IB Physics HL E 1 Structure of the Atom Questions
Practise IB Physics HL E.1 by analysing atomic spectra, nuclear structure, mass defect and binding-energy evidence in extended problems.
- Syllabus
- First assessment 2025
- Course
- Physics HL
- Level
- HL
Practise IB Physics HL E.1 by analysing atomic spectra, nuclear structure, mass defect and binding-energy evidence in extended problems.
Rutherford constructed a model of the atom based on the results of the alpha particle scattering experiment. Describe this model.
most of the mass of the atom is confined within a very small volume/nucleus
all the positive charge is confined within a very small volume/nucleus;
electrons orbit the nucleus in circular orbits
Bohr modified the Rutherford model by introducing the condition mvr=n2πh.
Outline the reason for this modification.
the electrons accelerate and so radiate energy;
they would therefore spiral into the nucleus/atoms would be unstable
electrons have discrete/only certain energy levels
the only orbits where electrons do not radiate are those that satisfy the Bohr condition mvr=n2πh
Calculate the electron's orbital radius in (c)(ii).
r=4π2×8.99×109×9.11×10−31×(1.6×10−19)2(6.63×10−34)2r=5.3×10−11 m
This question is in two parts. Part 1 is about kinematics and gravitation. Part 2 is about radioactivity.
Part 1 Kinematics and gravitation
A ball is released near the surface of the Moon at time t=0. The point of release is on a straight line between the centre of Earth and the centre of the Moon. The graph below shows the variation with time t of the displacement s of the ball from the point of release.
Part 2 Radioactivity
Radium-224 (88224Ra) is a radioactive nuclide that decays to form radon-220. Radon-220 is itself radioactive and undergoes a further decay. The table shows the series of radioactive nuclides that are formed as the decays proceed. The series ends with a stable isotope of lead.
For the final thallium nuclide, identify the
nucleon number.
208;
proton number.
81 ;
The diagram represents the three principal spectral lines in the visible region of the spectrum of atomic hydrogen.
The electron in the hydrogen atom can only occupy certain allowed energy levels.
Outline how the spectral lines provide evidence for the existence of these energy levels.
Spectral lines are discrete.
Therefore photon energy is discrete/quantized.
Photon energy is equivalent to the difference in energy that the electron has in each discrete level / Delta E = hf / OWTTE.
So electron levels are themselves discrete.
Determine the difference in energy between the two levels from which electron transitions give rise to the Hα and Hγ spectral lines respectively.
difference in energy =hc[λγ1−λα1];
=6.6×10−34×3×108×107×0.0780=1.54×10−19 J;
or
Eα=λαhc=3.02×10−19 J or 1.89 eV;
Eγ=λγhc=4.56×10−19 J or 2.85 eV;
difference in energy =1.54×10−19 J or 0.963 eV;
Rutherford constructed a model of the atom based on the results of the alpha particle scattering experiment. Describe this model.
most of the mass of the atom is confined within a very small volume/nucleus
all the positive charge is confined within a very small volume/nucleus;
electrons orbit the nucleus in circular orbits
The diagram represents the three principal spectral lines in the visible region of the spectrum of atomic hydrogen.
The electron in the hydrogen atom can only occupy certain allowed energy levels.
Outline how the spectral lines provide evidence for the existence of these energy levels.
Spectral lines are discrete.
Therefore photon energy is discrete/quantized.
Photon energy is equivalent to the difference in energy that the electron has in each discrete level / Delta E = hf / OWTTE.
So electron levels are themselves discrete.
This question is in two parts.
The energies of the principal energy levels in atomic hydrogen measured in eV are given by the expression
The visible lines in the spectrum correspond to electron transitions that end at n=2.
Show that the spectral line of wavelength λ=485 nm is the result of an electron transition from n=4.
energy difference between levels =λehc=4.85×10−7×1.6×10−196.63×10−34×3×108;
=2.55 eV;
[3.40−2.55]=0.85=n213.6 to give n2=16;
n=4;
Marking guidance:
Award [3] for reversed argument.
The diagram represents the three principal spectral lines in the visible region of the spectrum of atomic hydrogen.
The electron in the hydrogen atom can only occupy certain allowed energy levels.
Determine the difference in energy between the two levels from which electron transitions give rise to the Hα and Hγ spectral lines respectively.
difference in energy =hc[λγ1−λα1];
=6.6×10−34×3×108×107×0.0780=1.54×10−19 J;
or
Eα=λαhc=3.02×10−19 J or 1.89 eV;
Eγ=λγhc=4.56×10−19 J or 2.85 eV;
difference in energy =1.54×10−19 J or 0.963 eV;
The diagram shows a simplified energy-balance model for the Earth surface-atmosphere system.
The following data are given:
The primary energy source of the Sun is the proton-proton (p-p) chain of fusion reactions. Four protons and two electrons produce a helium nucleus together with neutrinos and gamma photons. The overall reaction is:
Outline how the presence of helium in the Sun can be confirmed empirically.
Any mention of absorption or emission spectrum «of the Sun» Observed lines/wavelengths/frequencies match those of helium OR
Helium lines/wavelengths/frequencies are present «in the spectrum»
Marking guidance:
Allow solar spectrum
State how the density of a nucleus varies with the number of nucleons in the nucleus.
it is constant
Show that the nuclear radius of phosphorus-31 (1531P) is about 4 fm .
R=1.20×10−15×3131=3.8×10−15 m
1532P is formed when a nucleus of deuterium (12H) collides with a nucleus of 1531P. The radius of a deuterium nucleus is 1.5 fm .
State the maximum distance between the centres of the nuclei for which the production of 1532P is likely to occur.
separation for interaction = 5.3 or 5.5 fm
In an experiment, alpha particles of initial kinetic energy 5.9 MeV are directed at stationary nuclei of lead-207. The distance of closest approach is about:
The radius of a nucleus of 82207 Pb is 7.1×10−15 m. Suggest why there will be no deviations from Rutherford scattering in the experiment described above.
The closest approach is «significantly» larger than the radius of the nucleus / far away from the nucleus/OWTTE.
«Therefore» the strong nuclear force will not act on the alpha particle.
In an experiment, alpha particles of initial kinetic energy 5.9 MeV are directed at stationary nuclei of lead (82207 Pb). Show that the distance of closest approach is about 4×10−14 m.
2e AND 82e seen
OR
3.2×10−19 «C» AND 1.312×10−17 «C» seen
d=5.9×106×e8.99×109×(2e)(82e)=3.998×10−14≈4×10−14<m≫
Must see either clear substitutions or answer to at least 4 s.f. for MP2.
This question is in two parts.
The energies of the principal energy levels in atomic hydrogen measured in eV are given by the expression
The visible lines in the spectrum correspond to electron transitions that end at n=2.
Calculate the energy of the level corresponding to n=2.
3.40 eV ;
Marking guidance:
Award [0] for omitted negative sign.