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Date May 2017 Marks available 1 Reference code 17M.3.hl.TZ1.29
Level HL Paper 3 Time zone TZ1
Command term Suggest 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]
a.

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]
b.

Calculate the rate constant, λ , 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]
d.

Markscheme

90Y → 90Zr + β

 

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

[1 mark]

a.

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]

b.

λ ( = ln 2 t 1 2 = 0.693 8.02   d a y ) = 8.64 × 10 2 / 0.0864  «day−1»

[1 mark]

c.i.

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

N «= 100 – 90» = 10% at time t

« ln ( 100 10 ) = 2.303 = 0.0864 t »

« t = 2.303 0.0864  da y 1 = » 26.7 «days»

 

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

 

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

10 = 100(0.5)n

« log ( 1 10 ) = n × log 0.5 »

« 1 = n ( 0.301 ) / n = 1 0.301 »

«t = 1 0.301 × 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]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.

Syllabus sections

Options » D: Medicinal chemistry » D.8 Nuclear medicine (HL only)
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