What two conclusions came from the Geiger–Marsden–Rutherford experiment?

Most alpha particles passed straight through, so the atom is mostly empty space; a few were deflected through large angles, so positive charge and most mass are concentrated in a tiny nucleus.

In $^{A}_{Z}\text{X}$, what do $A$ and $Z$ represent?

$A$ is the nucleon (mass) number; $Z$ is the proton (atomic) number. Neutrons $N = A - Z$.

Why does an emission spectrum consist of separate lines?

Electrons occupy discrete energy levels, so emitted photons can only have energies equal to level differences: $E = hf$.

How is an absorption spectrum formed?

White light passes through a cool gas; photons whose energies match level differences are absorbed, leaving dark lines in a continuous spectrum.

(HL) Equation for nuclear radius

$R = R_0 A^{1/3}$ with $R_0 \approx 1.2\ \text{fm}$. Nuclear density is therefore roughly constant, about $10^{17}\ \text{kg m}^{-3}$.

(HL) Bohr model energy levels of hydrogen

$E = -\frac{13.6}{n^2}\ \text{eV}$, from the condition $mvr = \frac{nh}{2\pi}$.

E.1 Structure of the atom — IB Physics notes and practice questions

How alpha scattering revealed a small dense nucleus, how nuclides are written, and how discrete atomic energy levels produce line spectra; at HL, nuclear size and density, departures from Rutherford scattering and the Bohr model. Study guide, 6 practice questions with markschemes and 6 flashcards. Free.

Geiger-Marsden-Rutherford experiment · nuclear notation · emission and absorption spectra · photons and discrete energy levels · (HL) nuclear radius and density · (HL) deviations from Rutherford scattering · (HL) Bohr model and quantised angular momentum

Study guide

Discovering the nucleus

In the Geiger–Marsden–Rutherford experiment, a narrow beam of alpha particles was aimed at very thin gold foil in a vacuum, and the scattered particles were counted at different angles.

Most alphas passed straight through, so most of the atom is empty space.
A few were deflected through large angles, so there is a concentrated region of positive charge.
A very small fraction bounced back (deflection over $90^\circ$), so this region holds most of the atom's mass and is tiny compared with the atom.

This replaced the earlier "plum pudding" picture with the nuclear model: a small, dense, positive nucleus surrounded by electrons. The nucleus is roughly $10^{-15}$ m across, while the atom is about $10^{-10}$ m.

Nuclear notation

A nuclide is written as $^{A}_{Z}\text{X}$:

$Z$, the proton number, fixes the element.
$A$, the nucleon number, counts protons plus neutrons.
The neutron number is $N = A - Z$.

Isotopes have the same $Z$ but different $N$, so they behave the same chemically but have different masses and nuclear stability.

Energy levels and spectra

Electrons in an atom can only occupy certain discrete energy levels. The levels have negative values because energy must be supplied to free the electron (the zero is an electron at rest far from the atom). The lowest level is the ground state.

When an electron drops from a higher level to a lower one, the atom emits a single photon whose energy equals the gap:

\[E = hf = \frac{hc}{\lambda} = E_{\text{upper}} - E_{\text{lower}}\]

An emission spectrum (from a hot, low-pressure gas) shows bright lines at specific wavelengths.
An absorption spectrum (white light through a cooler gas) shows dark lines at the same wavelengths, because the atoms absorb only photons whose energy exactly matches a gap.

Since each element has its own set of levels, its line spectrum works like a fingerprint. This is how astronomers identify elements in stars. An atom excited to level $n$ can emit up to $\frac{n(n-1)}{2}$ different photon energies on the way down.

Nuclear radius and density (HL)

Scattering experiments show the nuclear radius follows

\[R = R_0 A^{1/3}, \quad R_0 \approx 1.2\ \text{fm}\]

Since volume grows with $R^3 \propto A$, the volume is proportional to the number of nucleons. Mass is also proportional to $A$, so nuclear density is roughly the same for all nuclei, about $10^{17}\ \text{kg m}^{-3}$.

Deviations from Rutherford scattering (HL)

Rutherford's formula assumes only electrostatic repulsion. When alphas are given high enough energy, they get close enough to feel the short-range strong nuclear force, and the counts at large angles fall below the prediction. The energy at which this starts gives an estimate of nuclear size. High-energy electrons can also probe nuclei; they show a diffraction-like minimum because they behave as waves.

The Bohr model (HL)

Bohr proposed that the electron in hydrogen moves in circular orbits in which its angular momentum is quantised:

\[mvr = \frac{nh}{2\pi}\]

Combining this with the electrostatic force as the centripetal force gives allowed energies

\[E_n = -\frac{13.6}{n^2}\ \text{eV}\]

The model predicts hydrogen's spectrum well but fails for atoms with more than one electron, and it cannot explain why accelerating electrons do not radiate.

Worked example 1: photon from a transition

In hydrogen, an electron falls from $-3.40\ \text{eV}$ to $-13.6\ \text{eV}$. Find the wavelength of the photon.

$\Delta E = 10.2\ \text{eV} = 10.2 \times 1.60 \times 10^{-19} = 1.63 \times 10^{-18}\ \text{J}$
$\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{1.632 \times 10^{-18}} = 1.22 \times 10^{-7}\ \text{m}$ (122 nm, ultraviolet)

Worked example 2: nuclear radius (HL)

Find the radius of an aluminium-27 nucleus, with $R_0 = 1.20\ \text{fm}$.

$27^{1/3} = 3$, so $R = 1.20 \times 3 = 3.60\ \text{fm} = 3.60 \times 10^{-15}\ \text{m}$

Common mistakes

Thinking the deflected alphas hit electrons; electrons are far too light to turn them back.
Swapping $A$ and $Z$, or adding them to get the neutron number.
Forgetting to convert eV to J before using $E = hc/\lambda$.
Saying absorption lines are at different wavelengths from emission lines for the same gas.
Using $R \propto A$ instead of $R \propto A^{1/3}$.

Exam tips

Link each scattering observation to one conclusion; examiners mark them in pairs.
Draw an energy-level diagram with arrows when counting possible transitions.
Remember $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ and $1\ \text{fm} = 10^{-15}\ \text{m}$.
In density estimates, the answer should come out near $10^{17}\ \text{kg m}^{-3}$; use this as a check.

Cheat sheet

Equation	Symbols	Units
$N = A - Z$	$A$ nucleon number, $Z$ proton number	none
$E = hf = \frac{hc}{\lambda}$	$h = 6.63 \times 10^{-34}$ J s, $c = 3.00 \times 10^8$ m s$^{-1}$	J
$\Delta E = E_{\text{upper}} - E_{\text{lower}}$	photon energy equals level gap	J or eV
$1\ \text{eV} = 1.60 \times 10^{-19}$ J	electronvolt	J
(HL) $R = R_0 A^{1/3}$	$R_0 \approx 1.2$ fm	m
(HL) $\rho = \frac{Am_n}{\frac{4}{3}\pi R^3}$	$m_n$ nucleon mass	kg m$^{-3}$
(HL) $mvr = \frac{nh}{2\pi}$	$n$ integer, $r$ orbit radius	J s
(HL) $E_n = -\frac{13.6}{n^2}$ eV	hydrogen energy levels	eV

Number of lines from level $n$: $\frac{n(n-1)}{2}$. Nuclear density is about $10^{17}$ kg m$^{-3}$ for all nuclei.

Key definitions

Practice questions

SLPaper 1AFoundation

Question 1

(a) [1]

In the Geiger–Marsden–Rutherford experiment, alpha particles were fired at thin gold foil. Which observation led to the conclusion that most of the atom is empty space?

A. A very small number of alpha particles bounced back
B. Most alpha particles passed through with little or no deflection
C. Alpha particles were absorbed by the foil
D. The foil emitted electrons

Markscheme

1 B. If most alpha particles meet nothing that deflects them, most of the atom's volume contains no concentrated mass or charge.

SLPaper 1AFoundation

Question 2

(a) [1]

How many neutrons are in a nucleus of $^{56}_{26}\text{Fe}$?

A. 26
B. 30
C. 56
D. 82

Markscheme

1 B. Neutron number $N = A - Z = 56 - 26 = 30$.

SLPaper 2Challenging

Question 3

An atom of a gas has only three energy levels: $E_1 = -8.60\ \text{eV}$ (ground state), $E_2 = -3.10\ \text{eV}$ and $E_3 = -1.40\ \text{eV}$. The atom is excited to level $E_3$.

(a) [1]

State the number of different photon energies that could be emitted as the atom returns from level $E_3$ to the ground state.

Markscheme

1 3 (transitions $3 \to 2$, $2 \to 1$ and $3 \to 1$)

(b) [3]

Calculate the wavelength of the highest-energy photon that could be emitted.

Markscheme

1 Largest energy gap: $-1.40 - (-8.60) = 7.20\ \text{eV} = 7.20 \times 1.60 \times 10^{-19} = 1.15 \times 10^{-18}\ \text{J}$
1 $\lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{1.15 \times 10^{-18}}$
1 $\lambda = 1.7 \times 10^{-7}\ \text{m}$ (173 nm)

(c) [2]

A photon of energy $6.00\ \text{eV}$ is incident on an atom in the ground state. Explain whether the photon can be absorbed.

Markscheme

1 Possible absorption energies from the ground state are $5.50\ \text{eV}$ and $7.20\ \text{eV}$ only
1 $6.00\ \text{eV}$ matches no energy difference, so the photon is not absorbed (it passes through)

Equation	Symbols	Units
\(N = A - Z\)	\(A\) nucleon number, \(Z\) proton number	none
\(E = hf = \frac{hc}{\lambda}\)	\(h = 6.63 \times 10^{-34}\) J s, \(c = 3.00 \times 10^8\) m s\(^{-1}\)	J
\(\Delta E = E_{\text{upper}} - E_{\text{lower}}\)	photon energy equals level gap	J or eV
\(1\ \text{eV} = 1.60 \times 10^{-19}\) J	electronvolt	J
(HL) \(R = R_0 A^{1/3}\)	\(R_0 \approx 1.2\) fm	m
(HL) \(\rho = \frac{Am_n}{\frac{4}{3}\pi R^3}\)	\(m_n\) nucleon mass	kg m\(^{-3}\)
(HL) \(mvr = \frac{nh}{2\pi}\)	\(n\) integer, \(r\) orbit radius	J s
(HL) \(E_n = -\frac{13.6}{n^2}\) eV	hydrogen energy levels	eV

E.1 Structure of the atom

Study guide

Discovering the nucleus

Nuclear notation

Energy levels and spectra

Nuclear radius and density (HL)

Deviations from Rutherford scattering (HL)

The Bohr model (HL)

Worked example 1: photon from a transition

Worked example 2: nuclear radius (HL)

Common mistakes

Exam tips

Cheat sheet

Key definitions

Practice questions

Question 1

Question 2

Question 3

Key facts