Date | May 2017 | Marks available | 2 | Reference code | 17M.3.hl.TZ1.29 |
Level | HL | Paper | 3 | Time zone | TZ1 |
Command term | Calculate | 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.
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.
Calculate the rate constant, \(\lambda \), in day−1, for the decay of iodine-131 using section 1 of the data booklet.
Calculate the time, in days, for 90% of the sample to decay.
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.
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]
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:
«Nt = 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]
Anode (negative electrode): C2H5OH + H2O → CH3COOH + 4H+ + 4e–
Cathode (positive electrode): O2 + 4H+ + 4e– → 2H2O
[2 marks]