Question
Nuclear medicine can be used to diagnose and treat diseases.
(a) Deduce the nuclear equation for the beta decay of cobalt-60.
(b) Explain how Targeted Alpha Therapy (TAT) works and why it is used for treating cancers that have spread throughout the body.
Answer/Explanation
Answer:
(a) \(^{60}_{27} Co \rightarrow ^{60}_{28} Ni + ^0_{-1} \beta \)
(b) radiation source delivered directly to cancer cells
by a carrier drug/protein/antibody
several sites in body can be targeted «at same time»
Question
Two radiation treatments for cancer are Targeted Alpha Therapy, TAT, and Boron Neutron Capture Therapy, BNCT.
(a) Describe which cancers are treated by TAT and BNCT and the particles used in each treatment.
TAT:
BNCT:
(b) Suggest two reasons why technetium-99m is the most commonly used radioisotope in nuclear medicine.
(c) A freshly prepared solution for radiodiagnostics contains \(1.0 × 10^{-7}\) mol \(dm^{-3}\) of technetium-99m. Determine the concentration, in mol \(dm^{-3}\), of technetium-99m remaining in the solution 22.00 hours later. Use section 1 of the data booklet.
Half-life of technetium-99m = 6.01h
Answer/Explanation
Answer:
(a) TAT:
spread to multiple sites AND alpha particles
OR
«cancers of the» blood/leukaemia AND alpha particles
BNCT:
head/neck/brain «cancers» AND neutrons
(b) Any two of:
gamma radiation AND easily traced
weak/low-energy/low-frequency «gamma» radiation «low risk to patient»
short half-life «low risk to patient»
binds to range of biologically-active substances
«gamma-radiation of approximately» same frequency as X-rays «so can be
detected using existing X-ray equipment»
easily obtained
(c) «\(N_t = N_0(0.5)^{t/t_{1/2}}\) =» \(1.0 × 10^{–7}\) mol \(dm^{–3}\) × (0.5)22.00/6.01
«Nt =» \(7.9 × 10^{–9}\) «mol \(dm^{–3}\)»
Question
Technetium-99m is the most commonly used isotope for diagnostic medicine.
a. State the type of radiation technetium-99m emits.
b. Discuss the properties that make a radioisotope suitable for diagnosis.
c. Describe the proper disposal of low-level radioactive waste in hospitals.
d. 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.
▶️Answer/Explanation
Markscheme
a. gamma/ $/$
b. Any three of:
«easily» detected/traced
$O R$
“gamma-radiation of approximately» same frequency as X-rays «so can be detected using existing X-ray equipment»
short/intermediate half-life «hence does not remain in body for long time»
weak ionizing radiation «less harmful»
OR
low amount of radiation produced «so less harmful»
$O R$
energy of photons is low
form «variety of» compounds that are absorbed by «different» organs
OR
“chemically» binds to many biologically active compounds
excreted quickly «from body»
c. store until material becomes inactive/radiation levels drop
dispose with other waste
OR
dispose in landfills
Only award M2 if M1 correct.
Accept “dispose by incineration” for M2.
d. Alternative 1:
$$
\begin{aligned}
& « N=N_0(0.5) \frac{t}{t_{1 / 2}}=» 1.00 \times 10^{-11} \times(0.5) \frac{48.0}{6.03} \\
& « \mathrm{~N}=» 4.02 \times 10^{-14}\langle\mathrm{~mol} » \checkmark
\end{aligned}
$$
Alternative 2:
$$
\begin{aligned}
& \ll \lambda=\frac{\ln 2}{6.03}=» 0.115 \ll \mathrm{hr}^{-1} » \\
& « N=N_0 e^{-\lambda t}=1.00 \times 10^{-11} \times e^{-0.115 \times 48}=» 4.01 \times 10^{-14}\langle\mathrm{~mol} »
\end{aligned}
$$
Award [2] for correct final answer.