IB Physics SL E.1 Structure of the Atom Question Bank

Practise IB Physics SL E.1 by modelling nuclei, isotopes, spectra, binding energy and the evidence for atomic structure.

Exam points

Relate proton, neutron and electron structure to isotopes, nuclide notation and charge.
Interpret line spectra and energy-level transitions as evidence for quantized atomic structure.
Calculate mass defect or binding-energy relationships and compare nuclear stability evidence.

Covered syllabus points

E.1.1—Rutherford experiment E.1.2—Nuclear notation E.1.3—Atomic energy levels E.1.4—Atomic transitions E.1.5—Photon energy E.1.6—Spectra and composition

[Maximum number: 4]

This question is in two parts. Part 1 is about the nuclear model of the atom and radioactive decay. Part 2 is about waves.

Part 1 Nuclear model of the atom and radioactive decay

Outline how the evidence supplied by the Geiger-Marsden experiment supports the nuclear model of the atom.

[Maximum number: 1]

Tritium may be produced by bombarding a nucleus of the isotope lithium-7 with a high-energy neutron. The reaction equation for this interaction is

{ }_{3}^{7} \mathrm{Li}+{ }_{0}^{1} \mathrm{n} \rightarrow{ }_{1}^{3} \mathrm{H}+{ }_{Z}^{4} \mathrm{X}+{ }_{0}^{1} \mathrm{n} .

Identify the proton number Z of X.

[Maximum number: 10]

This question is about atomic energy levels.

(a)

Outline a laboratory procedure for producing and observing the atomic absorption spectrum of a gas.

[ 3 ]

(b)

Describe the appearance of an atomic absorption spectrum.

[ 1 ]

(c)

Explain why the spectrum in (a) provides evidence for quantization of energy in atoms.

[ 3 ]

(d)

The principal energy levels of the hydrogen atom in electronvolt ( eV ) are given by

E_{n}=\frac{13.6}{n^{2}}

where n is a positive integer.
Determine the wavelength of the absorption line that corresponds to an electron transition from the energy level given by n=1 to the level given by n=3.

[ 3 ]

$E=13.6\left[\frac{1}{1^{2}}-\frac{1}{3^{2}}\right]=12.1 \mathrm{eV}$ ;
$12.1 \mathrm{eV}=12.1 \times 1.6 \times 10^{-19} \mathrm{~J}=\frac{h c}{\lambda}$ ;
$\lambda=102 \mathrm{~nm}$ or 103 nm ;

Marking guidance:

Award [3] for a bald correct answer.

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Covered syllabus points

E.1 Structure of the atom question 1

E.1 Structure of the atom question 2

E.1 Structure of the atom question 3

Question (a)

Question (b)

Question (c)

Question (d)