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Date	May 2017	Marks available	2	Reference code	17M.3.hl.TZ1.29
Level	HL	Paper	3	Time zone	TZ1
Command term	Formulate	Question number	29	Adapted from	N/A

Question

Nuclear radiation is dangerous because of its ability to damage cells but it can also be used in nuclear medicine.

Iodine-131 is released in nuclear explosions but is used in scanners for thyroid cancer. The half-life of iodine-131 is 8.02 days.

Yttrium-90 is used in treating certain cancers.

Formulate a nuclear equation for the beta decay of yttrium-90.

[1]

Lutetium-177 is a common isotope used for internal radiation therapy.

Suggest why lutetium-177 is an ideal isotope for the treatment of certain cancers based on the type of radiation emitted.

[1]

Calculate the rate constant, $\lambda$ , in day⁻¹, for the decay of iodine-131 using section 1 of the data booklet.

[1]

c.i.

Calculate the time, in days, for 90% of the sample to decay.

[2]

c.ii.

A breathalyser measures the blood alcohol content from a breath sample. Formulate half-equations for the reactions at the anode (negative electrode) and the cathode (positive electrode) in a fuel cell breathalyser.

[2]

Markscheme

⁹⁰Y → ⁹⁰Zr + β^–

Accept β, e or e^–.
Accept ⁹⁰Y → ⁹⁰Zr + β^–+ v

[1 mark]

beta-radiation/emission AND targets tumour/cancer cells
OR
beta-radiation/emission AND limited damage to healthy cells/tissues
OR
beta-radiation/emission AND produces «small amount of» gamma-rays «for visualizing tumours/monitoring treatment»

[1 mark]

$\lambda \left( { = \frac{{\ln 2}}{{{t_{\frac{1}{2}}}}} = \frac{{0.693}}{{8.02{\text{ }}day}}} \right) = 8.64 \times {10^{ - 2}}/0.0864$ «day⁻¹»

[1 mark]

c.i.

ALTERNATIVE 1:
«N₀ = initial amount = 100%»

N «= 100 – 90» = 10% at time t

« $\ln \left( {\frac{{100}}{{10}}} \right) = 2.303 = 0.0864t$ »

« $t = \frac{{2.303}}{{0.0864{\text{ da}}{{\text{y}}^{ - 1}}}} =$ » 26.7 «days»

Accept 26.6 or 27 «days»
Award [2] for correct final answer.

ALTERNATIVE 2:
«N_t = N₀(0.5)ⁿ where n = number of half-lives»

10 = 100(0.5)ⁿ

« $\log \left( {\frac{1}{{10}}} \right) = n \times \log \,0.5$ »

« $- 1 = n\left( { - 0.301} \right)/n = \frac{1}{{0.301}}$ »

«t $= \frac{1}{{0.301}} \times 8.02 =$ » 26.6 «days»

Accept 26.7 or 27 «days»

Award [2] for correct final answer.

[2 marks]

c.ii.

Anode (negative electrode): C₂H₅OH + H₂O → CH₃COOH + 4H⁺ + 4e^–

Cathode (positive electrode): O₂ + 4H⁺ + 4e^– → 2H₂O

[2 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Syllabus sections

Options » D: Medicinal chemistry » D.9 Drug detection and analysis (HL only)

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