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What evidence established the nuclear model of the atom, and how do we describe nuclides and nuclear reactions?

Describe the evidence for the nuclear atom, represent nuclides and isotopes, and balance nuclear reaction and decay equations

A focused answer to the H2 Physics learning outcome on the nuclear atom. The alpha-scattering evidence for a small dense nucleus, nuclide notation and isotopes, the alpha, beta and gamma emissions, and balancing nuclear equations.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to describe the experimental evidence for the nuclear model of the atom, to represent nuclides and isotopes in standard notation, and to balance nuclear reaction and decay equations by conserving nucleon and proton numbers. This frames the structure of matter that the rest of nuclear physics builds on.

The answer

Evidence for the nuclear atom

The alpha-scattering experiment fired alpha particles at a thin gold foil. Two observations were decisive:

  • Most alpha particles passed straight through with little or no deflection, showing that the atom is mostly empty space.
  • A very small fraction were deflected through large angles, some bouncing almost straight back, showing that the positive charge and nearly all the mass are concentrated in a tiny, dense region.

These results overturned the earlier "plum pudding" picture and established the nuclear model: a small, dense, positively charged nucleus surrounded by electrons, with most of the atom empty.

Nuclide notation

A nuclide is a specific nuclear species, written:

ZAX^{A}_{Z}\text{X}

where XX is the chemical symbol, ZZ is the proton (atomic) number (the number of protons, which fixes the element), and AA is the nucleon (mass) number (the total number of protons and neutrons). The number of neutrons is AZA - Z.

Isotopes

Isotopes of an element have the same proton number ZZ but different nucleon numbers AA, that is, the same number of protons but different numbers of neutrons. They are chemically identical but differ in mass and may differ in nuclear stability. Carbon-12 and carbon-14 are isotopes of carbon.

Radioactive emissions

Unstable nuclei emit radiation to become more stable:

  • Alpha (α\alpha): a helium nucleus 24He^4_2\text{He}; reduces AA by 44 and ZZ by 22.
  • Beta-minus (β\beta^-): an electron 1  0e^{\ \ 0}_{-1}\text{e} emitted when a neutron becomes a proton; AA unchanged, ZZ increases by 11.
  • Gamma (γ\gamma): a high-energy photon; no change in AA or ZZ, just a loss of excess energy.

Balancing nuclear equations

In any nuclear reaction or decay, two quantities are conserved:

  • the total nucleon number AA (the top numbers balance),
  • the total proton number ZZ (the bottom numbers balance, conserving charge).

Use these two conservation rules to identify an unknown product.

Examples in context

Example 1. Smoke detectors. A common smoke detector contains a tiny americium source that emits alpha particles, ionising the air in a chamber and allowing a small current to flow. Smoke disrupts this current, triggering the alarm. The alpha emitter is chosen because alpha particles are easily stopped and stay safely inside the device.

Example 2. Decay chains. A heavy nucleus like uranium-238 decays through a long series of alpha and beta emissions, each step balanced by conserving nucleon and proton numbers, until it reaches stable lead-206. Tracking the chain is just repeated application of the two conservation rules.

Try this

Q1. State the two key observations of the alpha-scattering experiment and what each shows. [2 marks]

  • Cue. Most pass straight through (atom mostly empty space); a few deflect sharply (small, dense, positive nucleus).

Q2. Write the nuclide notation for an isotope of oxygen with 88 protons and 1010 neutrons. [1 mark]

  • Cue. 818O^{18}_{8}\text{O} (nucleon number 8+10=188 + 10 = 18).

Q3. A polonium-210 nucleus (84210Po^{210}_{84}\text{Po}) emits an alpha particle. Identify the daughter nucleus by its nucleon and proton numbers. [2 marks]

  • Cue. Nucleon number 2104=206210 - 4 = 206; proton number 842=8284 - 2 = 82 (lead), giving 82206Pb^{206}_{82}\text{Pb}.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksIn an alpha-scattering experiment, a thin gold foil is bombarded with alpha particles. (a) State the two key observations. (b) Explain what each observation reveals about the structure of the atom.
Show worked answer →

(a) The two key observations: most alpha particles passed straight through the foil with little or no deflection; a very small fraction were deflected through large angles, some almost straight back.

(b) Most passing straight through shows that the atom is mostly empty space. The rare large-angle deflections show that the positive charge and almost all the mass are concentrated in a very small, dense region (the nucleus), since only a concentrated positive charge could repel an alpha particle so strongly.

Markers reward both observations (most undeflected, a few large-angle), and the two deductions (atom mostly empty space; a small, dense, positively charged nucleus).

Original4 marksA nucleus of radium-226 (88226Ra^{226}_{88}\text{Ra}) decays by emitting an alpha particle. (a) Write the balanced nuclear equation, identifying the daughter nucleus by its proton number. (b) State how the nucleon and proton numbers are conserved.
Show worked answer →

(a) An alpha particle is 24He^4_2\text{He}. The equation is 88226Ra 86222X+ 24He^{226}_{88}\text{Ra} \rightarrow\ ^{222}_{86}\text{X} +\ ^4_2\text{He}. The daughter has proton number 8686 (radon, Rn\text{Rn}): 88226Ra 86222Rn+ 24He^{226}_{88}\text{Ra} \rightarrow\ ^{222}_{86}\text{Rn} +\ ^4_2\text{He}.

(b) Nucleon number is conserved: 226=222+4226 = 222 + 4. Proton number (charge) is conserved: 88=86+288 = 86 + 2.

Markers reward the alpha particle as 24He^4_2\text{He}, the daughter with nucleon number 222222 and proton number 8686, and the explicit conservation of both nucleon and proton numbers.

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