Chapter 2 · Structure of Atom

2.1 Subatomic Particles and Early Atomic Models

Cathode rays, oil drops and gold foil: how a handful of experiments revealed the electron, the proton, the neutron and the nucleus — and how we label every atom by its atomic number and mass number.

Section 1.2 described Dalton's atom: a tiny, hard, indivisible ball, different for each element. That picture explained the laws of chemical combination beautifully — and it was wrong in one important way. By the end of the nineteenth century, experiments on electricity passing through gases, on electrolysis and on radioactivity had shown that atoms can lose or gain tiny charged pieces. Atoms have an inside.

This section tells the story of how that inside was mapped: the electron in 1897, the nucleus in 1911, the neutron in 1932. The experiments are worth learning properly, not only because Board papers ask about them, but because the reasoning — measure a deflection, deduce a charge-to-mass ratio; fire particles at a target, deduce what is inside — is exactly how physicists and chemists still probe matter, from hospital mass spectrometers to the Large Hadron Collider. We finish with the vocabulary every later chapter uses: atomic number, mass number, isotopes and isobars.

Cathode Rays and the Discovery of the Electron

Gases are normally very poor conductors of electricity. In the 1850s–1870s, Michael Faraday, William Crookes and others sealed two metal electrodes into a glass tube, pumped out most of the gas and applied a very high voltage (several thousand volts). At very low pressure the gas stops glowing and a beam travels from the negative electrode (the cathode) towards the positive electrode (the anode). If the anode has a hole in it, the beam passes through and makes the glass — or a coating of zinc sulphide placed behind the anode — glow where it strikes. The old television sets and computer monitors that used picture tubes were exactly such tubes, with the beam steered across a fluorescent screen.

Careful experiments established the following properties of cathode rays:

The last point is the crucial one. Whatever the electrodes and whatever the gas, the same particles come out — so they must be part of every atom.

📘 Definition 2.1 — Cathode rays and the electron
Cathode rays are streams of negatively charged particles that travel from the cathode to the anode in a discharge tube at very low pressure and high voltage. The particles are called electrons. Electrons are identical whatever their source, and they are a basic constituent of all atoms.

Thomson's charge-to-mass ratio

In 1897 J. J. Thomson measured how much a beam of cathode rays is bent by electric and magnetic fields applied at right angles to the beam and to each other. The deflection of a charged particle from its path is larger when

Thomson's method had two stages. First he applied both fields and adjusted them until the beam went straight on: the electric force $eE$ and the magnetic force $evB$ then cancel, which gives the speed of the electrons, $v=E/B$. Then he switched the magnetic field off and measured how far the electric field alone moved the spot. Because the deflection depends only on the ratio of charge to mass, the experiment gives $e/m_{e}$ — not $e$ and $m_{e}$ separately. The modern value is

$$\frac{e}{m_{e}}=1.758\,820\times10^{11}\ \mathrm{C\,kg^{-1}}.$$

This ratio is about $1800$ times larger than that of the lightest ion known at the time (the hydrogen ion in electrolysis). Either the electron carries a huge charge, or it is very light. Millikan's experiment settled which.

Interactive 1 — Thomson's e/m Experiment

Electrons are accelerated by the voltage $V$, pass between charged plates (electric field $E$, pushing them up towards the + plate) inside a magnetic field $B$ (pushing them down), and hit the screen. Spot positions: A = electric field only, C = magnetic field only, B = both. Adjust $B$ until the spot returns to the centre — then $v=E/B$, and the widget works out $e/m_{e}$ from the deflection at A, just as Thomson did.

2.08.00.100
speed from $V$ (m/s)—
net force on beam—
spot on screen (mm)—
$v=E/B$ (m/s)—
$e/m_{e}$ found (C/kg)—

Millikan's oil-drop experiment: the charge of the electron

Between 1906 and 1914 R. A. Millikan sprayed a fine mist of oil droplets into a chamber between two horizontal metal plates. X-rays passing through the chamber knocked electrons out of air molecules, and some of these electrons stuck to the droplets. Watching one droplet through a microscope, Millikan first timed its steady fall under gravity (which gives its size and mass), then switched on an electric field across the plates and measured how much the field slowed, stopped or reversed the fall. From this he could calculate the charge on the droplet. Repeating the measurement on many drops, he found a striking pattern.

📐 Rule 2.1 — Quantisation of electric charge
The charge $q$ on any oil drop (and on any object) is always a whole-number multiple of a smallest charge $e$: $$q=n\,e,\qquad n=1,2,3,\ldots,\qquad e=1.602\,176\times10^{-19}\ \mathrm{C}.$$ The electron carries the charge $-e$. Combining this with Thomson's ratio gives the electron's mass: $$m_{e}=\frac{e}{e/m_{e}}=\frac{1.602\,176\times10^{-19}\ \mathrm{C}}{1.758\,820\times10^{11}\ \mathrm{C\,kg^{-1}}}=9.1094\times10^{-31}\ \mathrm{kg}.$$
✏️ Example 1 — The mass and charge of electrons
Using $e/m_{e}=1.7588\times10^{11}\ \mathrm{C\,kg^{-1}}$, $e=1.602\times10^{-19}\ \mathrm{C}$ and $N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$, find (a) the mass of one electron; (b) the mass of one mole of electrons; (c) the total charge of one mole of electrons; (d) the number of electrons that together carry a charge of $1.00\ \mathrm{C}$.
(a) $m_{e}=\dfrac{e}{e/m_{e}}=\dfrac{1.602\times10^{-19}\ \mathrm{C}}{1.7588\times10^{11}\ \mathrm{C\,kg^{-1}}}=9.108\times10^{-31}\ \mathrm{kg}$ (four significant figures, like the data; the precise value is $9.109\times10^{-31}\ \mathrm{kg}$).
(b) $9.108\times10^{-31}\ \mathrm{kg}\times6.022\times10^{23}\ \mathrm{mol^{-1}}=5.485\times10^{-7}\ \mathrm{kg\,mol^{-1}}$, i.e. about $0.55\ \mathrm{mg}$ per mole — a mole of electrons weighs about $\tfrac{1}{1840}$ as much as a mole of hydrogen atoms.
(c) $1.602\times10^{-19}\ \mathrm{C}\times6.022\times10^{23}\ \mathrm{mol^{-1}}=9.647\times10^{4}\ \mathrm{C\,mol^{-1}}$. (This quantity, about $96\,500\ \mathrm{C}$ per mole of electrons, is the Faraday constant you will meet in electrochemistry.)
(d) $n=\dfrac{q}{e}=\dfrac{1.00\ \mathrm{C}}{1.602\times10^{-19}\ \mathrm{C}}=6.24\times10^{18}$ electrons.
✏️ Example 2 — Finding $e$ from oil-drop data
In an oil-drop experiment five droplets carry charges of $4.80$, $8.00$, $11.21$, $6.40$ and $3.20$ (all $\times10^{-19}\ \mathrm{C}$). Show that the data are consistent with charge quantisation and find the best value of the elementary charge from them.
Idea Look for the largest charge that divides every value a whole number of times. The smallest charge, $3.20$, is not necessarily $e$ itself — it may be $2e$.
Differences Differences between values: $4.80-3.20=1.60$, $6.40-4.80=1.60$, $8.00-6.40=1.60$, $11.21-8.00=3.21$. The common step is about $1.60\times10^{-19}\ \mathrm{C}$.
Multiples Dividing each charge by $1.60$: $3.00,\ 5.00,\ 7.01,\ 4.00,\ 2.00$ — whole numbers $n=3,5,7,4,2$. ✓ The data are quantised.
Best value $e=\dfrac{\sum q}{\sum n}=\dfrac{(4.80+8.00+11.21+6.40+3.20)\times10^{-19}}{3+5+7+4+2}=\dfrac{33.61\times10^{-19}}{21}=1.60\times10^{-19}\ \mathrm{C}$.

Protons, Neutrons and the Particles of the Atom

Atoms are electrically neutral, so if they contain negative electrons they must also contain an equal amount of positive charge. In 1886 E. Goldstein used a discharge tube with a perforated cathode and saw a glow behind the cathode: rays moving the opposite way, towards and through the cathode. These canal rays (or anode rays) are positively charged ions formed when cathode rays knock electrons out of gas atoms. Unlike cathode rays,

The lightest positive ion, with the largest $e/m$, came from hydrogen: a hydrogen atom that has lost its one electron. This particle, with charge $+e$ and mass about $1836$ times the electron's, was named the proton (Rutherford, around 1919). Atomic masses, however, were too large to be accounted for by protons alone — helium has two protons but a mass of about four units. In 1932 James Chadwick bombarded a thin sheet of beryllium with α-particles and detected electrically neutral particles with a mass slightly larger than the proton's — the neutron:

$$\mathrm{{}^{9}_{4}Be+{}^{4}_{2}He\rightarrow{}^{12}_{6}C+{}^{1}_{0}n}$$
📘 Definition 2.2 — The fundamental particles of the atom

Atoms are built from three fundamental particles. The proton and the neutron are found in the nucleus and are together called nucleons.

ParticleSymbolCharge / CRelative chargeMass / kgMass / uApprox. mass / u
electron$\mathrm{e^{-}}$$-1.602\,176\times10^{-19}$$-1$$9.109\,38\times10^{-31}$$0.000\,54$$0$
proton$\mathrm{p}$$+1.602\,176\times10^{-19}$$+1$$1.672\,62\times10^{-27}$$1.007\,27$$1$
neutron$\mathrm{n}$$0$$0$$1.674\,93\times10^{-27}$$1.008\,67$$1$
✏️ Example 3 — How different are the proton and the electron?
(a) How many times heavier than an electron is a proton? (b) Find the charge-to-mass ratio of the proton and compare it with the electron's $1.7588\times10^{11}\ \mathrm{C\,kg^{-1}}$. (c) A hydrogen atom has mass $1.6735\times10^{-27}\ \mathrm{kg}$. What percentage of this mass is due to its electron? Use the masses in Definition 2.2.
(a) $\dfrac{m_{p}}{m_{e}}=\dfrac{1.67262\times10^{-27}}{9.10938\times10^{-31}}=1836$.
(b) $\dfrac{e}{m_{p}}=\dfrac{1.602176\times10^{-19}\ \mathrm{C}}{1.67262\times10^{-27}\ \mathrm{kg}}=9.579\times10^{7}\ \mathrm{C\,kg^{-1}}$ — smaller than the electron's ratio by the same factor $1836$, because the charges are equal in size.
(c) $\dfrac{9.109\times10^{-31}}{1.6735\times10^{-27}}\times100\%=0.0544\%$. Practically all of an atom's mass is in its nucleus.

The Thomson and Rutherford Models of the Atom

Knowing the particles is not the same as knowing how they are arranged. In 1898 Thomson proposed that an atom is a sphere of radius about $10^{-10}\ \mathrm{m}$ in which the positive charge is spread out uniformly, with the electrons embedded in it so as to give the most stable arrangement — like seeds in a watermelon (or plums in a pudding, hence the name "plum pudding model"). In this model the mass of the atom is also spread uniformly through the sphere. It explained why atoms are neutral, but nothing more, and it did not survive its first real test.

Rutherford's α-particle scattering experiment

In 1909–1911 Ernest Rutherford, with Hans Geiger and Ernest Marsden, fired a narrow beam of fast α-particles (helium nuclei, $\mathrm{He^{2+}}$, emitted by a radioactive source) at a very thin gold foil, about $100\ \mathrm{nm}$ thick. A circular screen coated with zinc sulphide surrounded the foil, and every α-particle that struck it produced a tiny flash of light that could be counted through a microscope. If Thomson were right, the positive charge and mass of each gold atom would be too thinly spread to turn a heavy, fast α-particle by more than a fraction of a degree. The results were:

  1. most α-particles passed straight through the foil undeflected;
  2. a small fraction were deflected through small angles;
  3. a very few (about $1$ in $20\,000$) were deflected through large angles, some bouncing almost straight back.

Rutherford said it was "as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you". Something very small, very heavy and strongly positive must be inside each atom.

📘 Definition 2.3 — Rutherford's nuclear model of the atom
  1. The positive charge and almost all the mass of an atom are concentrated in an extremely small central region, the nucleus.
  2. The nucleus is tiny: its radius is about $10^{-15}\ \mathrm{m}$, whereas the radius of the atom is about $10^{-10}\ \mathrm{m}$. Most of the atom is empty space.
  3. Electrons move around the nucleus at high speed in circular paths called orbits, rather as planets go round the Sun; the attraction between the nucleus and the electrons holds them together.
  4. The atom is neutral: the number of electrons equals the number of unit positive charges on the nucleus.

The three observations follow at once. An α-particle passes through mostly empty space, so it is usually not deflected; one that passes fairly close to a nucleus is pushed sideways by its positive charge and turns slightly; only an α-particle heading almost straight at a nucleus — a rare event, because the nucleus is a tiny target — is turned back. Interactive 2 lets you fire α-particles at a nucleus and compare the two models.

Interactive 2 — Rutherford's Gold-Foil Experiment

α-particles are fired from the left at random distances from the axis. In the nuclear model all the positive charge sits in a tiny nucleus, and an α-particle aimed close to it is turned through a large angle; in Thomson's model the same charge is spread through the whole atom (dashed circle) and every particle goes almost straight through. The picture is zoomed far in on one atom (the real nucleus is about $10^{5}$ times smaller than the atom, so large deflections are much rarer than here). Change the nuclear charge $Z$ to compare foils of different metals.

α-particles fired0
deflected less than 10°0
deflected 10°–90°0
deflected more than 90°0
last deflection—
✏️ Example 4 — How empty is an atom?
(a) Take the radius of an atom as $1.0\times10^{-10}\ \mathrm{m}$ and that of its nucleus as $1.0\times10^{-15}\ \mathrm{m}$. What fraction of the atom's volume does the nucleus occupy? (b) If the nucleus were enlarged to the size of a cricket ball of radius $3.6\ \mathrm{cm}$, how large would the atom be? (c) A carbon-12 nucleus has mass $1.99\times10^{-26}\ \mathrm{kg}$ and radius $2.7\times10^{-15}\ \mathrm{m}$. Estimate its density and compare it with water ($1.0\times10^{3}\ \mathrm{kg\,m^{-3}}$).
(a) Volume scales as radius cubed: $\dfrac{V_{\text{nucleus}}}{V_{\text{atom}}}=\left(\dfrac{10^{-15}}{10^{-10}}\right)^{3}=(10^{-5})^{3}=1\times10^{-15}$. The nucleus fills one part in a thousand million million of the atom.
(b) The ratio of radii is $10^{5}$, so the atom's radius would be $3.6\ \mathrm{cm}\times10^{5}=3.6\times10^{5}\ \mathrm{cm}=3.6\ \mathrm{km}$ — a cricket ball at the centre of a sphere several kilometres across.
(c) $V=\tfrac43\pi r^{3}=\tfrac43\pi(2.7\times10^{-15}\ \mathrm{m})^{3}=8.2\times10^{-44}\ \mathrm{m^{3}}$, so $\rho=\dfrac{m}{V}=\dfrac{1.99\times10^{-26}\ \mathrm{kg}}{8.2\times10^{-44}\ \mathrm{m^{3}}}=2.4\times10^{17}\ \mathrm{kg\,m^{-3}}$.
Compare $\dfrac{2.4\times10^{17}}{1.0\times10^{3}}\approx2\times10^{14}$: nuclear matter is about $10^{14}$ times denser than water. A teaspoon of it would weigh as much as a mountain.

What the Rutherford model could not explain

⚠️ Drawbacks of Rutherford's model

1. Stability. An electron moving in a circle is constantly changing direction, so it is accelerating. Classical electromagnetic theory (Maxwell) says that an accelerating charge radiates energy as electromagnetic waves. The orbiting electron should therefore lose energy and spiral into the nucleus — calculations give a lifetime of about $10^{-8}\ \mathrm{s}$. Real atoms are stable. (A planet does not have this problem because it is not charged.)

2. Line spectra. A spiralling electron would emit a continuous range of frequencies, but heated atoms emit only a few sharp lines (Section 2.2).

3. Electron arrangement. The model says nothing about how electrons are distributed around the nucleus or what their energies are — the very things that decide chemical behaviour.

Section 2.2 shows how Bohr repaired the model by quantising the electron's energy, and Section 2.3 how quantum mechanics replaced orbits by orbitals.

Atomic Number, Mass Number, Isotopes and Isobars

Rutherford's experiments, followed by Moseley's X-ray measurements in 1913, showed that the positive charge on the nucleus — the number of protons — is what makes an element what it is. Every carbon atom has six protons; every atom with six protons is carbon.

📘 Definition 2.4 — Atomic number and mass number

The atomic number $Z$ of an element is the number of protons in the nucleus of each of its atoms; in a neutral atom it also equals the number of electrons. The mass number $A$ is the total number of protons and neutrons (nucleons) in the nucleus:

$$A=Z+\text{number of neutrons}.$$

A particular kind of atom (a nuclide) is written ${}^{A}_{Z}\mathrm{X}$, with the mass number as a left superscript and the atomic number as a left subscript: $\mathrm{{}^{35}_{17}Cl}$, $\mathrm{{}^{12}_{6}C}$. Because the symbol already fixes $Z$, the subscript is often omitted: $\mathrm{{}^{35}Cl}$, carbon-12.

📐 Rule 2.2 — Counting the particles in an atom or ion
For the species ${}^{A}_{Z}\mathrm{X}^{q}$ with ionic charge $q$ (written as a signed whole number, e.g. $q=+3$ or $-2$): $$\text{protons}=Z,\qquad\text{neutrons}=A-Z,\qquad\text{electrons}=Z-q.$$ A cation ($q\gt0$) has lost electrons; an anion ($q\lt0$) has gained them. Forming an ion never changes the nucleus, so $Z$, $A$ and the number of neutrons stay the same.
✏️ Example 5 — Counting protons, neutrons and electrons
(a) Complete the table for $\mathrm{{}^{35}_{17}Cl^{-}}$, $\mathrm{{}^{56}_{26}Fe^{3+}}$, $\mathrm{{}^{238}_{92}U}$ and $\mathrm{{}^{18}_{8}O^{2-}}$: protons, neutrons, electrons. (b) A species has $18$ electrons, $20$ neutrons and a charge of $+1$. Identify it and write its symbol.
$\mathrm{{}^{35}_{17}Cl^{-}}$ protons $17$; neutrons $35-17=18$; electrons $17-(-1)=18$.
$\mathrm{{}^{56}_{26}Fe^{3+}}$ protons $26$; neutrons $56-26=30$; electrons $26-3=23$.
$\mathrm{{}^{238}_{92}U}$, $\mathrm{{}^{18}_{8}O^{2-}}$ Uranium: $92$, $146$, $92$. Oxide ion: $8$ protons, $10$ neutrons, $8-(-2)=10$ electrons.
(b) Electrons $=Z-q\Rightarrow18=Z-1\Rightarrow Z=19$, which is potassium. $A=Z+\text{neutrons}=19+20=39$. The species is $\mathrm{{}^{39}_{19}K^{+}}$.
📘 Definition 2.5 — Isotopes and isobars

Isotopes are atoms of the same element — the same atomic number $Z$ — with different mass numbers $A$, because they contain different numbers of neutrons. Hydrogen has three: protium $\mathrm{{}^{1}_{1}H}$ (no neutron, $99.985\%$ of natural hydrogen), deuterium $\mathrm{{}^{2}_{1}H}$ or D (one neutron, $0.015\%$) and tritium $\mathrm{{}^{3}_{1}H}$ or T (two neutrons, radioactive, only traces). Carbon has $\mathrm{{}^{12}C}$, $\mathrm{{}^{13}C}$ and radioactive $\mathrm{{}^{14}C}$; chlorine has $\mathrm{{}^{35}Cl}$ and $\mathrm{{}^{37}Cl}$.

Isobars are atoms of different elements — different $Z$ — with the same mass number $A$: for example $\mathrm{{}^{14}_{6}C}$ and $\mathrm{{}^{14}_{7}N}$, or $\mathrm{{}^{40}_{18}Ar}$, $\mathrm{{}^{40}_{19}K}$ and $\mathrm{{}^{40}_{20}Ca}$.

Chemical properties are decided by the number of electrons, which equals the number of protons; neutrons have almost no effect. So isotopes of an element have the same chemical properties (only the rates of their reactions differ slightly, an effect that is noticeable for the light hydrogen isotopes), while isobars, being different elements, behave quite differently. Isotopes are invaluable: carbon-14 dates archaeological remains, cobalt-60 is used in cancer radiotherapy, iodine-131 treats thyroid disease, and the ratio of $\mathrm{{}^{18}O}$ to $\mathrm{{}^{16}O}$ in polar ice records past climates. Section 1.2 showed how the abundances of isotopes give an element's average atomic mass. (A third, less common, term: atoms with the same number of neutrons, such as $\mathrm{{}^{14}_{6}C}$, $\mathrm{{}^{15}_{7}N}$ and $\mathrm{{}^{16}_{8}O}$, are called isotones.)

✏️ Example 6 — Isotopes or isobars?
For each pair, state whether the two are isotopes, isobars or neither, and give the number of neutrons in each: (a) $\mathrm{{}^{35}_{17}Cl}$ and $\mathrm{{}^{37}_{17}Cl}$; (b) $\mathrm{{}^{40}_{19}K}$ and $\mathrm{{}^{40}_{20}Ca}$; (c) $\mathrm{{}^{14}_{6}C}$ and $\mathrm{{}^{16}_{8}O}$; (d) $\mathrm{{}^{1}_{1}H}$ and $\mathrm{{}^{3}_{1}H}$. Which pairs have the same chemical properties?
(a) Same $Z=17$, different $A$: isotopes; neutrons $18$ and $20$.
(b) Same $A=40$, different $Z$: isobars; neutrons $21$ and $20$.
(c) Different $Z$ and different $A$: neither (both have $8$ neutrons, so they are isotones).
(d) Same $Z=1$: isotopes (protium and tritium); neutrons $0$ and $2$.
Chemistry Only the isotope pairs (a) and (d) have the same chemical properties, because they have the same number of electrons; $\mathrm{K}$ and $\mathrm{Ca}$ in (b) are different elements.

Exercises

I
Cathode rays, the electron and the fundamental particles. Use $e=1.602\times10^{-19}\ \mathrm{C}$, $m_{e}=9.109\times10^{-31}\ \mathrm{kg}$, $N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$ and $1\ \mathrm{u}=1.6605\times10^{-27}\ \mathrm{kg}$.
  • 1 Give two observations which show that cathode rays are made of negatively charged particles.
  • 2 Why was it so important that the properties of cathode rays do not depend on the cathode metal or the gas in the tube?
  • 3 How many electrons must be removed from a neutral metal sphere to give it a charge of $+3.204\times10^{-18}\ \mathrm{C}$?
  • 4 Three oil drops carry charges of $6.41$, $9.61$ and $12.82$ (all $\times10^{-19}\ \mathrm{C}$). Show that the data fit $e=1.602\times10^{-19}\ \mathrm{C}$ and find $n$ for each drop. Could these three drops alone also be explained by a smallest charge of $3.20\times10^{-19}\ \mathrm{C}$? What does this tell you about Millikan's method?
  • 5 What is the total charge on $1.00\ \mathrm{mmol}$ of electrons?
  • 6 An α-particle is a helium nucleus, $\mathrm{He^{2+}}$, of mass $6.645\times10^{-27}\ \mathrm{kg}$. Calculate its charge-to-mass ratio.
  • 7 A deuteron (the nucleus of deuterium) has mass $3.344\times10^{-27}\ \mathrm{kg}$. Compare its $e/m$ with that of the proton, $9.579\times10^{7}\ \mathrm{C\,kg^{-1}}$.
  • 8 Express the mass of a neutron, $1.67493\times10^{-27}\ \mathrm{kg}$, in unified atomic mass units.
1 They are deflected towards the positive plate of an electric field, and a magnetic field deflects them in the direction expected for a moving negative charge. (They also carry momentum — a light paddle wheel in their path turns.)
2 If the same particles come from every metal and every gas, they must be a constituent of all atoms — the first proof that atoms are divisible.
3 $n=\dfrac{q}{e}=\dfrac{3.204\times10^{-18}}{1.602\times10^{-19}}=20$ electrons.
4 $6.41/1.602=4.00$, $9.61/1.602=6.00$, $12.82/1.602=8.00$: $n=4,6,8$. ✓ But the same charges are also $2,3,4$ times $3.205\times10^{-19}\ \mathrm{C}$, so these three drops alone cannot tell $e$ from $2e$. Millikan needed many drops, including ones carrying an odd number of electrons, to fix the smallest unit.
5 $q=1.00\times10^{-3}\ \mathrm{mol}\times6.022\times10^{23}\ \mathrm{mol^{-1}}\times1.602\times10^{-19}\ \mathrm{C}=96.5\ \mathrm{C}$.
6 $\dfrac{q}{m}=\dfrac{2\times1.602\times10^{-19}\ \mathrm{C}}{6.645\times10^{-27}\ \mathrm{kg}}=4.822\times10^{7}\ \mathrm{C\,kg^{-1}}$ — about half the proton's value (twice the charge, about four times the mass).
7 $\dfrac{e}{m_{d}}=\dfrac{1.602\times10^{-19}}{3.344\times10^{-27}}=4.791\times10^{7}\ \mathrm{C\,kg^{-1}}$, almost exactly half the proton's: same charge, twice the mass.
8 $\dfrac{1.67493\times10^{-27}\ \mathrm{kg}}{1.6605\times10^{-27}\ \mathrm{kg\,u^{-1}}}=1.0087\ \mathrm{u}$.
II
Thomson's and Rutherford's models.
  • 1 Describe Thomson's model of the atom. What did it explain, and which experiment disproved it?
  • 2 State the three main observations of the α-particle scattering experiment and the conclusion drawn from each.
  • 3 Give two reasons why gold was a good choice of foil.
  • 4 The radius of an atom is about $10^{5}$ times that of its nucleus. If a model nucleus had a diameter of $1\ \mathrm{mm}$, what would the diameter of the model atom be?
  • 5 Why could a positive charge spread through the whole atom never turn an α-particle back?
  • 6 An aluminium-27 nucleus has a radius of about $3.6\times10^{-15}\ \mathrm{m}$ and the aluminium atom a radius of $143\ \mathrm{pm}$. What fraction of the atom's volume is occupied by its nucleus?
  • 7 Explain, using classical physics, why Rutherford's model predicts that atoms should collapse. Name one more observation it could not explain.
1 A sphere of radius about $10^{-10}\ \mathrm{m}$ with positive charge (and mass) spread uniformly, electrons embedded in it like seeds in a watermelon. It explained the neutrality of atoms; Rutherford's α-scattering experiment disproved it.
2 Most α-particles pass undeflected → most of the atom is empty space. A few are deflected through small angles → the positive charge is concentrated, and particles passing near it are repelled. A very few rebound → the positive charge and almost all the mass are packed into a tiny, heavy nucleus.
3 Gold is extremely malleable, so it can be beaten into a foil only a few hundred atoms thick (each α-particle meets few nuclei); and its nucleus is heavy and highly charged ($Z=79$), so deflections are large and the nucleus barely recoils.
4 $1\ \mathrm{mm}\times10^{5}=10^{5}\ \mathrm{mm}=100\ \mathrm{m}$ — a pinhead in the middle of a football ground.
5 Spread over a sphere $10^{-10}\ \mathrm{m}$ across, the charge produces only a weak electric field, and a heavy, fast α-particle passes through it with a deflection of a fraction of a degree. Only a charge concentrated in a tiny region gives the enormous repulsion needed to reverse it.
6 $\left(\dfrac{3.6\times10^{-15}}{1.43\times10^{-10}}\right)^{3}=(2.52\times10^{-5})^{3}=1.6\times10^{-14}$.
7 An orbiting electron accelerates (its direction changes continuously); an accelerating charge radiates energy, so the electron should lose energy and spiral into the nucleus in about $10^{-8}\ \mathrm{s}$. The model also cannot explain the line spectra of atoms (or say anything about how electrons are arranged and what their energies are).
III
Atomic number, mass number, isotopes and isobars.
  • 1 Find the numbers of protons, neutrons and electrons in $\mathrm{{}^{31}_{15}P}$.
  • 2 … in $\mathrm{{}^{24}_{12}Mg^{2+}}$.
  • 3 … in $\mathrm{{}^{32}_{16}S^{2-}}$.
  • 4 … in $\mathrm{{}^{80}_{35}Br}$.
  • 5 … in $\mathrm{{}^{107}_{47}Ag^{+}}$.
  • 6 Write the symbol of the species with $26$ protons, $30$ neutrons and $24$ electrons.
  • 7 An ion with a charge of $-3$ has $18$ electrons and $16$ neutrons. Identify it and write its symbol.
  • 8 An atom has mass number $75$, and the number of neutrons in it is $27.3\%$ greater than the number of protons. Identify the element.
  • 9 Are $\mathrm{{}^{3}_{1}H}$ and $\mathrm{{}^{3}_{2}He}$ isotopes or isobars?
  • 10 Classify the pairs $\mathrm{{}^{235}_{92}U}$, $\mathrm{{}^{238}_{92}U}$ and $\mathrm{{}^{40}_{18}Ar}$, $\mathrm{{}^{40}_{19}K}$.
  • 11 Explain why the isotopes of an element have the same chemical properties but slightly different physical properties (such as density).
  • 12 How many neutrons are present in $7.0\ \mathrm{mg}$ of carbon-14 (molar mass $14\ \mathrm{g\,mol^{-1}}$)?
  • 13 How many electrons are present in $1.80\ \mathrm{g}$ of water ($M=18.0\ \mathrm{g\,mol^{-1}}$)?
  • 14 Applied. Heavy water, $\mathrm{D_{2}O}$ ($M=20.0\ \mathrm{g\,mol^{-1}}$, with $\mathrm{{}^{16}O}$), is used as a moderator in some nuclear reactors. How many neutrons are there in one molecule, and in $2.00\ \mathrm{g}$ of it?
1 $p=15$, $n=31-15=16$, $e=15$.
2 $p=12$, $n=12$, $e=12-2=10$.
3 $p=16$, $n=16$, $e=16+2=18$.
4 $p=35$, $n=45$, $e=35$.
5 $p=47$, $n=60$, $e=46$.
6 $Z=26$ (iron), $A=26+30=56$, charge $=26-24=+2$: $\mathrm{{}^{56}_{26}Fe^{2+}}$.
7 $18=Z-(-3)\Rightarrow Z=15$ (phosphorus); $A=15+16=31$: $\mathrm{{}^{31}_{15}P^{3-}}$, the phosphide ion.
8 Let protons $=p$; neutrons $=1.273p$. Then $p+1.273p=75\Rightarrow p=\dfrac{75}{2.273}=33.0$. $Z=33$: arsenic, $\mathrm{{}^{75}_{33}As}$ ($42$ neutrons; check $42/33=1.273$ ✓).
9 Same $A=3$, different $Z$: isobars.
10 $\mathrm{{}^{235}U}$, $\mathrm{{}^{238}U}$: isotopes (same $Z=92$). $\mathrm{{}^{40}Ar}$, $\mathrm{{}^{40}K}$: isobars (same $A=40$).
11 Chemical behaviour depends on the number and arrangement of electrons, which is the same for all isotopes (same $Z$). Properties that depend on mass — density, rate of diffusion, and slightly the rates of reactions — differ because the nuclei have different masses.
12 $n(\text{atoms})=\dfrac{7.0\times10^{-3}\ \mathrm{g}}{14\ \mathrm{g\,mol^{-1}}}=5.0\times10^{-4}\ \mathrm{mol}$; each atom has $14-6=8$ neutrons, so $5.0\times10^{-4}\times8\times6.022\times10^{23}=2.4\times10^{21}$ neutrons.
13 $\dfrac{1.80}{18.0}=0.100\ \mathrm{mol}$ of $\mathrm{H_{2}O}$; each molecule has $2(1)+8=10$ electrons: $0.100\times10\times6.022\times10^{23}=6.02\times10^{23}$ electrons.
14 Each D has $1$ neutron and $\mathrm{{}^{16}O}$ has $8$: $10$ neutrons per molecule. $\dfrac{2.00}{20.0}=0.100\ \mathrm{mol}$, so $0.100\times10\times6.022\times10^{23}=6.02\times10^{23}$ neutrons.
IV
Harder problems.
  • 1 An experimenter on an imaginary planet measures the charges $7.5$, $12.5$, $17.5$ and $10.0$ (in units of $10^{-19}\ \mathrm{C}$) on four drops. What is the largest elementary charge consistent with these data, and what would one further drop have to show to prove that the true unit is smaller?
  • 2 Add the masses of a proton and an electron (Definition 2.2) and compare the total with the measured mass of a hydrogen atom, $1.6735\times10^{-27}\ \mathrm{kg}$.
  • 3 In a Thomson experiment the beam goes straight through when $E=2.0\times10^{4}\ \mathrm{V\,m^{-1}}$ and $B=1.0\times10^{-3}\ \mathrm{T}$. Find the speed of the electrons and their kinetic energy in joules.
  • 4 Medicine. Cobalt-60, $\mathrm{{}^{60}_{27}Co}$, used to treat cancer, decays into nickel-60, $\mathrm{{}^{60}_{28}Ni}$. Give the numbers of protons and neutrons in each, and name the relationship between them.
  • 5 Natural chlorine is $75.77\%$ $\mathrm{{}^{35}Cl}$ ($34.969\ \mathrm{u}$) and $24.23\%$ $\mathrm{{}^{37}Cl}$ ($36.966\ \mathrm{u}$). Find the average atomic mass of chlorine (Section 1.2).
1 All four are multiples of $2.5$: $3,5,7,4$ times $2.5\times10^{-19}\ \mathrm{C}$. The largest consistent unit is $2.5\times10^{-19}\ \mathrm{C}$. A drop carrying, say, $1.25\times10^{-19}\ \mathrm{C}$ (or any value that is not a multiple of $2.5$) would prove the unit is smaller.
2 $1.67262\times10^{-27}+0.00091\times10^{-27}=1.67353\times10^{-27}\ \mathrm{kg}$, which agrees with $1.6735\times10^{-27}\ \mathrm{kg}$ — a hydrogen atom is just one proton and one electron.
3 $v=\dfrac{E}{B}=\dfrac{2.0\times10^{4}}{1.0\times10^{-3}}=2.0\times10^{7}\ \mathrm{m\,s^{-1}}$; $\tfrac12m_{e}v^{2}=\tfrac12(9.109\times10^{-31})(2.0\times10^{7})^{2}=1.8\times10^{-16}\ \mathrm{J}$.
4 Co-60: $27$ protons, $33$ neutrons. Ni-60: $28$ protons, $32$ neutrons. Same $A$, different $Z$: they are isobars.
5 $\bar{A}=0.7577\times34.969+0.2423\times36.966=26.496+8.957=35.45\ \mathrm{u}$.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET Which pair consists of isobars? (A) $\mathrm{{}^{12}_{6}C}$, $\mathrm{{}^{14}_{6}C}$ (B) $\mathrm{{}^{40}_{18}Ar}$, $\mathrm{{}^{40}_{20}Ca}$ (C) $\mathrm{{}^{14}_{6}C}$, $\mathrm{{}^{16}_{8}O}$ (D) $\mathrm{{}^{1}_{1}H}$, $\mathrm{{}^{2}_{1}H}$

Test Isobars: same $A$, different $Z$. (A) and (D) have the same $Z$ (isotopes); (C) differ in both (they share $8$ neutrons — isotones).
Answer (B): both have $A=40$; $Z=18$ and $20$.

E2. JEE Main Numerical answer. The number of neutrons in $3.4\ \mathrm{g}$ of ammonia containing only $\mathrm{{}^{14}N}$ and $\mathrm{{}^{1}H}$ ($M=17\ \mathrm{g\,mol^{-1}}$) is $\dfrac{x}{10}N_{A}$. Find $x$.

Per molecule $\mathrm{{}^{14}_{7}N}$ has $7$ neutrons; $\mathrm{{}^{1}H}$ has none. So one $\mathrm{NH_{3}}$ molecule has $7$ neutrons.
Moles $n=\dfrac{3.4}{17}=0.20\ \mathrm{mol}$, so neutrons $=0.20\times7\,N_{A}=1.4\,N_{A}=\dfrac{14}{10}N_{A}$.
Answer $x=$ 14.

E3. Board · 3 marks State three observations of Rutherford's α-particle scattering experiment and the conclusion drawn from each about the structure of the atom.

1 mark Most α-particles passed through the gold foil undeflected → most of the space in an atom is empty.
1 mark A small fraction were deflected through small angles → the positive charge of the atom is not spread out but concentrated in a very small volume that repels α-particles passing close to it.
1 mark Very few (about $1$ in $20\,000$) were deflected through nearly $180^{\circ}$ → the positive charge and almost all the mass are concentrated in a tiny, dense nucleus (radius about $10^{-15}\ \mathrm{m}$ against about $10^{-10}\ \mathrm{m}$ for the atom).

E4. JEE Main NEET Assertion–reason. Assertion (A): The charge-to-mass ratio of cathode rays is the same whatever gas is used in the discharge tube. Reason (R): Cathode rays consist of electrons, which are present in the atoms of all elements. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Check A True: $e/m$ of cathode rays is always $1.76\times10^{11}\ \mathrm{C\,kg^{-1}}$. (It is the anode rays whose $e/m$ depends on the gas.)
Check R True, and it is exactly why A holds: identical particles from every source must have identical $e/m$.
Answer (A).

E5. JEE Advanced One or more options correct. Which statements are correct? (A) The charge-to-mass ratio of anode rays depends on the gas in the discharge tube. (B) The charge-to-mass ratio of an α-particle is about half that of a proton. (C) A proton is about $1836$ times heavier than an electron. (D) A beam of neutrons is deflected by an electric field, but less than a beam of protons.

(A) True — anode rays are the positive ions of the gas, so their $e/m$ changes with the gas.
(B) True — $\dfrac{2e}{\approx4\ \mathrm{u}}$ versus $\dfrac{e}{\approx1\ \mathrm{u}}$; precisely $\dfrac{4.82\times10^{7}}{9.58\times10^{7}}=0.503$.
(C) True — $m_{p}/m_{e}=1836$.
(D) False — neutrons are uncharged and are not deflected at all by electric or magnetic fields.
Answer (A), (B), (C).

E6. Board · 2 marks A species carries a charge of $+2$, contains $25$ electrons and has $32$ neutrons. Identify the element, give its mass number and write the complete symbol.

Protons electrons $=Z-q\Rightarrow25=Z-2\Rightarrow Z=27$: cobalt.
Answer $A=27+32=59$; the species is $\mathrm{{}^{59}_{27}Co^{2+}}$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How does a mass spectrometer — Thomson's deflection experiment grown up — separate isotopes, and how do isotope ratios let scientists check where a food, a drug or an athlete's hormone came from?

Why it matters. Isotope-ratio mass spectrometry is used by food-safety agencies in Europe and Asia to detect honey diluted with cheap cane syrup, by anti-doping laboratories to tell an athlete's own testosterone from a synthetic dose, and by geologists and climate scientists to date rocks and read ancient temperatures. All of it rests on the idea of this section: ions of different mass are deflected by different amounts.

What to do.

  1. In a magnetic-sector instrument, an ion of mass $m$, charge $q$ and speed $v$ moves in a magnetic field $B$ on a circle of radius $r=\dfrac{mv}{qB}$ (a result from Class 12 Physics, used here as a tool). Calculate $r$ for $\mathrm{{}^{35}Cl^{+}}$ ($34.969\ \mathrm{u}$) and $\mathrm{{}^{37}Cl^{+}}$ ($36.966\ \mathrm{u}$), each moving at $1.0\times10^{5}\ \mathrm{m\,s^{-1}}$ in a field of $0.50\ \mathrm{T}$ ($1\ \mathrm{u}=1.6605\times10^{-27}\ \mathrm{kg}$).
  2. After a half-circle the ions land at a distance $2r$ from the entrance slit. How far apart are the two landing points?
  3. The detector currents for the two isotopes are in the ratio $3.127:1$. Convert this into percentage abundances and use them to find the average atomic mass of chlorine.
  4. Find out (from a reliable source such as a national food-safety agency or the World Anti-Doping Agency) what isotope ratio is measured to detect added cane sugar in honey or synthetic testosterone in urine, and why plants or drugs from different sources have slightly different ratios.

Tools: Calculator or spreadsheet; Interactive 1 to recall how the deflection depends on charge and mass.

Hand in: A one-page note with your calculation, a labelled sketch of the spectrometer, and a paragraph explaining one real use of isotope ratios.

Going further. Why would a mass spectrometer struggle to separate $\mathrm{{}^{14}N_{2}^{+}}$ from $\mathrm{{}^{12}C^{16}O^{+}}$ (both of nominal mass $28$)? Look up how "high-resolution" instruments use the tiny differences in exact mass to tell them apart.

Instructor note: $r(\mathrm{{}^{35}Cl^{+}})=\dfrac{5.807\times10^{-26}\times1.0\times10^{5}}{1.602\times10^{-19}\times0.50}=7.25\ \mathrm{cm}$; $r(\mathrm{{}^{37}Cl^{+}})=7.66\ \mathrm{cm}$; landing points $2\Delta r\approx0.83\ \mathrm{cm}$ apart. Abundances $3.127/4.127=75.77\%$ and $24.23\%$, average atomic mass $35.45\ \mathrm{u}$. Honey testing uses $\mathrm{{}^{13}C/{}^{12}C}$ ($\delta^{13}\mathrm{C}$): cane and maize (C4 plants) are richer in $\mathrm{{}^{13}C}$ than the flowers bees visit; doping tests compare $\mathrm{{}^{13}C/{}^{12}C}$ of testosterone metabolites with that of the athlete's own reference steroids. $\mathrm{N_{2}}=28.006\ \mathrm{u}$ and $\mathrm{CO}=27.995\ \mathrm{u}$.

🧪 Self-check quiz

0/5 correct

Q1. Why is the charge-to-mass ratio of cathode rays the same for every gas used in the tube?

Q2. In Rutherford's experiment most α-particles passed straight through the gold foil. This shows that

Q3. How many neutrons are there in $\mathrm{{}^{56}_{26}Fe^{3+}}$?

Q4. Carbon-14 and nitrogen-14 are

Q5. Millikan's oil-drop experiment measured