Chapter 13 · Nuclei

13.1 Atomic Masses, Composition and Size of the Nucleus

Nuclei are weighed in atomic mass units, built from $Z$ protons and $N$ neutrons, and sorted into isotopes, isobars and isotones. Their radius grows only as the cube root of the number of nucleons — which means every nucleus has the same, enormous density.

A drop of blood from an athlete, a fragment of ice drilled from two kilometres under the Antarctic surface, a grain of meteorite: laboratories put all of them through the same machine, a mass spectrometer. It turns atoms into ions and lets a magnetic field sort them by mass, so precisely that it can tell oxygen-16 from oxygen-18, or a natural hormone from a synthetic one whose atoms differ by a single neutron. The ratio of heavy to light oxygen in old ice tells climate scientists the temperature of the Earth hundreds of thousands of years ago; the same kind of measurement tells geologists the age of a rock.

In Chapter 12 — Atoms we met the nucleus as a tiny, heavy, positively charged centre of the atom. In this section we look at it more closely. First we need a unit small enough to weigh atoms in, and a way of measuring their masses. Then we ask what the nucleus is made of — protons and neutrons — and how to label and classify nuclei. Finally we ask how big a nucleus is, and find a simple law, $R=R_0A^{1/3}$, with a striking consequence: nuclear matter has the same density in every nucleus, about $10^{14}$ times the density of water.

Atomic Masses and the Atomic Mass Unit

An atom of carbon has a mass of about $2\times10^{-26}$ kg. Numbers like that are awkward, so atomic and nuclear masses are measured in a unit chosen to make them close to whole numbers.

📘 Definition 13.1 — Atomic mass unit
The atomic mass unit (symbol $\text{u}$) is one-twelfth of the mass of a neutral atom of carbon-12 (six protons, six neutrons and six electrons): $$1\ \text{u}=\frac{\text{mass of one }{}^{12}\text{C atom}}{12}=1.660539\times10^{-27}\ \text{kg}\approx1.66\times10^{-27}\ \text{kg}.$$ By definition the atomic mass of ${}^{12}\text{C}$ is exactly $12\ \text{u}$. Atomic masses in tables are the masses of neutral atoms, electrons included.

On this scale the hydrogen atom has mass $1.007825$ u, the proton $m_p=1.007276$ u, the neutron $m_n=1.008665$ u and the electron only $m_e=0.000549$ u. Notice that $1\ \text{u}\approx$ the mass of one nucleon, and that $N_A$ atoms of mass $1$ u have a mass of $1$ g: $(6.022\times10^{23})(1.66\times10^{-24}\ \text{g})\approx1$ g. That is why the atomic mass in u and the molar mass in g/mol are the same number.

Measuring atomic masses — the mass spectrometer. Atomic masses are measured by turning atoms into positive ions, accelerating them, and sending them into a uniform magnetic field at right angles to their velocity. As you saw in Section 4.1, an ion of mass $m$ and charge $q$ moving with speed $v$ then moves on a circle of radius $$r=\frac{mv}{qB}.$$ Ions with the same speed and charge but different masses follow circles of different radii and land at different places on a detector. The positions give the masses; the number of ions arriving at each place gives the relative amounts. With ions accelerated through a potential difference $V$, $\tfrac12mv^2=qV$, and the radius becomes $r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$ — heavier ions still swing wider. When J. J. Thomson and F. W. Aston sent neon ions through such an instrument (1912–1919) they found two landing places, at masses close to $20$ u and $22$ u: atoms of the same element can have different masses. Natural neon is a mixture.

The atomic mass printed in a periodic table is the average over the natural mixture, weighted by the fraction (the natural abundance) of each kind of atom: $$\bar m=\sum_i f_i\,m_i,\qquad \sum_i f_i=1.$$

✏️ Example 1 — The atomic mass of chlorine
Natural chlorine is $75.76\%$ chlorine-35 (atomic mass $34.968853$ u) and $24.24\%$ chlorine-37 ($36.965903$ u). Find the average atomic mass of chlorine. Why is it so far from a whole number?
Weighted average $$\bar m=(0.7576)(34.968853)+(0.2424)(36.965903)$$ $=26.4924+8.9605=35.4529$ u $\approx35.45$ u.
Why not whole? Each kind of chlorine atom has a mass close to a whole number ($\approx35$ and $\approx37$), but the element is a mixture of the two in a ratio of about $3:1$, and the average of $35$ and $37$ weighted $3:1$ is $35.5$.
Check The answer must lie between $34.97$ and $36.97$ and nearer the more abundant one. ✓ Interactive 1 gives the same value.
✏️ Example 2 — Separating the isotopes of neon
Singly charged ions of neon-20 (mass $19.99$ u) and neon-22 ($21.99$ u), all moving at $1.0\times10^5$ m/s, enter a uniform magnetic field of $0.20$ T at right angles to it and travel half a circle before striking a photographic plate. Find the radius of each path and the distance between the two lines on the plate. Take $1\ \text{u}=1.66\times10^{-27}$ kg.
Setup $q=e=1.6\times10^{-19}$ C, $r=\dfrac{mv}{qB}$, with $qB=(1.6\times10^{-19})(0.20)=3.2\times10^{-20}$ C T.
Neon-20 $m=19.99\times1.66\times10^{-27}=3.318\times10^{-26}$ kg: $$r_{20}=\frac{(3.318\times10^{-26})(1.0\times10^5)}{3.2\times10^{-20}}\approx0.104\ \text{m}.$$
Neon-22 $m=3.650\times10^{-26}$ kg: $r_{22}\approx0.114$ m.
Separation Each ion lands one diameter from the entry slit, so the lines are $2(r_{22}-r_{20})=2(0.1141-0.1037)\approx0.021$ m $=2.1$ cm apart — easily resolved.
Check $r\propto m$ at fixed $v$: $r_{22}/r_{20}=21.99/19.99=1.10$. ✓
Interactive 1 — A Mass Spectrometer

Singly charged ions of the chosen element are accelerated through $1000$ V and enter a uniform magnetic field (shaded) through the slit at the left. Each ion follows a semicircle of radius $r=\tfrac1B\sqrt{2mV/q}$ — computed from its mass — and lands on the detector. Ions are picked at random in proportion to the natural abundances, so the counts build up into a mass spectrum. Change $B$ and watch every circle shrink together. The readouts compare the average mass from the counts with the abundance-weighted value.

0.25 T
radii (cm)—
ions detected—
average from counts (u)—
$\sum f_im_i$ (u)—

The Composition of the Nucleus

Rutherford's experiments showed that the nucleus carries a positive charge $+Ze$, where $Z$ is the number of electrons in the neutral atom. The obvious guess — that the nucleus is just $Z$ protons — fails at once: helium has $Z=2$ but a mass of about $4$ u, and uranium has $Z=92$ but a mass of about $238$ u. Something neutral and roughly as heavy as a proton must make up the rest.

In 1932 James Chadwick found it. When beryllium was bombarded with alpha particles it gave off a very penetrating, uncharged radiation that could knock protons out of paraffin wax with energies far too large for gamma rays. Chadwick showed, using conservation of energy and momentum in these collisions, that the radiation consisted of neutral particles with almost the same mass as the proton: neutrons. A free neutron is not stable — on its own it changes into a proton, an electron and an antineutrino in about a quarter of an hour on average — but inside a stable nucleus it lasts indefinitely.

📘 Definition 13.2 — Nucleons, atomic number, neutron number and mass number
A nucleus is made of protons (charge $+e$) and neutrons (charge $0$), together called nucleons. For a nucleus of element X:
  • the atomic number $Z$ is the number of protons (it fixes the element and its chemistry);
  • the neutron number $N$ is the number of neutrons;
  • the mass number $A=Z+N$ is the total number of nucleons.
A nucleus with a definite $Z$ and $N$ is called a nuclide and is written ${}^{A}_{Z}\text{X}$, for example ${}^{56}_{26}\text{Fe}$ ($26$ protons, $30$ neutrons). Since the symbol already fixes $Z$, it is often written simply ${}^{56}\text{Fe}$ or iron-56. The mass of the nuclide in u is close to — but not exactly — $A$.
📘 Definition 13.3 — Isotopes, isobars and isotones
  • Isotopes are nuclides with the same $Z$ but different $N$ (and so different $A$): the same element, different masses — ${}^{1}_{1}\text{H}$, ${}^{2}_{1}\text{H}$ (deuterium) and ${}^{3}_{1}\text{H}$ (tritium); ${}^{35}_{17}\text{Cl}$ and ${}^{37}_{17}\text{Cl}$.
  • Isobars are nuclides with the same $A$ but different $Z$: different elements with nearly the same mass — ${}^{3}_{1}\text{H}$ and ${}^{3}_{2}\text{He}$; ${}^{40}_{18}\text{Ar}$ and ${}^{40}_{20}\text{Ca}$.
  • Isotones are nuclides with the same $N$ but different $Z$ — ${}^{198}_{80}\text{Hg}$ and ${}^{197}_{79}\text{Au}$ (both $N=118$); ${}^{14}_{6}\text{C}$ and ${}^{16}_{8}\text{O}$ (both $N=8$).
A memory aid: isotopes share the proton number, isotones share the neutron number, and isobars share A (from the Greek barys, heavy).

Isotopes have the same number of electrons and therefore the same chemistry; they can be separated only by physical methods that depend on mass, such as the mass spectrometer. Isobars are different elements and behave completely differently in chemical reactions, even though their masses are almost equal.

Why not electrons in the nucleus? Before the neutron was known, some physicists suggested that a nucleus of mass number $A$ held $A$ protons and $A-Z$ electrons. Two facts rule this out. Confining an electron to a region $10^{-14}$ m across would, by the uncertainty principle, give it a kinetic energy of many MeV, far more than any nucleus could hold; and the spins of nuclei such as nitrogen-14 come out wrong. Electrons that emerge from nuclei in some decays are created at the moment of emission, not stored inside.

✏️ Example 3 — Counting particles
Find the number of protons, neutrons and electrons in (a) a neutral atom of ${}^{56}_{26}\text{Fe}$, (b) a neutral atom of ${}^{235}_{92}\text{U}$, (c) the ion ${}^{37}_{17}\text{Cl}^-$ and (d) the ion ${}^{24}_{12}\text{Mg}^{2+}$.
Rule Protons $=Z$, neutrons $N=A-Z$; electrons $=Z$ in a neutral atom, minus the positive charge number (or plus the negative one) for an ion. Ionisation never changes the nucleus.
(a), (b) Fe: $26$ p, $30$ n, $26$ e. U: $92$ p, $143$ n, $92$ e.
(c) $17$ p, $20$ n, $17+1=18$ e.
(d) $12$ p, $12$ n, $12-2=10$ e.
✏️ Example 4 — Isotope, isobar or isotone?
Classify each pair: (a) ${}^{12}_{6}\text{C}$ and ${}^{14}_{6}\text{C}$; (b) ${}^{40}_{19}\text{K}$ and ${}^{40}_{20}\text{Ca}$; (c) ${}^{2}_{1}\text{H}$ and ${}^{3}_{2}\text{He}$; (d) ${}^{13}_{6}\text{C}$ and ${}^{14}_{7}\text{N}$; (e) ${}^{3}_{1}\text{H}$ and ${}^{4}_{2}\text{He}$.
Method Write $(Z,\,N,\,A)$ for each and see which number is shared.
(a)–(c) (a) $(6,6,12)$ and $(6,8,14)$: same $Z$ — isotopes. (b) $(19,21,40)$ and $(20,20,40)$: same $A$ — isobars. (c) $(1,1,2)$ and $(2,1,3)$: same $N$ — isotones.
(d) $(6,7,13)$ and $(7,7,14)$: same $N=7$ — isotones.
(e) $(1,2,3)$ and $(2,2,4)$: same $N=2$ — isotones. (Not isobars: $A$ differs.)

The Size of the Nucleus

The alpha-scattering experiments of Chapter 12 — Atoms only gave an upper limit on the size of a nucleus: an alpha particle of a few MeV is turned back by the Coulomb repulsion of a gold nucleus while still a few times $10^{-14}$ m from its centre, so the nucleus must be smaller than that. To probe further you need projectiles that get closer. In the 1950s beams of fast electrons — which feel only the electric force and are not repelled by the nucleus — were scattered from many different nuclei, and the pattern of scattering revealed how the charge is spread out. Nuclear sizes are measured in femtometres (also called fermis): $1\ \text{fm}=10^{-15}$ m.

📐 Result 13.1 — The radius of a nucleus

To a good approximation a nucleus behaves like a sphere whose radius depends only on its mass number $A$: $$R=R_0A^{1/3},\qquad R_0\approx1.2\ \text{fm}=1.2\times10^{-15}\ \text{m}.$$ So the volume $\tfrac43\pi R^3=\tfrac43\pi R_0^3A$ is proportional to $A$: every nucleon added brings the same extra volume, as if the nucleus were a tightly packed cluster of hard balls. ($R_0$ is an experimental constant; values from $1.1$ to $1.3$ fm appear, depending on how the "edge" of the nucleus is defined. We use $1.2$ fm.)

Because $A^{1/3}$ grows so slowly, nuclear radii span only a small range: from $1.2$ fm for a single proton to about $7.4$ fm for uranium-238, even though the number of nucleons increases more than two-hundredfold. For two nuclei, $\dfrac{R_1}{R_2}=\left(\dfrac{A_1}{A_2}\right)^{1/3}$.

📐 Result 13.2 — Nuclear density is the same for all nuclei

A nucleus of mass number $A$ has mass $\approx A\,m$, where $m\approx1.66\times10^{-27}$ kg is the mass of one nucleon. Its density is $$\rho=\frac{Am}{\tfrac43\pi R_0^3A}=\frac{3m}{4\pi R_0^3}\approx2.3\times10^{17}\ \text{kg/m}^3,$$ independent of $A$: light and heavy nuclei have the same density. Nuclear matter is about $2\times10^{14}$ times denser than water. Neutron stars — collapsed cores of massive stars — are made of matter of roughly this density.

✏️ Example 5 — Radii of nuclei
Using $R_0=1.2$ fm, find the radii of (a) ${}^{4}\text{He}$, (b) ${}^{56}\text{Fe}$ and (c) ${}^{238}\text{U}$. (d) Find the ratio of the radii of ${}^{27}_{13}\text{Al}$ and ${}^{125}_{52}\text{Te}$ without a calculator.
(a) $4^{1/3}=1.587$: $R=1.2\times1.587\approx1.9$ fm.
(b) $56^{1/3}=3.826$: $R\approx4.6$ fm.
(c) $238^{1/3}=6.197$: $R\approx7.4$ fm. Uranium has $60$ times the nucleons of helium but only $3.9$ times the radius.
(d) $\dfrac{R_\text{Al}}{R_\text{Te}}=\left(\dfrac{27}{125}\right)^{1/3}=\dfrac35$. (Only $A$ matters — $Z$ plays no part.)
Check Interactive 2 shows $R=4.59$ fm for $Z=26$, $N=30$. ✓
✏️ Example 6 — How dense, and how empty?
(a) Calculate the density of nuclear matter and the mass of one cubic centimetre of it. (b) A gold atom (${}^{197}\text{Au}$) has a radius of about $0.144$ nm. What fraction of the atom's volume does its nucleus occupy? Take $R_0=1.2$ fm and $m=1.66\times10^{-27}$ kg per nucleon.
(a) Density $R_0^3=(1.2\times10^{-15})^3=1.728\times10^{-45}\ \text{m}^3$: $$\rho=\frac{3(1.66\times10^{-27})}{4\pi(1.728\times10^{-45})}\approx2.3\times10^{17}\ \text{kg/m}^3.$$
(a) One cm³ $1\ \text{cm}^3=10^{-6}\ \text{m}^3$, so $M=\rho V\approx2.3\times10^{11}$ kg — about $230$ million tonnes in a sugar cube.
(b) Gold nucleus $R=1.2\times197^{1/3}=1.2\times5.819\approx6.98$ fm. The volume fraction is $$\left(\frac{6.98\times10^{-15}}{1.44\times10^{-10}}\right)^3=(4.85\times10^{-5})^3\approx1.1\times10^{-13}.$$
Meaning The nucleus fills about one ten-millionth of a millionth of the atom yet holds more than $99.9\%$ of its mass. Ordinary matter is almost entirely empty space; nuclear matter is not.
Interactive 2 — Build a Nucleus

Choose the number of protons $Z$ and neutrons $N$. Left: the nucleus drawn to scale with $R=R_0A^{1/3}$ (protons red, neutrons blue; the faint ring is the size of uranium-238). Right: a patch of the chart of nuclides around your nucleus — the row holds its isotopes (same $Z$), the column its isotones (same $N$) and the diagonal its isobars (same $A$). The shaded band is an approximate guide to where stable nuclei lie; a nucleus far from it does not exist or falls apart at once. Press Play for a tour of real nuclei.

2630
nuclide—
radius $R$—
density—
$N/Z$—
⚠️ Common mistakes with atomic masses and nuclear size

1. Mixing up $A$ and $Z$. In ${}^{A}_{Z}\text{X}$ the top number is the mass number (nucleons), the bottom the atomic number (protons). Neutrons are $A-Z$, not $A$.

2. Thinking $A$ is the mass. $A$ is a whole number that counts nucleons; the atomic mass in u is close to $A$ but not equal to it ($^{56}\text{Fe}$: $55.934936$ u). The difference matters in Section 13.2.

3. $R\propto A$. It is the volume that is proportional to $A$; the radius goes as $A^{1/3}$. Doubling $A$ increases $R$ by only $2^{1/3}\approx1.26$.

4. Expecting heavier nuclei to be denser. Nuclear density is the same for all nuclei. Questions asking for the ratio of densities of two nuclei have the answer $1:1$.

5. Confusing isobars and isotones. Isobars: same A. Isotones: same N.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — a mass spectrometer with a velocity selector

Not required by the CBSE syllabus. The force on a moving charge in magnetic (and, for JEE Main and NEET, crossed electric) fields is listed by JEE Main, NEET and JEE Advanced; none names the mass spectrometer, which applies it.

For the ions to separate cleanly by mass they should all enter the magnetic field with the same speed. A velocity selector (crossed fields $E$ and $B_1$, Chapter 4 extension) passes undeflected only ions with $qE=qvB_1$, that is $v=E/B_1$. In a second field $B_2$ they then move on circles of radius $$r=\frac{mv}{qB_2}=\frac{mE}{qB_1B_2},\qquad\text{so}\qquad m=\frac{qB_1B_2\,r}{E}.$$ The radius is now strictly proportional to the mass, and the lines on the detector form an evenly spaced mass scale. For example, with $E=2.0\times10^4$ V/m and $B_1=B_2=0.20$ T, singly charged ions have $v=1.0\times10^5$ m/s — the speed used in Example 2. Velocity selectors are not in the current board syllabus, but JEE Main problems on charges in combined uniform $\mathbf E$ and $\mathbf B$ fields use exactly this idea.

Exercises

I
Atomic masses and the mass spectrometer. Take $1\ \text{u}=1.66\times10^{-27}$ kg and $e=1.6\times10^{-19}$ C.
  • 1 Express the mass of one carbon-12 atom in kilograms.
  • 2 Natural boron is $19.9\%$ boron-10 ($10.012937$ u) and $80.1\%$ boron-11 ($11.009305$ u). Find its average atomic mass.
  • 3 The average atomic mass of copper is $63.546$ u. Its isotopes are copper-63 ($62.929598$ u) and copper-65 ($64.927790$ u). Find the percentage abundance of copper-63.
  • 4 How many atoms are there in $1.0$ mg of carbon-12?
  • 5 What magnetic field makes singly charged neon-20 ions ($19.99$ u) moving at $1.0\times10^5$ m/s travel on a circle of radius $25$ cm?
  • 6 Singly charged ions of chlorine-35 ($34.97$ u) and chlorine-37 ($36.97$ u) are accelerated from rest through $2000$ V and enter a field of $0.30$ T at right angles. Find the radius of each path.
  • 7 Explain why isotopes can be separated in a mass spectrometer but not by chemical reactions.
1 $12\times1.66\times10^{-27}\approx1.99\times10^{-26}$ kg.
2 $(0.199)(10.012937)+(0.801)(11.009305)$ $=1.9926+8.8184\approx10.81$ u.
3 With fraction $f$ of copper-63: $62.929598f+64.927790(1-f)=63.546$, so $f=\dfrac{64.927790-63.546}{1.998192}\approx0.6915$: about $69.2\%$.
4 $\dfrac{1.0\times10^{-6}\ \text{kg}}{1.99\times10^{-26}\ \text{kg}}\approx5.0\times10^{19}$ atoms.
5 $B=\dfrac{mv}{qr}$ $=\dfrac{(19.99)(1.66\times10^{-27})(1.0\times10^5)}{(1.6\times10^{-19})(0.25)}$ $\approx0.083$ T.
6 $r=\dfrac1B\sqrt{\dfrac{2mV}{q}}$. Chlorine-35: $m=5.805\times10^{-26}$ kg, $\sqrt{2mV/q}=0.03809$ T m, $r\approx0.127$ m. Chlorine-37: $r\approx0.131$ m (larger by $\sqrt{36.97/34.97}\approx1.028$).
7 Isotopes have the same $Z$, hence the same electron arrangement and the same chemistry. They differ only in mass, and the mass spectrometer's radius $r=mv/qB$ depends directly on mass.
II
Composition of nuclei.
  • 1 Give the numbers of protons and neutrons in ${}^{16}_{8}\text{O}$, ${}^{40}_{20}\text{Ca}$, ${}^{208}_{82}\text{Pb}$ and ${}^{3}_{1}\text{H}$.
  • 2 How many neutrons are there in $12$ g of carbon-14? ($N_A=6.022\times10^{23}\ \text{mol}^{-1}$; take the molar mass as $14$ g/mol.)
  • 3 Classify ${}^{14}_{6}\text{C}$ and ${}^{14}_{7}\text{N}$.
  • 4 Classify ${}^{39}_{19}\text{K}$ and ${}^{40}_{20}\text{Ca}$.
  • 5 Classify ${}^{235}_{92}\text{U}$ and ${}^{238}_{92}\text{U}$.
  • 6 Write the symbol of (a) the isotone of ${}^{16}_{8}\text{O}$ with $Z=7$ and (b) the isobar of ${}^{16}_{8}\text{O}$ with $Z=7$.
  • 7 A nucleus has $A=27$ and $14$ neutrons. Identify the element and write the nuclide symbol.
  • 8 Why do isotopes of an element have identical chemical properties, while isobars do not?
  • 9 Give two pieces of evidence that a nucleus cannot be made of protons alone.
1 O: $8$ p, $8$ n. Ca: $20$ p, $20$ n. Pb: $82$ p, $126$ n. H-3: $1$ p, $2$ n.
2 $\dfrac{12}{14}$ mol of atoms, $8$ neutrons each: $\dfrac{12}{14}\times8\times6.022\times10^{23}\approx4.1\times10^{24}$ neutrons.
3 $(6,8,14)$ and $(7,7,14)$: same $A$ — isobars.
4 $(19,20,39)$ and $(20,20,40)$: same $N=20$ — isotones.
5 Same $Z=92$ — isotopes.
6 (a) $N=8$, $Z=7$: ${}^{15}_{7}\text{N}$. (b) $A=16$, $Z=7$: ${}^{16}_{7}\text{N}$.
7 $Z=27-14=13$: aluminium, ${}^{27}_{13}\text{Al}$.
8 Chemistry is decided by the electrons, whose number equals $Z$. Isotopes share $Z$; isobars have different $Z$, so they are different elements.
9 (i) The mass in u is about twice $Z$ or more (helium: charge $2e$, mass $4$ u). (ii) Neutrons are knocked out of nuclei directly (Chadwick's beryllium experiment). Either the charge-to-mass mismatch or the direct observation is enough.
III
Size and density of nuclei. Use $R_0=1.2$ fm and $1.66\times10^{-27}$ kg per nucleon.
  • 1 Find the radius of ${}^{16}\text{O}$.
  • 2 Find the radius of ${}^{107}\text{Ag}$.
  • 3 A nucleus has radius $3.6$ fm. Find its mass number.
  • 4 Find the ratio of the radii of nuclei with $A=64$ and $A=216$.
  • 5 Find the ratio of the nuclear densities of ${}^{208}\text{Pb}$ and ${}^{4}\text{He}$.
  • 6 Two nuclei have $A=8$ and $A=64$. Find the ratios of their volumes and of their surface areas.
  • 7 Find the mass number of the nucleus whose radius is twice that of ${}^{4}\text{He}$.
  • 8 A neutron star is a sphere of radius $10$ km with (roughly) nuclear density. Estimate its mass and compare it with the mass of the Sun, $2.0\times10^{30}$ kg.
1 $1.2\times16^{1/3}=1.2\times2.520\approx3.0$ fm.
2 $1.2\times107^{1/3}=1.2\times4.747\approx5.7$ fm.
3 $A=(R/R_0)^3=3^3=27$.
4 $(64/216)^{1/3}=4/6=2:3$.
5 $1:1$ — nuclear density does not depend on $A$.
6 Volume $\propto A$: $1:8$. $R\propto A^{1/3}$ gives radii $1:2$, so surface areas $1:4$.
7 $R$ doubles when $A$ is multiplied by $2^3=8$: $A=32$ (for example ${}^{32}\text{S}$).
8 $M=\tfrac43\pi(1.0\times10^4)^3(2.3\times10^{17})\approx9.6\times10^{29}$ kg — about half the mass of the Sun, squeezed into a city-sized ball.
IV
Applications and longer problems.
  • 1 Modelling. Natural lithium is $7.6\%$ lithium-6 ($6.015122$ u) and $92.4\%$ lithium-7 ($7.016003$ u). (a) Find its average atomic mass. (b) A sample of lithium bought for a battery laboratory has average atomic mass $6.970$ u. What percentage of it is lithium-6? Suggest why it might differ from natural lithium.
  • 2 Show that the number of nucleons per unit volume in a nucleus is $n=\dfrac{3}{4\pi R_0^3}$, and evaluate it in nucleons per $\text{fm}^3$.
  • 3 Using your answer to part 2, estimate the average distance between neighbouring nucleons, taking each nucleon to occupy a cube of volume $1/n$.
  • 4 In a mass spectrometer, singly charged ions of chlorine-35 and chlorine-37 are accelerated through the same potential difference and enter the same magnetic field. Find the ratio of the radii of their paths.
  • 5 If the whole Earth ($6.0\times10^{24}$ kg) were squeezed to nuclear density, what would its radius be?
  • 6 The radius of a nucleus is $R$. What is the radius of a nucleus with three times as many protons and three times as many neutrons, and how do their densities compare?
1 (a) $(0.076)(6.015122)+(0.924)(7.016003)\approx6.94$ u. (b) $f=\dfrac{7.016003-6.970}{7.016003-6.015122}\approx0.046$: about $4.6\%$ lithium-6. Lithium-6 is extracted for nuclear uses, so lithium left over from that process ("depleted" lithium) is richer in lithium-7.
2 $A$ nucleons fill $\tfrac43\pi R_0^3A$, so $n=\dfrac{A}{\tfrac43\pi R_0^3A}=\dfrac{3}{4\pi R_0^3}=\dfrac{3}{4\pi(1.728)}\approx0.14$ nucleons per $\text{fm}^3$.
3 $d=(1/n)^{1/3}=(7.24\ \text{fm}^3)^{1/3}\approx1.9$ fm — about the range of the nuclear force (Section 13.2).
4 $r=\dfrac1B\sqrt{\dfrac{2mV}{q}}\propto\sqrt m$: $\dfrac{r_{37}}{r_{35}}=\sqrt{\dfrac{36.97}{34.97}}\approx1.028$.
5 $R=\left(\dfrac{3M}{4\pi\rho}\right)^{1/3}$ $=\left(\dfrac{3(6.0\times10^{24})}{4\pi(2.3\times10^{17})}\right)^{1/3}$ $\approx1.8\times10^2$ m — a ball less than $400$ m across.
6 $A$ triples, so $R'=3^{1/3}R\approx1.44R$; the densities are equal.
🎯 Exam corner — atomic masses, composition and size of nuclei BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Define isotopes, isobars and isotones, giving one example of each. Which of the three have identical chemical properties?

Definitions Isotopes: same $Z$, different $A$ (${}^{12}_{6}\text{C}$, ${}^{14}_{6}\text{C}$). Isobars: same $A$, different $Z$ (${}^{40}_{18}\text{Ar}$, ${}^{40}_{20}\text{Ca}$). Isotones: same $N$, different $Z$ (${}^{14}_{6}\text{C}$, ${}^{16}_{8}\text{O}$, both $N=8$).
Answer Isotopes — same $Z$, so the same number and arrangement of electrons. Exam tip: always give $Z$ and $A$ on the symbols of your examples; a bare "C-12 and C-14" can lose the mark.

E2. Board · 3 marks Write the relation between the radius $R$ of a nucleus and its mass number $A$. Hence show that nuclear density is independent of $A$. Find the ratio of the radii of ${}^{64}_{29}\text{Cu}$ and ${}^{27}_{13}\text{Al}$.

Relation $R=R_0A^{1/3}$, $R_0\approx1.2$ fm.
Density Mass $\approx Am$ ($m$ = mass of a nucleon), volume $\tfrac43\pi R_0^3A$: $$\rho=\frac{Am}{\tfrac43\pi R_0^3A}=\frac{3m}{4\pi R_0^3},$$ which contains no $A$. Numerically $\rho\approx2.3\times10^{17}$ kg/m³.
Ratio $\dfrac{R_\text{Cu}}{R_\text{Al}}=\left(\dfrac{64}{27}\right)^{1/3}=\dfrac43$, i.e. $4:3$. Exam tip: choose the cube-root ratio first — the numbers are usually perfect cubes.

E3. JEE Main The radius of ${}^{27}_{13}\text{Al}$ is $3.6$ fm. The radius of ${}^{125}_{52}\text{Te}$ is approximately (A) $4.8$ fm (B) $6.0$ fm (C) $7.2$ fm (D) $3.0$ fm.

Concept $R\propto A^{1/3}$: $R_\text{Te}=3.6\times\left(\dfrac{125}{27}\right)^{1/3}=3.6\times\dfrac53$.
Answer (B) $6.0$ fm. Exam tip: (A) comes from using $Z$ instead of $A$ — the radius depends only on the mass number.

E4. NEET The ratio of the nuclear densities of ${}^{4}_{2}\text{He}$ and ${}^{238}_{92}\text{U}$ is (A) $1:1$ (B) $4:238$ (C) $(4/238)^{1/3}$ (D) $2:92$.

Concept $\rho=\dfrac{3m}{4\pi R_0^3}$ is the same for all nuclei.
Answer (A) $1:1$. Exam tip: mass and volume both grow in proportion to $A$; their ratio cannot change.

E5. JEE Advanced (One or more correct.) Which of the following statements are correct? (A) The volume of a nucleus is proportional to its mass number. (B) ${}^{3}_{1}\text{H}$ and ${}^{3}_{2}\text{He}$ are isotones. (C) A nucleus contains roughly $0.14$ nucleons per cubic femtometre, whatever its size. (D) The atomic mass of every nuclide in u is exactly equal to its mass number.

Options (A) $V=\tfrac43\pi R_0^3A$: true. (B) $N=2$ and $N=1$; they share $A=3$ — isobars, not isotones: false. (C) $n=\dfrac{3}{4\pi R_0^3}=\dfrac{3}{4\pi(1.2)^3}\approx0.14\ \text{fm}^{-3}$: true. (D) Only ${}^{12}\text{C}$ is exactly $12$ u, by definition; e.g. ${}^{56}\text{Fe}$ is $55.934936$ u: false.
Answer (A), (C). Exam tip: in multi-correct questions test each statement on its own — the "almost true" ones such as (D) are the traps.

E6. Board · Case study A geology laboratory uses a mass spectrometer to measure boron in a mineral. Singly charged ions of boron-10 ($10.012937$ u) and boron-11 ($11.009305$ u), all with the same speed, enter a uniform magnetic field at right angles to it. The detector records $19.9\%$ of the ions as boron-10 and $80.1\%$ as boron-11. (i) (2 marks) Find the average atomic mass of the boron. (ii) (1 mark) Which ions travel on the larger circle, and why? (iii) (1 mark) Find the ratio of the radii of the two paths.

(i) $\bar m=(0.199)(10.012937)$ $+\,(0.801)(11.009305)$ $\approx10.81$ u.
(ii) Boron-11: $r=\dfrac{mv}{qB}$, and with equal $v$, $q$ and $B$ the heavier ion has the larger radius.
(iii) $\dfrac{r_{11}}{r_{10}}=\dfrac{11.009305}{10.012937}\approx1.10$. Exam tip: at equal speed $r\propto m$; at equal accelerating voltage $r\propto\sqrt m$ — read which one the question fixes.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How well does the simple law $R=R_0A^{1/3}$ describe real nuclei, and is $R_0$ really the same for light and heavy nuclei?

Why it matters. Nuclear radii are measured by firing high-energy electrons at targets (at laboratories in the USA, Germany, Japan and elsewhere) and, more recently, by laser spectroscopy of rare isotopes at facilities such as CERN's ISOLDE. These measurements test models of the nuclear force and are needed to interpret neutron-star observations. Experiments report the root-mean-square charge radius $r_\text{rms}$; a uniform sphere of radius $R$ has $r_\text{rms}=\sqrt{3/5}\,R$, so the equivalent sharp-edged radius is $R=\sqrt{5/3}\,r_\text{rms}\approx1.29\,r_\text{rms}$.

What to do. Use these measured rms charge radii (fm): ${}^{12}\text{C}$ $2.47$, ${}^{16}\text{O}$ $2.70$, ${}^{40}\text{Ca}$ $3.48$, ${}^{56}\text{Fe}$ $3.74$, ${}^{120}\text{Sn}$ $4.65$, ${}^{208}\text{Pb}$ $5.50$.

  1. Convert each to the equivalent radius $R=1.29\,r_\text{rms}$.
  2. Tabulate $A^{1/3}$ and $R$, and plot $R$ against $A^{1/3}$. Is the graph a straight line through the origin?
  3. Calculate $R/A^{1/3}$ for each nucleus. How does it change from light to heavy nuclei?
  4. Light nuclei have a relatively thick "skin" compared with their size. Use this idea to explain the trend you found.

Tools: Calculator or spreadsheet; graph paper or plotting software.

Hand in: The table, the graph, your values of $R/A^{1/3}$ and a short paragraph on whether "$R_0=1.2$ fm" is a good rule.

Going further. Fit a straight line $R=a+bA^{1/3}$ to the data. Does allowing an intercept $a$ make the fit better? What would a non-zero $a$ mean physically?

Instructor note: $R\approx3.19,\ 3.48,\ 4.49,\ 4.82,\ 6.00,\ 7.10$ fm; $A^{1/3}=2.29,\ 2.52,\ 3.42,\ 3.83,\ 4.93,\ 5.92$; $R/A^{1/3}\approx1.39,\ 1.38,\ 1.31,\ 1.26,\ 1.22,\ 1.20$ fm. The graph is close to a straight line but does not quite pass through the origin; $R_0$ falls towards $1.2$ fm for heavy nuclei because the surface region (where density falls off over about $2$ fm) is a larger fraction of a small nucleus. A fit $R=a+bA^{1/3}$ (least squares) gives $a\approx0.8$ fm and $b\approx1.06$ fm — a positive intercept, reflecting the surface thickness.

🧪 Self-check quiz

0/5 correct

Q1. One atomic mass unit is defined as

Q2. A nucleus with $A=64$ has a radius that is how many times the radius of a nucleus with $A=8$?

Q3. The nuclides ${}^{14}_{6}\text{C}$ and ${}^{16}_{8}\text{O}$ are

Q4. How does the density of nuclear matter depend on the mass number $A$?

Q5. The number of neutrons in a nucleus of ${}^{235}_{92}\text{U}$ is