1.2 Exam Practice
Exam-style practice questions on The Nuclear Atom
Section B: Data Analysis (Paper 1B Style)
Calculator and Data Booklet permitted. Show all working clearly.
Question 1: Mass Spectrometry Data Analyse HL
5 marksA mass spectrometer analysis of a sample of silicon produced the following data:
| Isotope | m/z | Relative abundance (%) |
|---|---|---|
| \({}^{28}\text{Si}\) | 28 | 92.23 |
| \({}^{29}\text{Si}\) | 29 | 4.67 |
| \({}^{30}\text{Si}\) | 30 | 3.10 |
(a) Define the term "isotopes". [2]
(b) Calculate the relative atomic mass of silicon using the data above. Give your answer to two decimal places. [2]
(c) Explain why the three isotopes of silicon have identical chemical properties. [1]
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(a) Atoms of the same element / with the same number of protons [1]
but with different numbers of neutrons / different mass numbers [1]
(b) \(A_r = (0.9223 \times 28) + (0.0467 \times 29) + (0.0310 \times 30)\)
\(= 25.8244 + 1.3543 + 0.9300 = \mathbf{28.11}\) [2]
Award [1] for correct substitution with an arithmetic error.
(c) They have the same electron configuration / same number of electrons in the same arrangement [1]
Section C: Structured Questions (Paper 2 Style)
Show all working. State answers with appropriate significant figures and units.
Question 2: Isotopes and Relative Atomic Mass Deduce
5 marksA sample of Gallium was analysed in a mass spectrometer and found to contain two naturally occurring isotopes: \({}^{69}\text{Ga}\) and \({}^{71}\text{Ga}\).
(a) Deduce the number of protons, neutrons, and electrons in a neutral atom of \({}^{71}\text{Ga}\). [1]
(b) State and explain why \({}^{69}\text{Ga}\) and \({}^{71}\text{Ga}\) exhibit identical chemical properties. [2]
(c) The relative abundance of \({}^{69}\text{Ga}\) is 60.1%, and the remainder is \({}^{71}\text{Ga}\). Calculate the relative atomic mass (\(A_r\)) of this sample of Gallium. Give your answer to two decimal places. [2]
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(a) Protons = 31, Neutrons = 40 (71 - 31), Electrons = 31. [1] (Award 1 mark only if all three are correct.)
(b) They have the same electron configuration [1]
Chemical properties are determined by how electrons interact / the arrangement of electrons [1]
(c) Abundance of \({}^{71}\text{Ga}\) = 39.9%
\(A_r = (0.601 \times 69) + (0.399 \times 71) = 41.469 + 28.329 = \mathbf{69.80}\) [2]
Question 3: Subatomic Particles in Ions Determine
4 marks(a) Complete the table below for the following species. [3]
| Species | Protons | Neutrons | Electrons |
|---|---|---|---|
| \({}^{35}\text{Cl}^{-}\) | ........ | ........ | ........ |
| \({}^{24}\text{Mg}^{2+}\) | ........ | ........ | ........ |
| \({}^{31}\text{P}^{3-}\) | ........ | ........ | ........ |
(b) Which two of the species above are isoelectronic? Justify your answer. [1]
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(a)
- \({}^{35}\text{Cl}^{-}\): 17 protons, 18 neutrons, 18 electrons [1]
- \({}^{24}\text{Mg}^{2+}\): 12 protons, 12 neutrons, 10 electrons [1]
- \({}^{31}\text{P}^{3-}\): 15 protons, 16 neutrons, 18 electrons [1]
(b) \(\text{Cl}^{-}\) and \(\text{P}^{3-}\) are isoelectronic because both have 18 electrons [1]
Question 4: Emission Spectra Explain
4 marksWhen hydrogen gas is placed in a discharge tube and an electric current is passed through it, a line emission spectrum is observed.
(a) Explain how a line emission spectrum provides evidence for the existence of discrete energy levels in atoms. [3]
(b) Explain why the lines in the emission spectrum of hydrogen converge at higher frequencies. [1]
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(a)
- Electrons are excited to higher energy levels [1]
- When they fall back to lower energy levels, they emit photons of specific energy / frequency [1]
- Each line corresponds to a specific energy transition, which proves that energy levels are quantised / discrete (not continuous) [1]
(b) The energy levels get closer together as they get further from the nucleus / the energy difference between successive levels decreases [1]
Question 5: Mass Spectrometry – Fragmentation Patterns Analyse HL
5 marksThe mass spectrum of an organic compound with molecular formula C3H8O shows the following major peaks:
| m/z | Relative intensity (%) |
|---|---|
| 60 | 20 |
| 45 | 8 |
| 31 | 100 |
| 29 | 42 |
| 15 | 25 |
(a) Identify the molecular ion peak (M+) and state the molar mass of the compound. [1]
(b) The base peak at m/z = 31 corresponds to a fragment [CH3O]+ or [CH2OH]+. Suggest the structure of the compound and explain how this fragmentation pattern supports your answer. [2]
(c) Identify the fragment responsible for the peak at m/z = 29 and write the fragmentation equation. [2]
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(a) M+ = m/z = 60, so molar mass = 60 g mol−1 [1]
(b) Propan-1-ol (CH3CH2CH2OH) [1]
The base peak at 31 corresponds to loss of C2H5 (mass 29) from the molecular ion, leaving [CH2OH]+, which is consistent with the primary alcohol structure [1]
(c) m/z = 29 corresponds to [C2H5]+ or [CHO]+ [1]
CH3CH2CH2OH+ → [C2H5]+ + CH2OH [1]