IChO 1999 · Theory problem 3

The thorium decay series

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One of naturally occurring radioactive decay series begins with \(\ce{^{232}_{90}Th}\) and ends with a stable \(\ce{^{208}_{82}Pb}\).

3.1 How many beta (\(\beta^-\)) decays are there in this series? Show by calculation.

3.2 How much energy in MeV is released in the complete chain?

3.3 Calculate the rate of production of energy (power) in watts (\(1\ \mathrm{W}=\mathrm{J\,s^{-1}}\)) produced by 1.00 kilogram of \(\ce{^{232}Th}\) (\(t_{½}=1.40\times10^{10}\) years).

3.4 \(\ce{^{228}Th}\) is a member of the thorium series. What volume in \(\mathrm{cm^3}\) of helium at \(0\ ^\circ\mathrm{C}\) and 1 atm collected when 1.00 gram of \(\ce{^{228}Th}\) (\(t_{1/2}=1.91\) years) is stored in a container for 20.0 years. The half-lives of all intermediate nuclides are short compared to the half-life of \(\ce{^{228}Th}\).

3.5 One member of thorium series, after isolation, is found to contain \(1.50\times10^{10}\) atoms of the nuclide and decays at the rate of 3440 disintegrations per minute. What is the half-life in years?

The necessary atomic masses are :

\[ \ce{^{4}_{2}He}=4.00260\ \mathrm{u},\quad \ce{^{208}_{82}Pb}=207.97664\ \mathrm{u},\quad \ce{^{232}_{90}Th}=232.03805\ \mathrm{u}; \quad\text{and }1\mathrm{u}=931.5\ \mathrm{MeV} \]

\[ 1\ \mathrm{MeV}=1.602\times10^{-13}\ \mathrm{J} \]

\[ N_A=6.022\times10^{23}\ \mathrm{mol^{-1}} \]

The molar volume of an ideal gas at \(0\ ^\circ\mathrm{C}\) and 1 atm is \(22.4\ \mathrm{dm^3\,mol^{-1}}\).

Official English text of the problem.

Answers

  • 3.1: \(4\) beta-minus decays.
  • 3.2: \(42.67\ \mathrm{MeV}\) per complete decay chain.
  • 3.3: \(2.79\times10^{-5}\ \mathrm{W}\), assuming secular equilibrium of the complete chain.
  • 3.4: \(4.91\times10^2\ \mathrm{cm^3}\) of helium.
  • 3.5: \(5.75\ \mathrm{years}\).

Solution

3.1

An alpha decay decreases the mass number by 4 and the atomic number by 2. A beta-minus decay leaves the mass number unchanged and increases the atomic number by 1.

Let \(a\) and \(b\) be the numbers of alpha and beta-minus decays, respectively. Conservation of mass number gives

\[ 232=208+4a \quad\Longrightarrow\quad a=6. \]

The atomic-number change therefore satisfies

\[ 82=90-2(6)+b \quad\Longrightarrow\quad \boxed{b=4}. \]

The overall nuclear equation is

\[ \ce{^{232}_{90}Th -> ^{208}_{82}Pb + 6 ^{4}_{2}He + 4 ^{0}_{-1}e} +4\bar{\nu}_e, \]

where \(\bar{\nu}_e\) denotes an electron antineutrino.

3.2

Using the given neutral-atom masses, the mass defect is

\[ \begin{aligned} \Delta m &=M(\ce{^{232}Th})-M(\ce{^{208}Pb})-6M(\ce{^{4}He})\\ &=232.03805-207.97664-6(4.00260)\\ &=0.04581\ \mathrm{u}. \end{aligned} \]

No separate beta-electron mass term is needed: the electron masses cancel when the nuclear mass balance is expressed using neutral-atom masses.

Consequently,

\[ Q=\Delta m c^2 =(0.04581\ \mathrm{u}) (931.5\ \mathrm{MeV\,u^{-1}}) =\boxed{42.67\ \mathrm{MeV}}. \]

In joules, retaining guard digits,

\[ Q=(42.672015)(1.602\times10^{-13}) =6.83606\times10^{-12}\ \mathrm{J}. \]

3.3

Using the full-chain energy assumes that the daughter nuclides are in secular equilibrium: the number of completed chains per second then equals the activity of the thorium-232.

The number of thorium-232 atoms is

\[ N= \frac{1000\ \mathrm{g}}{232.03805\ \mathrm{g\,mol^{-1}}} (6.022\times10^{23}\ \mathrm{mol^{-1}}) =2.59526\times10^{24}. \]

Taking one year as 365 days,

\[ t_{1/2} =(1.40\times10^{10})(365)(24)(3600) =4.41504\times10^{17}\ \mathrm{s}. \]

The decay constant and activity are

\[ \lambda=\frac{\ln2}{t_{1/2}}, \]

\[ A=\lambda N =\frac{(\ln2)(2.59526\times10^{24})} {4.41504\times10^{17}\ \mathrm{s}} =4.07448\times10^6\ \mathrm{s^{-1}}. \]

Thus the total rate of nuclear energy release is

\[ \begin{aligned} P&=AQ\\ &=(4.07448\times10^6\ \mathrm{s^{-1}}) (6.83606\times10^{-12}\ \mathrm{J})\\ &=\boxed{2.79\times10^{-5}\ \mathrm{W}}. \end{aligned} \]

3.4

The decrease in mass number from 228 to 208 requires

\[ a=\frac{228-208}{4}=5 \]

alpha decays per completed chain. Atomic-number balance also gives two beta-minus decays, so the overall nuclear equation is

\[ \ce{^{228}_{90}Th -> ^{208}_{82}Pb + 5 ^{4}_{2}He + 2 ^{0}_{-1}e} +2\bar{\nu}_e. \]

Each alpha particle ultimately becomes one helium-4 atom, \(\ce{^{4}He}\), SMILES [4He]. Thus each completed chain produces five helium atoms.

The fraction of thorium-228 that decays during 20.0 years is

\[ f=1-\exp(-\lambda t) =1-2^{-t/t_{1/2}} =1-2^{-20.0/1.91} =0.9992955. \]

Because the intermediate half-lives are short, we approximate each decayed thorium-228 atom as having completed its chain. Using \(M(\ce{^{228}Th})\approx228\ \mathrm{g\,mol^{-1}}\),

\[ n_{\mathrm{He}} =5\left(\frac{1.00\ \mathrm{g}} {228\ \mathrm{g\,mol^{-1}}}\right)(0.9992955) =0.0219144\ \mathrm{mol}. \]

At the specified temperature and pressure,

\[ \begin{aligned} V_{\mathrm{He}} &=(0.0219144\ \mathrm{mol}) (22.4\times10^3\ \mathrm{cm^3\,mol^{-1}})\\ &=490.882\ \mathrm{cm^3}\\ &=\boxed{4.91\times10^2\ \mathrm{cm^3}}. \end{aligned} \]

3.5

For an isolated radionuclide, the activity is \(A=\lambda N\). Therefore,

\[ t_{1/2}=\frac{\ln2}{\lambda} =\frac{N\ln2}{A}. \]

Substitution gives

\[ t_{1/2} =\frac{(1.50\times10^{10})(\ln2)} {3440\ \mathrm{min^{-1}}} =3.02244\times10^6\ \mathrm{min}. \]

Converting to years,

\[ t_{1/2} =\frac{3.02244\times10^6} {60\times24\times365}\ \mathrm{years} =\boxed{5.75\ \mathrm{years}}. \]

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.