Composition, isotopes and size of the nucleus
Nuclear medicine, nuclear power stations and the dating of ancient objects all start with the structure of the nucleus. Link to earlier classes : protons, neutrons and isotopes from Class 9 and Chapter 12.
The nucleus is made of protons and neutrons (together called nucleons). The atomic number Z is the number of protons, the mass number A is the number of nucleons, and the number of neutrons is N = A − Z. Isotopes have the same Z but different A. The atomic mass unit is 1 u = 1.66 × 10⁻²⁷ kg, equal to 931.5 MeV ÷ c².
Rule 1: Size of the nucleus — R = R₀ A^(1/3) with R₀ = 1.2 fm = 1.2 × 10⁻¹⁵ m. The volume is therefore proportional to A.
- Easy : Aluminium, A = 27 ⇒ R = 1.2 × 3 = 3.6 fm.
- Medium : A nucleus with A = 216 has a radius twice that with A = 27, since (216 ÷ 27)^(1/3) = 2.
- Exam : For a nucleus with 238 nucleons, N = 238 − 92 = 146 neutrons (uranium, Z = 92).
Rule 2: Nuclear density — Mass is proportional to A and volume is proportional to A, so the density is the same for all nuclei: ρ = 1.66 × 10⁻²⁷ A ÷ (4/3 π R₀³ A) ≈ 2.3 × 10¹⁷ kg m⁻³.
- Easy : This is about 10¹⁴ times the density of water.
- Medium : One cubic centimetre of such matter would have a mass of about 2.3 × 10¹¹ kg, that is 230 million tonnes.
- Exam : The constant density shows that nucleons are packed in much the same way in every nucleus.
A = Z + N; R = R₀A^(1/3); ρ ≈ 2.3 × 10¹⁷ kg m⁻³ (same for all nuclei).
Saying that isotopes have different numbers of protons. They differ in neutrons.
Memory clue : "Iso-tope: same Z, same place in the periodic table."
Spot the difference
| Term | Same | Different |
|---|---|---|
| Isotopes | Z | A, N |
| Isobars | A | Z |
| Isotones | N | Z, A |
Practice questions
- 1Define isotopes.
- 2Write the relation between A, Z and N.
- 3Write 1 u in kilogram.
- 4Find the radius of an aluminium nucleus (A = 27).
- 5Find the ratio of the radii of nuclei with A = 216 and A = 27.
- 6Find the nuclear density.
- 7How many neutrons are there in uranium-238 (Z = 92)?
- 8Why is the nuclear density the same for all nuclei?
Show answers (try them yourself first)
- 1Nuclei with the same atomic number and different mass numbers.
- 2A = Z + N.
- 31.66 × 10⁻²⁷ kg.
- 4R = 1.2 × 27^(1/3) = 3.6 fm.
- 52 : 1.
- 6ρ ≈ 2.3 × 10¹⁷ kg m⁻³.
- 7N = 238 − 92 = 146.
- 8Both the mass and the volume are proportional to A, so A cancels.
Mass–energy and nuclear binding energy
A small amount of mass lost in a nuclear reaction gives an enormous amount of energy, the principle of a nuclear power station. Link to earlier classes : conservation of energy and Einstein's relation.
Einstein showed that mass and energy are equivalent: E = m c². The mass of a nucleus is always less than the sum of the masses of its separate nucleons. This difference is the mass defect Δm, and the energy equivalent of Δm is the binding energy, the energy needed to separate the nucleus completely into nucleons.
Rule 1: Mass defect and binding energy — Δm = Z m_p + (A − Z) m_n − m(nucleus) and BE = Δm × 931.5 MeV (with Δm in u). The binding energy per nucleon is BE ÷ A.
- Easy : 1 u of mass is equivalent to 931.5 MeV.
- Medium : For helium-4, Δm = 2 × 1.007825 + 2 × 1.008665 − 4.002603 = 0.030377 u (using atomic masses of hydrogen), so BE = 28.3 MeV.
- Exam : BE per nucleon of helium-4 = 28.3 ÷ 4 = 7.07 MeV.
Rule 2: Energy from mass — The energy equivalent of a mass m is m c².
- Easy : Converting 1 g of mass completely gives 10⁻³ × (3 × 10⁸)² = 9 × 10¹³ J.
- Medium : In chemical reactions the mass change is too small to detect, but in nuclear reactions it is significant.
- Exam : The larger the binding energy per nucleon, the more stable the nucleus.
In the figure : Binding energy per nucleon against mass number: it rises quickly, peaks near iron (about 8.8 MeV) and falls slowly for heavy nuclei.
E = mc²; BE = Δm × 931.5 MeV; stable nuclei have a high BE per nucleon.
Taking the nuclear mass to be the sum of the nucleon masses. It is smaller.
Memory clue : "Mass defect in u times 931.5 gives the binding energy in MeV."
Spot the difference
| Nucleus | BE per nucleon |
|---|---|
| ⁴He | about 7.1 MeV |
| ⁵⁶Fe | about 8.8 MeV (maximum) |
| ²³⁸U | about 7.6 MeV |
Practice questions
- 1State Einstein's mass–energy relation.
- 2Define the mass defect.
- 3Define the binding energy of a nucleus.
- 4Find the mass defect of helium-4 for the masses 1.007825 u (H), 1.008665 u (n) and 4.002603 u (He).
- 5Find the binding energy of helium-4.
- 6Find its binding energy per nucleon.
- 7How many MeV is 1 u equivalent to?
- 8How much energy is released when 1 g of matter is completely converted into energy?
Show answers (try them yourself first)
- 1E = mc².
- 2The difference between the sum of the masses of the nucleons and the mass of the nucleus.
- 3The energy needed to split the nucleus into free nucleons.
- 4Δm = 0.030377 u.
- 5BE = 0.030377 × 931.5 ≈ 28.3 MeV.
- 6≈ 7.07 MeV per nucleon.
- 7931.5 MeV.
- 89 × 10¹³ J.
Nuclear force and nuclear stability
A magnet holds a nail firmly only when it is very close, and the grip disappears with a small gap; the nuclear force is like this, but far stronger. Link to earlier classes : Coulomb's law from Chapter 1.
The positively charged protons in a nucleus repel one another strongly, yet the nucleus stays together because of the nuclear force, a strong attraction between nucleons. It is the strongest of the basic forces but acts only over a very short distance.
Rule 1: Properties of the nuclear force — It is (i) much stronger than the electrostatic force, (ii) short-ranged (a few fm; it drops to nearly zero beyond about 2.5 fm), (iii) attractive at about 0.8 fm to 2 fm and strongly repulsive at smaller distances, (iv) independent of charge (p–p, n–n and p–n forces are about equal), and (v) saturated, since each nucleon attracts only its near neighbours.
- Easy : Two protons 1 fm apart repel with a Coulomb force of 9 × 10⁹ × (1.6 × 10⁻¹⁹)² ÷ (10⁻¹⁵)² = 230 N.
- Medium : The gravitational force between them is only about 2 × 10⁻³⁴ N, completely negligible.
- Exam : The nuclear force must exceed 230 N at that distance, otherwise the nucleus would fly apart.
Rule 2: Stability — Light nuclei are stable with N ≈ Z. Heavy nuclei need N > Z because the Coulomb repulsion has a long range and builds up as Z grows, while the nuclear force saturates.
- Easy : Most stable nuclei have a binding energy of about 8 MeV per nucleon.
- Medium : Iron-56 has the highest binding energy per nucleon, about 8.8 MeV.
- Exam : Very heavy nuclei beyond bismuth (Z = 83) are all unstable.
Nuclear force: strong, short-range, charge-independent, saturated.
Thinking that the nuclear force follows an inverse-square law like gravity.
Memory clue : "Strong glue, tiny reach, no charge preference."
Spot the difference
| Force | Range | Relative strength |
|---|---|---|
| Nuclear | about 2 fm | strongest |
| Electrostatic | long (1/r²) | about 1/100 of nuclear |
| Gravitational | long (1/r²) | extremely weak |
Practice questions
- 1What is the approximate range of the nuclear force?
- 2Is the nuclear force dependent on the charge of the nucleons?
- 3Which is stronger at 1 fm, the nuclear or the electrostatic force between two protons?
- 4What is the typical binding energy per nucleon of stable nuclei?
- 5Which nucleus has the highest binding energy per nucleon?
- 6Why do heavy nuclei need more neutrons than protons?
- 7Find the Coulomb force between two protons 1 fm apart.
- 8What can you say about the nuclear force at that distance?
Show answers (try them yourself first)
- 1A few fm (about 2 fm).
- 2No, it is independent of charge.
- 3The nuclear force.
- 4About 8 MeV.
- 5Iron-56 (about 8.8 MeV).
- 6The long-range Coulomb repulsion grows with Z while the nuclear force saturates, so extra neutrons add attraction without repulsion.
- 7F ≈ 230 N.
- 8It is attractive and larger than 230 N, so it holds the protons together.
Radioactivity and the decay law
Archaeologists estimate the age of ancient wood and bones from the carbon-14 in them; doctors use radioactive tracers in scans. Link to earlier classes : exponential change and logarithms from mathematics.
Radioactivity is the spontaneous decay of unstable nuclei with the emission of alpha (a helium nucleus, ⁴He), beta (an electron or positron) or gamma (high-energy photon) radiation. Each decay of a particular nucleus is random, but a large number of nuclei decay in a predictable way.
Rule 1: Decay law — The number of nuclei left falls as N = N₀ e^(−λ t). The half-life is T½ = 0.693 ÷ λ, the mean life is τ = 1 ÷ λ, and the activity is R = λ N (unit becquerel, 1 Bq = 1 decay per second). After n half-lives, N = N₀ ÷ 2ⁿ.
- Easy : T½ = 5 days ⇒ λ = 0.693 ÷ 5 = 0.139 day⁻¹, and after 15 days (3 half-lives) one-eighth remains.
- Medium : 80 g of a sample is left as 10 g after 3 half-lives.
- Exam : A sample of 2.0 × 10²⁰ nuclei with T½ = 10 h has λ = 0.693 ÷ 36 000 = 1.93 × 10⁻⁵ s⁻¹ and activity R = 3.9 × 10¹⁵ Bq.
Rule 2: Types of decay — Alpha decay lowers Z by 2 and A by 4 (²²⁶Ra → ²²²Rn + ⁴He). Beta-minus decay raises Z by 1 and leaves A unchanged (n → p + e⁻ + antineutrino). Gamma emission leaves Z and A unchanged.
- Easy : Carbon-14 (T½ = 5730 years) decays by beta emission to nitrogen-14.
- Medium : When 25 % of the carbon-14 of a sample remains, it has gone through two half-lives, which is about 11 460 years.
- Exam : Gamma rays are the most penetrating radiation; alpha particles are stopped by paper.
In the figure : The exponential decay curve: the number of nuclei halves after each half-life.
N = N₀e^(−λt); T½ = 0.693/λ; R = λN; after n half-lives N = N₀/2ⁿ.
Believing that all the nuclei are gone after two half-lives. One quarter is still left.
Memory clue : "Each half-life halves what is left: 1, ½, ¼, ⅛."
Spot the difference
| Radiation | Nature | Penetration |
|---|---|---|
| Alpha | ⁴He nucleus | least (paper) |
| Beta | electron | moderate (aluminium sheet) |
| Gamma | photon | greatest (thick lead) |
Practice questions
- 1State the radioactive decay law.
- 2Define the half-life.
- 3Write the SI unit of activity.
- 4A sample has T½ = 5 days. What fraction is left after 15 days?
- 5How much of an 80 g sample remains after 3 half-lives?
- 6Find λ for T½ = 5 days.
- 7A sample of 2.0 × 10²⁰ nuclei has T½ = 10 h. Find its activity.
- 8A wood sample has 25 % of its original carbon-14 (T½ = 5730 years). Find its age.
Show answers (try them yourself first)
- 1The rate of decay is proportional to the number of nuclei present: N = N₀e^(−λt).
- 2The time in which half of the nuclei decay.
- 3Becquerel (Bq).
- 41/8.
- 510 g.
- 6λ = 0.139 day⁻¹.
- 7R = λN ≈ 3.9 × 10¹⁵ Bq.
- 8Two half-lives: 2 × 5730 = 11 460 years.
Nuclear fission and fusion
A nuclear power station produces electricity by fission; the Sun shines by fusion. Link to earlier classes : binding energy per nucleon from the second section.
In nuclear fission, a heavy nucleus such as uranium-235 absorbs a slow neutron and splits into two medium-sized nuclei, releasing about 200 MeV and 2 to 3 more neutrons. These neutrons can cause further fissions, a chain reaction. In nuclear fusion, two light nuclei join to form a heavier one, again releasing energy.
Rule 1: Fission and the reactor — A reactor uses a moderator (heavy water or graphite) to slow the neutrons and control rods (cadmium or boron) to absorb the extra ones and keep the reaction steady. The energy comes from the binding energy per nucleon rising from about 7.6 MeV to about 8.5 MeV.
- Easy : Complete fission of 1 kg of uranium-235 releases 1000 ÷ 235 × 6.022 × 10²³ × 200 MeV = 8.2 × 10¹³ J.
- Medium : Coal gives about 3 × 10⁷ J per kg, so this equals about 2.7 million kg of coal.
- Exam : An uncontrolled chain reaction is an atom bomb; a controlled one is a reactor.
Rule 2: Fusion — In the Sun, four hydrogen nuclei fuse to form helium (the proton–proton cycle): 4 ¹H → ⁴He + 2e⁺ + 2ν + 26 MeV. It needs temperatures of about 10⁷ K to overcome the Coulomb repulsion.
- Easy : The energy per nucleon is 26 ÷ 4 = 6.5 MeV in fusion, but only 200 ÷ 236 = 0.85 MeV in fission.
- Medium : Controlled fusion on the Earth is still an experiment in progress, using strong magnetic fields to hold the hot plasma.
- Exam : Fusion gives more energy per unit mass than fission, and its fuel (hydrogen) is plentiful.
Fission: heavy splits, 200 MeV. Fusion: light joins, needs about 10⁷ K.
Mixing up the roles: the moderator slows neutrons, the control rods absorb them.
Memory clue : "Fission splits, Fusion Fuses; moderator slows, rods absorb."
Spot the difference
| Feature | Fission | Fusion |
|---|---|---|
| Reaction | heavy nucleus splits | light nuclei join |
| Example | uranium-235 | hydrogen to helium |
| Condition | slow neutrons | very high temperature |
Practice questions
- 1What is a nuclear chain reaction?
- 2What is the function of a moderator?
- 3What is the function of control rods?
- 4Find the energy from the complete fission of 1 kg of uranium-235 at 200 MeV per fission.
- 5Find the mass of coal (3 × 10⁷ J kg⁻¹) giving the same energy.
- 6Compare the energy per nucleon in fission and in the proton–proton fusion cycle.
- 7Why does fusion need a very high temperature?
- 8What is the source of the Sun's energy?
Show answers (try them yourself first)
- 1A self-sustaining reaction in which neutrons released by fission cause further fissions.
- 2It slows the fast neutrons.
- 3They absorb extra neutrons and control the rate of the reaction.
- 4E ≈ 8.2 × 10¹³ J.
- 5About 2.7 × 10⁶ kg.
- 6About 0.85 MeV per nucleon in fission and 6.5 MeV per nucleon in fusion.
- 7The nuclei must overcome their mutual Coulomb repulsion to come within the range of the nuclear force.
- 8Fusion of hydrogen into helium (about 26 MeV released for each helium nucleus formed).