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Atomic and nuclear physics - TCE Physics (Tasmania)

Syllabus dot point

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

TCEPhysicsUnit 4: Revolutions in Modern Physics9 min read

Quick answer

Electrons occupy discrete energy levels, and light is emitted or absorbed when they jump between levels with photon energy ΔE=hf\Delta E = hf. Nuclei decay by alpha, beta and gamma emission, and the energy released comes from a mass defect via E=mc2E = mc^2.

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=Ehigh−Elow=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.

Spectra prove energy levels are discrete

The fact that gases emit only sharp spectral lines, not a continuous range of colours, is direct evidence that electron energies inside atoms are quantised rather than continuous.

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-life

The half-life of a radioactive isotope is the time taken for half the radioactive nuclei in a sample to decay, a constant value independent of the starting amount.

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=Δm c2E = \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.

Worked example: energy from a mass defect

A nuclear reaction has a mass defect Δm=3.0×10−29 kg\Delta m = 3.0 \times 10^{-29}\ \text{kg}. Find the energy released, using c=3.0×108 m s−1c = 3.0 \times 10^8\ \text{m s}^{-1}.

Step 1: Identify the mass-energy equivalence formula

The mass defect is the small amount of mass that disappears when nucleons bind together. Einstein's relation E=Δm c2E = \Delta m\, c^2 converts that missing mass directly into energy, so all that is needed is to substitute the given values. The factor c2c^2 is extremely large, which is why even a tiny mass defect releases a substantial amount of energy.

Step 2: Substitute and evaluate

Squaring cc first gives c2=(3.0×108)2=9.0×1016 m2 s−2c^2 = (3.0 \times 10^8)^2 = 9.0 \times 10^{16}\ \text{m}^2\text{ s}^{-2}. Multiplying by the mass defect then converts the missing mass into joules of released energy:

E=(3.0×10−29)(3.0×108)2=(3.0×10−29)(9.0×1016)=2.7×10−12 J.E = (3.0 \times 10^{-29})(3.0 \times 10^8)^2 = (3.0 \times 10^{-29})(9.0 \times 10^{16}) = 2.7 \times 10^{-12}\ \text{J}.

Final answer: The reaction releases about 2.7×10−12 J2.7 \times 10^{-12}\ \text{J}.

Treating half-life as if decay were linear

A frequent error is assuming that after two half-lives nothing remains, or halving by subtraction. After each half-life the remaining amount is multiplied by one half: one half, then one quarter, then one eighth, and so on. Use N=N0(1/2)t/t1/2N = N_0(1/2)^{t/t_{1/2}} rather than subtracting equal amounts.

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=1.66×10−27 kg1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}, or use 931.5 MeV/u931.5\ \text{MeV/u}) before applying E=Δm c2E = \Delta m\,c^2.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks
Calculate 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=56.46340−55.9349375=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.

Practice2 marks
Some 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.

Practice2 marks
The 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 =E3−E2=−1.511−(−3.400)=1.889 eV=1.889×1.6×10−19=3.02×10−19 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=(6.63×10−34)(3.0×108)3.02×10−19=6.58×10−7 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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