths must fit round the orbit, \(2\pi r = n\lambda\). That is the same equation.

\(mvr\) is the angular momentum of a particle in a circular orbit. The derivation is not required, but you should be able to use the rule.

Nature of science

Where the Bohr model fails.

IT WORKS FORIT FAILS FOR

Hydrogen: one electron

≠

Atoms with two or more electrons

The wavelengths of the lines

≠

Why some lines are brighter than others

Assuming the quantization rule

≠

Explaining why the rule holds

Replaced by quantum mechanics: the electron is described by a wave, not an orbit.

E.1 STRUCTURE OF THE ATOM · ADDITIONAL HL

Why does a second electron break it?

Bohr’s calculation has one electron attracted by one nucleus. Add another electron and the two repel each other, which the model has no way to include. It gives the wrong levels even for helium.

What else it gets wrong

It mixes classical orbits with a quantum rule that has no justification. It cannot say how likely each transition is, so it cannot give line intensities. And an electron with a definite orbit has an exact position and momentum together, which later physics showed is impossible.

Why still teach it?

It was the first model to give discrete levels from a calculation, and its formula for hydrogen is exactly what quantum mechanics gives too.

Quick check · HL

The nucleus and Bohr.

01Radius of a nucleus with \(A = 64\)? \((R_{0} = 1.20 \,\mathrm{fm})\)

\(1.20 \times 64^{1/3} = 4.80 \,\mathrm{fm}\)

02Why do all nuclei have about the same density?

Mass \(\propto A\) and volume \(\propto R^{3} \propto A\)

03At closest approach, what has the kinetic energy become?

Electric potential energy

04Why does Rutherford’s formula fail at high energy?

The alpha reaches the nucleus: the strong force acts

05Energy of the \(n = 5\) level of hydrogen?

\(\frac{-13.6}{25} = -0.544 \,\mathrm{eV}\)

06What did Bohr say is quantized?

Angular momentum: \(mvr = \frac{nh}{2\pi}\)

PRESS → TO REVEAL EACH ANSWER

Summary

Key words.

NucleusTiny, dense, positiveNearly all the mass

Nucleon number\(A\): protons + neutrons

Proton number\(Z\): fixes the element

Energy levelAn allowed energy of the atomDiscrete

PhotonPacket of light energy\(E = hf\)

TransitionJump between levels\(E_\gamma = hf = |\Delta E_\mathrm{atom}|\)

Emission spectrumBright lines from a hot gas

Absorption spectrumDark lines on a continuous spectrum

Nuclear radius\(R = R_{0}A^{1/3}\)Density the same for all

Closest approach\(E_{\mathrm{k}}\) → electric \(E_{\mathrm{p}}\)

Bohr levels\(E = -\frac{13.6}{n^{2}} \,\mathrm{eV}\)Hydrogen only

Quantization\(mvr = \frac{nh}{2\pi}\)

E.1 STRUCTURE OF THE ATOM · SL + HL