IChO 2001 · Theory problem 4

Thorium, uranium and lead in monazite

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Figure 1

Beach Sand Mineral in Kerala

Beach sand mineral, monazite, is a rich source of thorium, available in large quantities in the state of Kerala in India. A typical monazite sample contains about 9 % \(\ce{ThO2}\) and 0.35 % \(\ce{U3O8}\). \(\ce{^{208}Pb}\) a \(\ce{^{206}Pb}\) are the stable end-products in the radioactive decay series of \(\ce{^{232}Th}\) and \(\ce{^{238}U}\), respectively. All the lead (\(\ce{Pb}\)) found in monazite is of radiogenic origin.

The isotopic atom ratio \(\ce{^{208}Pb}/\ce{^{232}Th}\), measured mass spectrometrically, in a monazite sample was found to be 0.104. The half-lives of \(\ce{^{232}Th}\) and \(\ce{^{238}U}\) \(1.41\times10^{10}\) years and \(4.47\times10^9\) years, respectively. Assume that \(\ce{^{208}Pb}\), \(\ce{^{206}Pb}\), \(\ce{^{232}Th}\) and \(\ce{^{238}U}\) remained entirely in the monazite sample since the formation of monazite mineral.

4.1 Calculate the age (time elapsed since its formation) of the monazite sample.

4.2 Estimate the isotopic atom ratio \(\ce{^{206}Pb}/\ce{^{238}U}\) in the monazite sample.

4.3 Thorium-232 is a fertile material for nuclear energy. In thermal neutron irradiation, it absorbs a neutron and the resulting isotope forms \(\ce{^{233}U}\) by successive \(\beta^-\) decays. Write the nuclear reactions for the formation of \(\ce{^{233}U}\) from \(\ce{^{232}Th}\).

In nuclear fission of \(\ce{^{233}U}\) a complex mixture of radioactive fission products is formed. The fission product \(\ce{^{101}Mo}\) initially undergoes radioactive decay as shown below:

Figure 2

4.4 A freshly prepared radiochemically pure sample of \(\ce{^{101}Mo}\) contains 5000 atoms of \(\ce{^{101}Mo}\) initially. How many atoms of

i) \(\ce{^{101}Mo}\)

ii) \(\ce{^{101}Tc}\)

iii) \(\ce{^{101}Ru}\)

will be present in the sample after 14.6 min?

Official English text of the problem.

Answers

  • 4.1: \(\boxed{2.01\times10^9\ \text{years}}\).
  • 4.2: \(\boxed{N(\ce{^{206}Pb})/N(\ce{^{238}U})=0.366}\).
  • 4.3: Neutron capture produces thorium-233, which undergoes two successive \(\beta^-\) decays through protactinium-233 to uranium-233: \[ \ce{^{232}_{90}Th + ^1_0n -> ^{233}_{90}Th + \gamma} \] \[ \ce{^{233}_{90}Th -> ^{233}_{91}Pa + ^0_{-1}e}+\bar{\nu}_e \] \[ \ce{^{233}_{91}Pa -> ^{233}_{92}U + ^0_{-1}e}+\bar{\nu}_e \]
  • 4.4: Expected populations, rounded to whole atoms: (i) 2500 atoms of \(\ce{^{101}Mo}\); (ii) 1720 atoms of \(\ce{^{101}Tc}\); (iii) 780 atoms of \(\ce{^{101}Ru}\).

Solution

4.1 — Age of the monazite

Take the radiogenic lead to have accumulated since mineral formation, with no initial lead and negligible accumulation of intermediate decay-chain nuclides. Each parent nucleus that completes its decay chain produces one lead nucleus.

Let \(P_0\) be the initial number of parent atoms, \(P\) the present number, and \(D\) the number of radiogenic daughter atoms. Radioactive decay gives

\[ P=P_0e^{-\lambda t}, \qquad D=P_0-P, \qquad \lambda=\frac{\ln 2}{t_{1/2}}. \]

Consequently,

\[ \frac{D}{P} =\frac{P_0-P}{P} =e^{\lambda t}-1, \]

and hence

\[ t=\frac{t_{1/2}}{\ln 2} \ln\left(1+\frac{D}{P}\right). \]

For the thorium-232–lead-208 system,

\[ t= \frac{1.41\times10^{10}\ \mathrm{years}}{\ln 2} \ln(1+0.104) =2.0126364\times10^9\ \mathrm{years}. \]

Thus,

\[ \boxed{t=2.01\times10^9\ \mathrm{years}}. \]

4.2 — Lead-206/uranium-238 atom ratio

The uranium system has aged for the same time. Therefore,

\[ \frac{N(\ce{^{206}Pb})}{N(\ce{^{238}U})} =e^{\lambda_{\mathrm U}t}-1, \qquad \lambda_{\mathrm U} =\frac{\ln 2}{4.47\times10^9\ \mathrm{years}}. \]

Using the unrounded age from 4.1,

\[ \frac{N(\ce{^{206}Pb})}{N(\ce{^{238}U})} = \exp\left( \frac{\ln 2\,(2.0126364\times10^9\ \mathrm{years})} {4.47\times10^9\ \mathrm{years}} \right)-1 =0.366281. \]

Hence,

\[ \boxed{\frac{N(\ce{^{206}Pb})}{N(\ce{^{238}U})}=0.366}. \]

The oxide mass percentages are unnecessary for these calculations because the required quantities are daughter/parent atom ratios.

4.3 — Formation of uranium-233

Neutron capture increases the mass number by one without changing the atomic number. Thorium-232 therefore forms thorium-233:

\[ \boxed{\ce{^{232}_{90}Th + ^1_0n -> ^{233}_{90}Th + \gamma}}. \]

In each \(\beta^-\) decay, a neutron converts into a proton, emitting an electron and an electron antineutrino. The mass number remains unchanged, while the atomic number increases by one.

Thus, thorium-233 (\(Z=90\)) first forms protactinium-233 (\(Z=91\)):

\[ \boxed{\ce{^{233}_{90}Th -> ^{233}_{91}Pa + ^0_{-1}e}+\bar{\nu}_e}. \]

Protactinium-233 then forms uranium-233 (\(Z=92\)):

\[ \boxed{\ce{^{233}_{91}Pa -> ^{233}_{92}U + ^0_{-1}e}+\bar{\nu}_e}. \]

Here \(\bar{\nu}_e\) denotes the electron antineutrino. Each equation conserves mass number and electric charge.

4.4 — Successive decay of molybdenum-101 and technetium-101

Let \(N_1,N_2,N_3\) denote the numbers of \(\ce{^{101}Mo}\), \(\ce{^{101}Tc}\), and \(\ce{^{101}Ru}\) atoms, respectively. Initially,

\[ N_1(0)=N_0=5000, \qquad N_2(0)=N_3(0)=0. \]

The decay constants are

\[ \lambda_1=\frac{\ln 2}{14.6\ \mathrm{min}} =0.0474758\ \mathrm{min^{-1}}, \]

\[ \lambda_2=\frac{\ln 2}{14.3\ \mathrm{min}} =0.0484718\ \mathrm{min^{-1}}. \]

(i) Molybdenum-101

After one molybdenum half-life,

\[ N_1(14.6\ \mathrm{min}) =N_0e^{-\lambda_1t} =5000\left(\frac12\right). \]

Thus,

\[ \boxed{N_1=2500\ \text{atoms}}. \]

(ii) Technetium-101

Technetium is simultaneously produced by molybdenum decay and consumed by its own decay:

\[ \frac{dN_2}{dt} =\lambda_1N_1-\lambda_2N_2 =\lambda_1N_0e^{-\lambda_1t}-\lambda_2N_2. \]

With \(N_2(0)=0\), integration gives

\[ N_2(t) = N_0\frac{\lambda_1}{\lambda_2-\lambda_1} \left(e^{-\lambda_1t}-e^{-\lambda_2t}\right). \]

At \(t=14.6\ \mathrm{min}\), this becomes

\[ N_2 = 5000\frac{14.3}{14.6-14.3} \left(\frac12-2^{-14.6/14.3}\right) =1720.3295. \]

Therefore,

\[ \boxed{N_2\approx1720\ \text{atoms}}. \]

(iii) Ruthenium-101

Each decay step preserves one heavy nucleus, and ruthenium is the terminal product in the given chain. Thus,

\[ N_1+N_2+N_3=N_0. \]

Therefore,

\[ N_3=5000-2500-1720.3295=779.6705, \]

so

\[ \boxed{N_3\approx780\ \text{atoms}}. \]

These are expected populations; individual radioactive decays are stochastic.

The solution was checked against the official solution.