IChO 1993 · Theory problem 1

Radioactive iodine-131 in nuclear medicine

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\(\ce{^{131}I}\) is a radioactive isotope of iodine (\(\ce{e-}\) emitter) used in nuclear medicine for analytical procedures to determine thyroid endocrine disorders by scintigraphy. The decay rate constant, \(k\), of \(\ce{^{131}I}\) is \(9.93\times10^{-7}\ \mathrm{s^{-1}}\).

Questions:

1.1 Write the decay reaction of \(\ce{^{131}I}\).

1.2 Calculate the half-life of \(\ce{^{131}I}\) expressed in days.

1.3 Calculate the time necessary (expressed in days) for a sample of \(\ce{^{131}I}\) to reduce its activity to 30 % of the original value.

1.4 Knowing that a Geiger counter detects activities of the order of \(10^{-4}\ \mu\mathrm{c}\), calculate the minimum amount of \(\ce{^{131}I}\) (in grams) which could be detected by this counter.

1 Curie (c) is the amount of a radioisotope that produces \(3.7\times10^{10}\) disintegrations \(\mathrm{s^{1}}\).

Official English text of the problem.

Answers

  • 1.1 \[ \ce{^{131}_{53}I -> ^{131}_{54}Xe + ^{0}_{-1}e}+\bar{\nu}_e. \] The daughter is xenon-131 (atomic SMILES: [131Xe]).
  • 1.2 \(t_{1/2}=\boxed{8.08\ \text{days}}\).
  • 1.3 \(t=\boxed{14.0\ \text{days}}\).
  • 1.4 \(m_{\min}\approx\boxed{8.1\times10^{-16}\ \mathrm{g}}\).

Solution

1.1

In beta-minus decay, a neutron in the nucleus becomes a proton, emitting an electron and an electron antineutrino, \(\bar{\nu}_e\). The mass number therefore remains unchanged, while the atomic number increases by one.

Iodine has atomic number 53, so the daughter has atomic number 54 and is xenon:

\[ \boxed{\ce{^{131}_{53}I -> ^{131}_{54}Xe + ^{0}_{-1}e}+\bar{\nu}_e}. \]

The mass numbers balance as \(131=131+0\), and the nuclear charges balance as \(53=54-1\). The antineutrino has zero electric charge and zero nucleon number.

The isotope atoms are iodine-131, [131I], and xenon-131, [131Xe]; these atomic SMILES identify the isotopes, while the equation describes the nuclear transformation.

1.2

Radioactive decay follows the first-order law

\[ -\frac{dN}{dt}=kN, \qquad N(t)=N_0e^{-kt}, \]

where \(N(t)\) is the number of undecayed nuclei. At the half-life, \(N/N_0=1/2\), giving

\[ t_{1/2}=\frac{\ln 2}{k}. \]

Thus,

\[ t_{1/2} =\frac{0.693147}{9.93\times10^{-7}\ \mathrm{s^{-1}}} =6.9803\times10^5\ \mathrm{s}. \]

Using \(1\ \text{day}=86400\ \mathrm{s}\),

\[ \boxed{t_{1/2}=8.08\ \text{days}}. \]

1.3

The activity is \(A=kN\), so it decreases by the same fraction as the number of radioactive nuclei:

\[ \frac{A(t)}{A_0}=e^{-kt}. \]

For an activity equal to \(30\%\) of the original value,

\[ 0.30=e^{-kt}, \qquad t=\frac{-\ln(0.30)}{k}. \]

Therefore,

\[ t =\frac{1.203973}{9.93\times10^{-7}\ \mathrm{s^{-1}}} =1.21246\times10^6\ \mathrm{s}, \]

and

\[ \boxed{t=\frac{1.21246\times10^6}{86400}\ \text{days} =14.0\ \text{days}}. \]

1.4

Convert the minimum detectable activity to disintegrations per second:

\[ A_{\min} =10^{-4}\ \mu\mathrm{c} =10^{-10}\ \mathrm{c}. \]

Using the given curie conversion,

\[ A_{\min} =10^{-10}\times3.7\times10^{10}\ \mathrm{s^{-1}} =3.7\ \mathrm{s^{-1}} =3.7\ \mathrm{Bq}. \]

Since \(A=kN\), the corresponding number of radioactive nuclei is

\[ N_{\min} =\frac{A_{\min}}{k} =\frac{3.7\ \mathrm{s^{-1}}}{9.93\times10^{-7}\ \mathrm{s^{-1}}} =3.726\times10^6. \]

Taking the molar mass of iodine-131 as \(M\approx131\ \mathrm{g\,mol^{-1}}\) and using Avogadro’s constant \(N_{\mathrm A}=6.022\times10^{23}\ \mathrm{mol^{-1}}\),

\[ m_{\min} =\frac{N_{\min}}{N_{\mathrm A}}M =\frac{3.726\times10^6}{6.022\times10^{23}\ \mathrm{mol^{-1}}} \left(131\ \mathrm{g\,mol^{-1}}\right). \]

Hence,

\[ \boxed{m_{\min}\approx8.1\times10^{-16}\ \mathrm{g}}. \]

The solution was checked against the official solution.