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How do atoms emit light and how do nuclei release energy?

Explain atomic energy levels and spectra and analyse nuclear decay and mass-energy in reactions.

Bohr energy levels and line spectra, radioactive decay and half-life, and how mass defect and binding energy explain the energy released in nuclear reactions.

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

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What this dot point is asking

This dot point combines the quantised atom with the physics of the nucleus, both central revolutions in modern physics.

Atomic energy levels and spectra

In the Bohr model, electrons in an atom can only exist in certain allowed energy levels, not anywhere in between. An electron moves to a higher level by absorbing exactly the right amount of energy and falls to a lower level by emitting a photon. The photon's energy equals the difference between the two levels:

ΔE=EhighElow=hf=hcλ\Delta E = E_\text{high} - E_\text{low} = hf = \frac{hc}{\lambda}

Because only specific energy differences are possible, atoms emit and absorb light at specific frequencies, producing line spectra. An emission spectrum is a set of bright coloured lines; an absorption spectrum is dark lines on a continuous background. These line spectra act as fingerprints that identify elements, which is how astronomers determine the composition of distant stars.

Radioactive decay

Unstable nuclei decay to become more stable, emitting radiation. The three common types are:

  • Alpha (α\alpha): emission of a helium nucleus (24He^4_2\text{He}); mass number falls by 44, atomic number by 22.
  • Beta-minus (β\beta^-): a neutron becomes a proton, emitting an electron and an antineutrino; atomic number rises by 11, mass number unchanged.
  • Gamma (γ\gamma): a high-energy photon released as the nucleus drops to a lower energy state; no change in mass or atomic number.

Decay is random for any single nucleus, but a large sample decays predictably. The half-life t1/2t_{1/2} is the time for half the nuclei in a sample to decay:

N=N0(12)t/t1/2N = N_0\left(\frac{1}{2}\right)^{t/t_{1/2}}

Half-lives span an enormous range, from fractions of a second to billions of years, and this is the basis of radiometric dating. Carbon-14, with a half-life of about 57305730 years, lets archaeologists date once-living material, while uranium isotopes with half-lives of billions of years date rocks and the age of the Earth. The activity of a sample (decays per second) falls in step with the number of remaining nuclei, so it too halves every half-life.

Mass defect and binding energy

The mass of a nucleus is slightly less than the total mass of its separate protons and neutrons. This difference is the mass defect Δm\Delta m. The missing mass is the energy that was released when the nucleons bound together, given by mass-energy equivalence:

E=Δmc2E = \Delta m\, c^2

This binding energy is what holds the nucleus together. The binding energy per nucleon peaks around iron (56Fe^{56}\text{Fe}), which is the most stable nucleus. In nuclear reactions the products have a larger total binding energy per nucleon than the reactants, so mass is converted to energy. Fusion of light nuclei (below iron) and fission of heavy nuclei (above iron) both move toward iron and release energy, which is the source of the energy in nuclear power and in stars.

When balancing nuclear equations, conserve both the mass number (top) and the atomic number (bottom) on each side. For energy questions, convert atomic mass units to kilograms if needed (1 u=1.66×1027 kg1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}, or use 931.5 MeV/u931.5\ \text{MeV/u}) before applying E=Δmc2E = \Delta m\,c^2.

Exam-style practice questions

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

TCE 20244 marksCalculate the mass defect (in atomic mass units) of 56Fe^{56}\text{Fe} given its atomic mass is 55.9349375 u55.9349375\ \text{u}, then calculate the binding energy per nucleon. Use mH=1.007825 um_H = 1.007825\ \text{u} and mn=1.008665 um_n = 1.008665\ \text{u}.
Show worked answer →

The mass defect is the difference between the total mass of the separate constituents (26 hydrogen atoms, which include the electrons, plus 30 neutrons) and the actual atomic mass.

Sum=(26)(1.007825)+(30)(1.008665)=26.20345+30.25995=56.46340 u.\text{Sum} = (26)(1.007825) + (30)(1.008665) = 26.20345 + 30.25995 = 56.46340\ \text{u}.

Δm=56.4634055.9349375=0.52846 u.\Delta m = 56.46340 - 55.9349375 = 0.52846\ \text{u}.

Binding energy =Δm×931.5 MeV/u=(0.52846)(931.5)=492.3 MeV= \Delta m \times 931.5\ \text{MeV/u} = (0.52846)(931.5) = 492.3\ \text{MeV}.
Per nucleon: 492.356=8.79 MeV per nucleon\dfrac{492.3}{56} = 8.79\ \text{MeV per nucleon}.

56Fe^{56}\text{Fe} has one of the highest binding energies per nucleon, which is why it is so stable. Markers want hydrogen-atom masses (to account for electrons), conversion at 931.5 MeV/u931.5\ \text{MeV/u}, and division by 5656.

TCE 20242 marksSome nuclear batteries use the beta-minus decay of 63Ni^{63}\text{Ni} (Z=28Z = 28) to 63Cu^{63}\text{Cu} (Z=29Z = 29). Write the decay equation and show that mass and charge are conserved.
Show worked answer →

In beta-minus decay a neutron turns into a proton, emitting an electron (the beta particle) and an antineutrino. The mass number stays the same while the atomic number rises by one.

2863Ni 2963Cu+ 10e+νˉe.^{63}_{28}\text{Ni} \rightarrow\ ^{63}_{29}\text{Cu} + \ ^{0}_{-1}e + \bar{\nu}_e.

Mass number is conserved (63=6363 = 63) and charge is conserved (28=29+(1)28 = 29 + (-1)). Markers want the daughter 63Cu^{63}\text{Cu}, the emitted electron and the antineutrino, with both mass number and charge balancing.

TCE 20222 marksThe hydrogen energy levels are En=13.6n2 eVE_n = -\dfrac{13.6}{n^2}\ \text{eV}. Show that the wavelength of the emission between n=3n = 3 and n=2n = 2 is about 660 nm660\ \text{nm}.
Show worked answer →

E3=13.69=1.511 eVE_3 = -\dfrac{13.6}{9} = -1.511\ \text{eV} and E2=13.64=3.400 eVE_2 = -\dfrac{13.6}{4} = -3.400\ \text{eV}.

Photon energy =E3E2=1.511(3.400)=1.889 eV=1.889×1.6×1019=3.02×1019 J= E_3 - E_2 = -1.511 - (-3.400) = 1.889\ \text{eV} = 1.889 \times 1.6 \times 10^{-19} = 3.02 \times 10^{-19}\ \text{J}.

Wavelength: λ=hcE=(6.63×1034)(3.0×108)3.02×1019=6.58×107 m=658 nm\lambda = \dfrac{hc}{E} = \dfrac{(6.63\times10^{-34})(3.0\times10^8)}{3.02\times10^{-19}} = 6.58 \times 10^{-7}\ \text{m} = 658\ \text{nm}.

This is about 660 nm660\ \text{nm}, the red Balmer (H-alpha) line. Markers want the level difference, conversion to joules, then λ=hcΔE\lambda = \dfrac{hc}{\Delta E}.

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