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Date	November 2017	Marks available	2	Reference code	17N.3.hl.TZ0.21
Level	HL	Paper	3	Time zone	TZ0
Command term		Question number	21	Adapted from	N/A

Question

Lutetium-177 is used in radiotherapy. It emits beta radiation when it decays.

State a nuclear equation to show the decay of lutetium-177.

[2]

a.

The half-life of lutetium-177 is 6.73 days. Determine the percentage of a sample of lutetium-177 remaining after 14.0 days.

[2]

b.

Explain the low environmental impact of most medical nuclear waste.

[2]

c.

Markscheme

${}_{71}^{177}{\text{Lu}} \to {}_{71}^{177}{\text{Hf}} + {}_{ - 1}^0{\text{e}}$ ${}_{71}^{177}{\text{Lu}} \to {}_{71}^{177}{\text{Hf}} + {}_{ - 1}^0{\text{e}}$ «+ $\nu$ $\nu$ »

Hf

correct A and Z AND beta product

Accept “β/ β^–/e/e^–” for “ ${}_{-1}^{~0}{\text{e}}$ ${}_{-1}^{~0}{\text{e}}$ ”.

Accept “¹⁷⁷Lu → ¹⁷⁷Hf + e^– «+ $\nu$ $\nu$ »”.

a.

number of half-lives = $\frac{t}{{{t_{^{\frac{1}{2}}}}}}$ $\frac{t}{{{t_{^{\frac{1}{2}}}}}}$ = 2.08

OR

$\frac{{N\left( t \right)}}{{{N_0}}} = {0.5^{\frac{{14.0}}{{6.73}}}}$ $\frac{{N\left( t \right)}}{{{N_0}}} = {0.5^{\frac{{14.0}}{{6.73}}}}$

OR

$\lambda =$ $\lambda =$ « $\frac{{{\text{ln}}2}}{{{t_{^{\frac{1}{2}}}}}} = \frac{{{\text{ln}}2}}{{6.73}} =$ $\frac{{{\text{ln}}2}}{{{t_{^{\frac{1}{2}}}}}} = \frac{{{\text{ln}}2}}{{6.73}} =$ » 0.103 «day^–1»

OR

$\frac{{N\left( t \right)}}{{{N_0}}} = {e^{ - 0.103\, \times 14.0}}$ $\frac{{N\left( t \right)}}{{{N_0}}} = {e^{ - 0.103\, \times 14.0}}$

23.6 «%»

Award [2] for correct final answer.

b.

Any two of:

emits weak ionising radiation

OR

low activity/radioactivity

can be stored until material becomes inactive AND then disposed with normal waste

«isotopes» have short lives

OR

exist for a short period of time

Award [1 max] for “low-level waste/LLW”.

[Max 2 Marks]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

Syllabus sections

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

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17M.3.hl.TZ1.29a:
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20N.3.hl.TZ0.17d:
Technetium-99m has a half-life of $6.03$ $6.03$ hours. Calculate the amount of $1.00 \times 10^{- 11} mol$ $1.00 \times 10^{- 11} mol$ of technetium-99m remaining after $48.0$ $48.0$ hours.
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19M.3.hl.TZ1.24b(ii):
Outline the disposal of LLW.
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18N.3.hl.TZ0.22b.ii:
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19N.3.hl.TZ0.27c:
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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.
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Determine the percentage of technetium-99m remaining after 24.0 hours.
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18N.3.hl.TZ0.22a:
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Explain how TAT is relatively safe to use in the treatment of dispersed cancers.
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Options » D: Medicinal chemistry