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Date	May 2017	Marks available	1	Reference code	17M.3.hl.TZ2.26
Level	HL	Paper	3	Time zone	TZ2
Command term	Identify	Question number	26	Adapted from	N/A

Question

Targeted Alpha Therapy (TAT) is a technique that involves using alpha-radiation to treat leukemia and other dispersed cancers.

Yttrium-90 and lutetium-177 are used in radiotherapy.

Explain why alpha-radiation is particularly suitable for this treatment.

[2]

a.i.

Outline how the alpha-radiation in TAT is directed to cancer cells.

[1]

a.ii.

Identify the type of radiation emitted by these two radioisotopes.

[1]

b.i.

State an equation for the one-step decay of yttrium-90.

[1]

b.ii.

The half-life of lutetium-177 is 6.75 days. Calculate the percentage remaining after 27 days.

[1]

b.iii.

Markscheme

more damaging than other radiation types
OR
very damaging to «cancer» cells
OR
high ionizing density «of alpha particles»

absorbed within a very short range of emission
OR
causes little damage to surrounding tissues

Accept “high ionizing power «of alpha particles»” for M1.

Accept “low penetrating power «of alpha particles»” for M2.

[2 marks]

a.i.

«radioactive isotope/radionuclide/alpha-emitter» administered using carrier drug/protein/antibodies

[1 mark]

a.ii.

beta/β «radiation»

[1 mark]

b.i.

${}_{39}^{90}Y \to {}_{40}^{90}Zr + \beta$

Accept " $_{ - 1}^{\;\,0}e/e/{e^ - }$ " OR " $_{ - 1}^{\;\,0}\beta /{\beta ^ - }$ "

Accept ECF from (b) (i) if incorrect radiation identified, eg, $_{39}^{90}Y \to _{37}^{86}Rb + _2^4He$

[1 mark]

b.ii.

ALTERNATIVE 1:
«4 half-lives»
6.25 «%»

ALTERNATIVE 2:
«N_t = N₀ ${\left( {0.5} \right)^{\frac{t}{{{t_{1/2}}}}}} = 100{\left( {0.5} \right)^{\frac{{27}}{{6.75}}}} =$ » 6.25 «%»

[1 mark]

b.iii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

Syllabus sections

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

Show 33 related questions

17M.3.hl.TZ1.29b:
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.
17M.3.hl.TZ2.26a.ii:
Outline how the alpha-radiation in TAT is directed to cancer cells.
16N.3.hl.TZ0.28b: Molybdenum-99 has a half-life of 66 hours, while technetium-99m has a half-life of 6 hours....
16N.3.hl.TZ0.28c: Outline two reasons, other than its half-life, why technetium-99m is so useful in medical...
18M.3.hl.TZ1.19b:
Outline how Targeted Alpha Therapy (TAT) is used for treating cancers that have spread throughout the body.
17M.3.hl.TZ1.29a:
Yttrium-90 is used in treating certain cancers.

Formulate a nuclear equation for the beta decay of yttrium-90.
20N.3.hl.TZ0.17d:
Technetium-99m has a half-life of $6.03$ hours. Calculate the amount of $1.00 \times 10^{- 11} mol$ of technetium-99m remaining after $48.0$ hours.
19M.3.hl.TZ1.24b(ii):
Outline the disposal of LLW.
19N.3.hl.TZ0.27a: State two common side effects of radiotherapy.
17M.3.hl.TZ2.26b.iii:
The half-life of lutetium-177 is 6.75 days. Calculate the percentage remaining after 27 days.
18N.3.hl.TZ0.22b.ii:
Suggest why the percentage of technetium-99m remaining in the human body two days after injection will be lower than that calculated in (b)(i).
18M.3.hl.TZ2.26a:
Phosphorous-32 undergoes beta decay. Formulate a balanced nuclear equation for this process.
17M.3.hl.TZ2.26a.i:
Explain why alpha-radiation is particularly suitable for this treatment.
19N.3.hl.TZ0.27b: Explain why technetium-99m is the most common radioisotope used in nuclear medicine.
17M.3.hl.TZ1.29c.i:
Calculate the rate constant, $\lambda$ , in day⁻¹, for the decay of iodine-131 using section 1 of the data booklet.
17M.3.hl.TZ2.26b.ii:
State an equation for the one-step decay of yttrium-90.
19N.3.hl.TZ0.27c:
25.0 μg of iodine-131, with a half-life of 8.00 days, was left to decay.

Calculate the mass of iodine-131, in μg, remaining after 32.0 days. Use section 1 of the data booklet.
17N.3.hl.TZ0.21b:
The half-life of lutetium-177 is 6.73 days. Determine the percentage of a sample of lutetium-177 remaining after 14.0 days.
18M.3.hl.TZ2.26b:
The half-life of phosphorus-32 is 14.3 days. Calculate the mass, in g, of ³²P remaining after 57.2 days if the initial sample contains 2.63 × 10⁻⁸ mol. Use table 1 of the data booklet and M_r = 31.97 g mol⁻¹.
19M.3.hl.TZ1.24a(i):
Determine the percentage of technetium-99m remaining after 24.0 hours.
19M.3.hl.TZ2.26a:
Evaluate the suitability of technetium-99m for this use.
18M.3.hl.TZ1.19a:
Describe how ionizing radiation destroys cancer cells.
18N.3.hl.TZ0.22a:
Alpha particles are more damaging to human cells than any other nuclear radiation and yet they are used in targeted alpha therapy (TAT).

Explain how TAT is relatively safe to use in the treatment of dispersed cancers.
17M.3.hl.TZ1.29c.ii:
Calculate the time, in days, for 90% of the sample to decay.
18M.3.hl.TZ2.26c:
Explain the targeted alpha therapy (TAT) technique and why it is useful.
19M.3.hl.TZ2.26b:
Calculate the percentage of technetium-99m remaining after 10.0 hours. Use section 1 of the data booklet.
16N.3.hl.TZ0.28a:
Deduce equations for the following nuclear reactions:

(i) Molybdenum-98 absorbs a neutron.

(ii) The isotope produced in (a) (i) decays into technetium-99m.
17N.3.hl.TZ0.21a:
State a nuclear equation to show the decay of lutetium-177.
19M.3.hl.TZ1.24a(ii):
Technetium-99 decays further, emitting beta radiation. Formulate the equation for the decay of technetium-99.
19M.3.hl.TZ1.24c:
Magnetic resonance imaging (MRI) is an application of NMR technology using radiowaves.

Suggest why MRI is much less dangerous than imaging techniques such as X-rays and radiotracers. Use section 3 of the data booklet.
18N.3.hl.TZ0.22b.i:
Technetium-99m ( ${}_{43}^{99{\text{m}}}{\text{Tc}}$ ) has a half-life of 6.0 hours. Calculate the percentage of ${}_{43}^{99{\text{m}}}{\text{Tc}}$ remaining in a sample of the radioisotope after two days.
20N.3.hl.TZ0.17b: Discuss the properties that make a radioisotope suitable for diagnosis.
20N.3.hl.TZ0.17a:
State the type of radiation technetium-99m emits.

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Options » D: Medicinal chemistry

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