Isotopes of an element have:

The same number of protons but different numbers of neutrons

Different numbers of electrons only

Properties of the Nucleus | HSC Physics Module 8

In short: The nucleus holds protons and neutrons. An unstable nucleus decays by giving out alpha, beta or gamma radiation, and its half-life is the time for half of it to decay. Fission (splitting heavy nuclei) and fusion (joining light nuclei) both release energy, from the mass defect through $E = m c^{2}$ .

The nucleus

A nucleus is written $_{Z}^{A} X$ :

$Z$ = atomic number = number of protons
$A$ = mass number = protons + neutrons
Isotopes of an element have the same $Z$ but different $A$ (different numbers of neutrons).

The strong nuclear force holds protons and neutrons together. It is very strong but only acts over very short distances, about the width of a nucleus. It overcomes the electric repulsion between protons.

Radioactive decay

Unstable nuclei give off radiation to become more stable.

Type	What it is	Charge	Stopped by	Change to nucleus
Alpha (α)	helium nucleus $_{2}^{4} He$	+2	paper	$A$ − 4, $Z$ − 2
Beta minus (β⁻)	fast electron $_{- 1}^{0} e$	−1	a few mm of aluminium	$A$ same, $Z$ + 1
Beta plus (β⁺)	positron $_{+ 1}^{0} e$	+1	a few mm of aluminium	$A$ same, $Z$ − 1
Gamma (γ)	high-energy photon	0	thick lead or concrete	no change in $A$ or $Z$

In beta minus decay, a neutron turns into a proton, giving out an electron and an antineutrino. In beta plus decay, a proton turns into a neutron, giving out a positron and a neutrino.

In every nuclear equation, mass numbers balance and atomic numbers balance.

Half-life

Radioactive decay is random: you cannot say when one nucleus will decay. But with huge numbers of nuclei, the pattern is predictable.

The half-life $t_{1/2}$ is the time for half of the radioactive nuclei in a sample to decay.

Time	Fraction left
0	1
1 half-life	½
2 half-lives	¼
3 half-lives	⅛

The number left after time $t$ is:

$N_{t} = N_{0} e^{- λ t} λ = \frac{ln 2}{t _{1/2}}$

where $λ$ is the decay constant.

Mass defect and binding energy

Weigh a nucleus and it has less mass than the protons and neutrons that make it up, weighed separately. The missing mass is the mass defect $Δ m$ .

That mass was released as energy when the nucleus formed. To pull the nucleus apart, you would need to put the same energy back. This is the binding energy:

$E = Δ m c^{2}$

In atomic mass units, use $1 u = 931.5$ MeV.

Binding energy per nucleon

Dividing the binding energy by the number of nucleons shows how tightly bound each nucleon is. Plotting this against mass number gives a curve that:

rises steeply for light nuclei,
peaks near iron-56 (the most stable nucleus),
falls slowly for heavy nuclei.

Energy is released by any reaction that moves nuclei towards the peak:

Fusion: joining light nuclei (like hydrogen into helium).
Fission: splitting heavy nuclei (like uranium) into middle-sized ones.

Fission

A slow neutron hits a uranium-235 nucleus, which splits into two smaller nuclei and gives out 2 or 3 neutrons and a lot of energy:

$_{92}^{235} U +_{0}^{1} n \to_{56}^{141} Ba +_{36}^{92} Kr + 3_{0}^{1} n + energy$

The new neutrons can split more uranium nuclei: a chain reaction.

Uncontrolled: the number of fissions grows very fast. This is a nuclear bomb.
Controlled: in a reactor, control rods absorb spare neutrons so that, on average, exactly one neutron from each fission causes another. A moderator (water or graphite) slows the neutrons down so uranium-235 captures them more easily.

Enrico Fermi built the first controlled nuclear reactor in 1942.

The Nucleus

Notation

Z protons, A nucleons
Isotopes
Strong nuclear force

Decay

Alpha: A − 4, Z − 2
Beta minus: Z + 1
Beta plus: Z − 1
Gamma: photon

Half-life

Random but predictable
N = N₀ e^(−λt)
λ = ln 2 / t½

Binding energy

Mass defect
E = Δmc²
1 u = 931.5 MeV
Peak at iron

Energy

Fusion: light nuclei
Fission: heavy nuclei
Chain reaction
Control rods, moderator

Year 12Energy released in a reaction

For any nuclear reaction: energy released $= (mass before - mass after) \times c^{2}$ .

Alpha decay of polonium-210:

$^{210} Po$ (209.982874 u) $\to$ $^{206} Pb$ (205.974465 u) $+$ $^{4} He$ (4.002603 u)

$Δ m = 209.982874 - 209.977068 = 0.005806$ u, so $E = 0.005806 \times 931.5 = 5.41$ MeV.

The products have less mass but a greater total binding energy: they are more tightly bound. Most of the released energy goes to the alpha particle as kinetic energy.

Using the decay constant. Starting with $N_{0}$ nuclei, the number left after a time that is not a whole number of half-lives needs the exponential. For a half-life of 2.6 hours, $λ = \frac{0.693}{2.6} = 0.267$ per hour. After 6.0 hours, $\frac{N}{N _{0}} = e^{- 0.267 \times 6.0} = 0.20$ , so 20% is left.

Activity (decays per second, in becquerels) also halves every half-life. A detector also picks up background radiation, which must be subtracted before using the half-life.

Spotting the decay. If $Z$ drops by 1 and $A$ stays the same (phosphorus-30 to silicon-30), a positron and a neutrino were given off (beta plus).

Background and percentages. Subtract the background first: 340 Bq read with 20 Bq background is 320 Bq from the source; after 4 half-lives it is 20 Bq, so the detector reads 40 Bq. After 6 half-lives, $(\frac{1}{2})^{6} = 1.6%$ remains, so 98.4% has decayed.

From a graph. $ln \frac{N}{N _{0}} = - λ t$ , so a graph of $ln \frac{N}{N _{0}}$ against $t$ is a straight line with gradient $- λ$ . Then $t_{1/2} = \frac{ln 2}{λ}$ : a gradient of $- 0.0502$ s⁻¹ gives 13.8 s.

Dating. Solve $\frac{N}{N _{0}} = e^{- λ t}$ for $t$ : wood with 30% of its carbon-14 is $t = \frac{ln ( 1/0.30 )}{ln 2} \times 5730 \approx 9950$ years old. Falling to $\frac{1}{8}$ always takes exactly 3 half-lives.

Sharing the energy. A nucleus at rest splits with equal and opposite momentum, so the lighter particle gets most of the kinetic energy: $K_{α} = E \frac{M}{m + M}$ . For polonium-210, $5.41 \times \frac{206}{210} = 5.31$ MeV.

Decay chains. From U-238 to Pb-206: $A$ drops by 32, and only alpha decays change $A$ , so there are $\frac{32}{4} = 8$ alphas. These alone would lower $Z$ by 16, but it only drops by 10, so 6 beta minus decays raised it back.

Fuel use. A 1000 MW reactor for a day releases $8.64 \times 1 0^{13}$ J. Each fission gives $173 \times 1.602 \times 1 0^{- 13} = 2.77 \times 1 0^{- 11}$ J, so about $3.1 \times 1 0^{24}$ fissions, which is about 1.2 kg of U-235.

Quick summary

$Z$ = protons, $A$ = protons + neutrons. Isotopes differ in neutron number.
Alpha: $A - 4$ , $Z - 2$ . Beta minus: $Z + 1$ . Beta plus: $Z - 1$ . Gamma: no change.
Half-life is the time for half the nuclei to decay. $N = N_{0} e^{- λ t}$ and $λ = \frac{ln 2}{t _{1/2}}$ .
Mass defect $Δ m$ gives the binding energy $E = Δ m c^{2}$ (1 u = 931.5 MeV).
Binding energy per nucleon peaks at iron. Fusion of light nuclei and fission of heavy nuclei both release energy.
Fission neutrons can cause a chain reaction, controlled in reactors by control rods and a moderator.

Quiz

13 questions with instant marking. It is free, and your scores are saved to your account so you can carry on from any device.

Turn on JavaScript to sign in and take the quiz.

Try 3 questions now

The number of protons in a nucleus is its:
1. A Atomic number
2. B Mass number
3. C Half-life
4. D Binding energy
Show answer
A: Atomic number
Z is the atomic number.
Isotopes of an element have:
1. A Different numbers of protons
2. B The same number of protons but different numbers of neutrons
3. C Different numbers of electrons only
4. D The same mass number
Show answer
B: The same number of protons but different numbers of neutrons
Same element means same Z. Different isotopes have different A.
An alpha particle is:
1. A An electron
2. B A helium nucleus
3. C A photon
4. D A neutron
Show answer
B: A helium nucleus
Alpha is ⁴₂He: 2 protons and 2 neutrons.

Properties of the Nucleus

The nucleus

Radioactive decay

Half-life

Mass defect and binding energy

Binding energy per nucleon

Fission

Quick summary

Quiz

Feedback on this chapter

Physics chapters