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IB Physics – Theme E: Nuclear and Quantum Physics

Atomic structure, quantum physics (HL), radioactive decay, fission, fusion and stars.

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12 questions  ·  39 marks  ·  answers included  ·  IB Physics SL & HL (2016 curriculum, questions marked HL for Higher Level only)
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This worksheet covers:

  • Atomic structure
  • Radioactive decay
  • Nuclear energy
  • Fission/fusion and stellar processes (SL/HL) plus the photoelectric effect
  • Matter waves
  • The Bohr model (HL)

A calculator and the IB Physics data booklet ($h=6.63\times10^{-34}\,\text{J s}$, $c=3.00\times10^8\,\text{m s}^{-1}$, $e=1.60\times10^{-19}\,\text{C}$, $1\,\text{u}=931.5\,\text{MeV}/c^2$) are assumed throughout; allow about 55 minutes.

Questions

Section A – Atomic structure and radioactivity

SL and HL
1

State the number of protons, neutrons and electrons in a neutral atom of $^{235}_{92}\text{U}$.

Show answer
Protons $=92$, neutrons $=235-92=143$, electrons $=92$.
[2]
2

Nuclide X has 6 protons and 8 neutrons.(a) Write the full nuclide notation for X.(b) State the name of this nuclide and explain why it is an isotope of carbon.

Show answer
(a) $^{14}_{6}\text{C}$.(b) Carbon-14; it has the same proton (atomic) number as carbon (6) but a different number of neutrons/nucleon number, so it is an isotope of carbon.
[3]
3

A radioactive source emits alpha, beta and gamma radiation.(a) State one property that distinguishes alpha particles from beta particles in terms of penetrating power.(b) Write a nuclear equation for the alpha decay of $^{226}_{88}\text{Ra}$.

Show answer
(a) Alpha particles are stopped by a few cm of air or a sheet of paper, while beta particles need a few mm of aluminium to be stopped (alpha is far less penetrating).(b) $^{226}_{88}\text{Ra} \rightarrow \, ^{222}_{86}\text{Rn} + \,^{4}_{2}\text{He}$.
[3]
4

The isotope $^{226}_{88}\text{Ra}$ has mass 226.025403 u and decays to $^{222}_{86}\text{Rn}$ (mass 222.017576 u) by alpha emission (alpha particle mass 4.002603 u). Calculate the energy released in this decay, in MeV.

Show answer
Mass defect $\Delta m = 226.025403-(222.017576+4.002603)=0.005224\,\text{u}$. Energy released $=0.005224\times931.5=4.87\,\text{MeV}$.
[3]

Section B – Half-life, fission, fusion and stars

Exam-style, SL and HL
5

The isotope iodine-131 has a half-life of 8.02 days.(a) Define the term half-life.(b) A hospital sample initially contains $4.0\times10^{15}$ atoms of I-131. Calculate the number of atoms remaining after 24.06 days.

Show answer
(a) Half-life is the time taken for the activity (or number of undecayed nuclei) of a radioactive sample to fall to half of its original value.(b) $24.06\,\text{days} = 3$ half-lives, so fraction remaining $=(1/2)^3=0.125$. Atoms remaining $=0.125\times4.0\times10^{15}=5.0\times10^{14}$.
[4]
6

Explain, in terms of binding energy per nucleon, why energy is released in both nuclear fission of $^{235}\text{U}$ and nuclear fusion of hydrogen isotopes.

Show answer
Binding energy per nucleon is low for very heavy nuclei (like $\text{U}$) and for very light nuclei (like $\text{H}$), and is greatest for medium-mass nuclei (around iron). Fission splits a heavy nucleus into medium-mass fragments, and fusion joins light nuclei into a medium-mass nucleus; in both cases the products have greater binding energy per nucleon than the reactants, so energy is released.
[3]
7

A fission reaction is $^{235}_{92}\text{U} + \,^{1}_{0}\text{n} \rightarrow \,^{141}_{56}\text{Ba} + \,^{92}_{36}\text{Kr} + x\,^{1}_{0}\text{n}$.(a) Find the value of $x$.(b) State one condition needed for a self-sustaining chain reaction to occur.

Show answer
(a) Nucleon number balance: $235+1=141+92+x \Rightarrow x=3$.(b) The mass of fissile material must exceed the critical mass, so that on average at least one neutron from each fission goes on to cause a further fission (accept: sufficient concentration/enrichment of $^{235}\text{U}$, or neutrons must be moderated/slowed).
[3]
8

Outline, using the Hertzsprung–Russell diagram and stellar evolution, the sequence of stages a star of mass similar to the Sun passes through from the main sequence to its final state.

Show answer
Main sequence star (fusing H to He in core) $\rightarrow$ core H exhausted, core contracts and outer layers expand and cool, star moves to red giant branch $\rightarrow$ He fusion begins in the core (and H fusion in a shell) $\rightarrow$ outer layers are expelled as a planetary nebula $\rightarrow$ the remaining core, no longer able to fuse and supported by electron degeneracy pressure, becomes a white dwarf, which then cools.
[4]

Section C – Quantum physics (HL)

Higher Level only
9

The work function of sodium is 2.28 eV. Light of wavelength 400 nm is incident on a sodium surface.(a) Calculate the energy of a photon of this light, in eV.(b) Show that photoelectrons are emitted, and calculate their maximum kinetic energy in eV.

Show answer
(a) $E=\dfrac{hc}{\lambda}=\dfrac{6.63\times10^{-34}\times3.00\times10^{8}}{400\times10^{-9}}=4.97\times10^{-19}\,\text{J} = 3.11\,\text{eV}$.(b) Since $3.11\,\text{eV} > 2.28\,\text{eV}$ (the work function), photoelectrons are emitted. $E_{k,\text{max}}=3.11-2.28=0.83\,\text{eV}$.
[4]
10

(a) State the photoelectric effect evidence that light behaves as a particle (photon) rather than purely as a wave.(b) Calculate the threshold wavelength for sodium (work function 2.28 eV), above which no photoelectrons are emitted.

Show answer
(a) Emission of electrons is instantaneous (no time delay) and occurs only above a threshold frequency independent of intensity; increasing intensity increases the number of photoelectrons but not their maximum kinetic energy — this cannot be explained by a purely wave model, which predicts energy build-up over time and emission at any frequency given enough intensity.(b) $\lambda_0=\dfrac{hc}{\phi}=\dfrac{6.63\times10^{-34}\times3.00\times10^8}{2.28\times1.60\times10^{-19}}=5.45\times10^{-7}\,\text{m}=545\,\text{nm}$.
[3]
11

An electron is accelerated from rest through a potential difference of 150 V.(a) Calculate the kinetic energy gained by the electron, in J.(b) Calculate the de Broglie wavelength of the electron.

Show answer
(a) $E_k=eV=1.60\times10^{-19}\times150=2.40\times10^{-17}\,\text{J}$.(b) $v=\sqrt{2E_k/m_e}=7.26\times10^{6}\,\text{m s}^{-1}$, so $\lambda=\dfrac{h}{m_ev}=\dfrac{6.63\times10^{-34}}{9.11\times10^{-31}\times7.26\times10^{6}}=1.00\times10^{-10}\,\text{m}$ (100 pm).
[4]
12

In the Bohr model of the hydrogen atom, electrons occupy discrete energy levels and emission spectra consist of sharp lines rather than a continuum.(a) Explain, in terms of electron transitions, how a line emission spectrum is produced.(b) Explain why this observation supports the idea of quantised (discrete) atomic energy levels.

Show answer
(a) An electron in an excited atom falls from a higher energy level to a lower one, emitting a photon whose energy equals the difference between the two energy levels; each possible transition between levels produces a photon of a specific frequency, seen as one spectral line.(b) Only certain photon energies (frequencies/wavelengths) are observed, producing a discrete line spectrum rather than a continuous range of frequencies; this can only occur if the electron energies within the atom are restricted to a discrete (quantised) set of allowed values, since a continuum of energy levels would give a continuous spectrum.
[3]

Related worksheets

More worksheets in IB Physics (SL & HL):

IB Physics – Theme A: Space, Time and MotionIB Physics – Theme B: The Particulate Nature of MatterIB Physics – Theme C: Wave BehaviourIB Physics – Theme D: FieldsIB Physics – Data-Based Questions (Paper 1B)

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