IChO 1994 · Theory problem 8

Radioactive decay of uranium and ruthenium-97

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Avogadro's Number: \(6.022\cdot10^{23}\)

8.1 An atom of \(\ce{^{238}U}\) disintegrates by a series of \(\alpha\)-decays and \(\beta^-\)-decays until it becomes \(\ce{^{206}Pb}\), which is stable.

i) How many \(\alpha\)-decays and how many \(\beta^-\)-decays does an atom starting as \(\ce{^{238}U}\) undergo before it becomes stable?

ii) One of the following ten nuclides is formed from a series of disintegrations starting at \(\ce{^{238}U}\). Which one ?

\(\ce{^{235}U}\), \(\ce{^{234}U}\), \(\ce{^{228}Ac}\), \(\ce{^{224}Ra}\), \(\ce{^{224}Rn}\), \(\ce{^{220}Rn}\), \(\ce{^{215}Po}\), \(\ce{^{212}Po}\), \(\ce{^{212}Pb}\), \(\ce{^{211}Pb}\).

8.2 In a thermal neutron-induced fission process, \(\ce{^{235}U}\) reacts with a neutron and breaks up into energetic fragments and (normally) 2-3 new neutrons.

We consider one single fission event:

Figure 1

Identify the fragment X.

8.3 The half-life of \(\ce{^{238}U}\) is \(4.5\times10^9\) years, the half-life of \(\ce{^{235}U}\) is \(7.0\times10^8\) years. Natural uranium consists of 99.28 % \(\ce{^{238}U}\) and 0.72 % \(\ce{^{235}U}\).

i) Calculate the ratio in natural U between the disintegration rates of these two uranium isotopes.

ii) A mineral contains 50 weight percent uranium. Calculate the disintegration rate of \(\ce{^{238}U}\) in 1.0 kg of this mineral.

8.4 We have the following radioactive sequence:

Figure 2

Halflives: \(\ce{^{97}Ru}\): 2.7 days; \(\ce{^{97}Tc}\): \(2.6\times10^6\) years

At \(t=0\) a radioactive source containing only \(\ce{^{97}Ru}\) has a disintegration rate of \(1.0\times10^9\) Bq.

i) What is the total disintegration rate of the source at \(t=6.0\) days?

ii) What is the total disintegration rate of the source at \(t=6000\) years?

Official English text of the problem.

Answers

  • 8.1(i): \(8\) \(\alpha\)-decays and \(6\) \(\beta^-\)-decays.
  • 8.1(ii): Uranium-234, \(\ce{^{234}U}\).
  • 8.2: Zirconium-97, \(\ce{^{97}_{40}Zr}\).
  • 8.3(i): \(R_{238}/R_{235}\approx21\), or approximately \(21:1\).
  • 8.3(ii): \(R_{238}=6.1\times10^6\,\mathrm{Bq}\).
  • 8.4(i): \(R_{\mathrm{total}}(6.0\,\mathrm{days})=2.1\times10^8\,\mathrm{Bq}\).
  • 8.4(ii): \(R_{\mathrm{total}}(6000\,\mathrm{years})=2.8\,\mathrm{Bq}\).

Solution

8.1(i)

An \(\alpha\)-decay reduces the mass number by \(4\) and the atomic number by \(2\). A \(\beta^-\)-decay leaves the mass number unchanged and increases the atomic number by \(1\).

Thus the number of \(\alpha\)-decays is

\[ n_\alpha=\frac{238-206}{4}=\boxed{8}. \]

Uranium has atomic number \(92\), and lead has atomic number \(82\). Therefore,

\[ 92-2n_\alpha+n_{\beta^-}=82, \]

giving

\[ n_{\beta^-}=82-92+16=\boxed{6}. \]

The net nuclear equation is

\[ \ce{^{238}_{92}U -> ^{206}_{82}Pb + 8 ^{4}_{2}He + 6 ^{0}_{-1}e} +6\bar{\nu}_e, \]

where \(\bar{\nu}_e\) denotes the electron antineutrino emitted in each \(\beta^-\)-decay.

8.1(ii)

After any number \(k\) of \(\alpha\)-decays, the mass number must be

\[ A=238-4k. \]

Since \(\beta^-\)-decay does not change \(A\), every member of this decay series has \(A\equiv2\pmod4\). Among the ten listed nuclides, only \(\ce{^{234}U}\) satisfies this condition.

It is produced through the sequence

\[ \ce{^{238}_{92}U -> ^{234}_{90}Th + ^{4}_{2}He}, \]

\[ \ce{^{234}_{90}Th -> ^{234}_{91}Pa + ^{0}_{-1}e} +\bar{\nu}_e, \]

\[ \ce{^{234}_{91}Pa -> ^{234}_{92}U + ^{0}_{-1}e} +\bar{\nu}_e. \]

Hence the answer is uranium-234, \(\ce{^{234}U}\).

8.2

Conservation of mass number gives

\[ 235+1=137+A_X+2, \]

so

\[ A_X=97. \]

Tellurium has atomic number \(52\). Conservation of nuclear charge gives

\[ 92=52+Z_X, \qquad Z_X=40. \]

Element \(40\) is zirconium. Thus \(X\) is zirconium-97, and the balanced fission equation is

\[ \boxed{\ce{^{235}_{92}U + ^{1}_{0}n -> ^{137}_{52}Te + ^{97}_{40}Zr + 2 ^{1}_{0}n}}. \]

8.3(i)

The disintegration rate, or activity, of a radioactive isotope is

\[ R=\lambda N=\frac{\ln2}{T_{1/2}}N. \]

Taking the given isotopic abundances as atom fractions,

\[ \frac{R_{238}}{R_{235}} = \frac{N_{238}}{N_{235}} \frac{T_{1/2,235}}{T_{1/2,238}} = \frac{0.9928}{0.0072} \frac{7.0\times10^8}{4.5\times10^9} =21.45. \]

Therefore,

\[ \boxed{R_{238}:R_{235}\approx21:1}. \]

8.3(ii)

The uranium mass in the mineral is

\[ m_U=(1.0\,\mathrm{kg})(0.50)=0.50\,\mathrm{kg}=500\,\mathrm{g}. \]

Approximating isotopic molar masses by their mass numbers, the mean molar mass of this uranium is

\[ \overline M_U = [0.9928(238)+0.0072(235)]\,\mathrm{g\,mol^{-1}} = 237.9784\,\mathrm{g\,mol^{-1}}. \]

The number of \(\ce{^{238}U}\) atoms is consequently

\[ N_{238} = 0.9928\, \frac{500\,\mathrm{g}}{237.9784\,\mathrm{g\,mol^{-1}}} (6.022\times10^{23}\,\mathrm{mol^{-1}}) = 1.2561\times10^{24}. \]

Using \(1\,\mathrm{year}=365\times24\times3600\,\mathrm{s}\),

\[ R_{238} = \frac{(\ln2)(1.2561\times10^{24})} {(4.5\times10^9)(365)(24)(3600)\,\mathrm{s}} = 6.1354\times10^6\,\mathrm{s^{-1}}. \]

Since \(1\,\mathrm{Bq}=1\,\mathrm{s^{-1}}\),

\[ \boxed{R_{238}=6.1\times10^6\,\mathrm{Bq}}. \]

This is the activity of \(\ce{^{238}U}\) itself.

8.4 — decay equations used in both parts

Let \(\lambda_R\) and \(\lambda_T\) be the decay constants of \(\ce{^{97}Ru}\) and \(\ce{^{97}Tc}\), respectively:

\[ \lambda_R = \frac{\ln2}{2.7(86400)\,\mathrm{s}} = 2.9713\times10^{-6}\,\mathrm{s^{-1}}, \]

\[ \lambda_T = \frac{\ln2}{(2.6\times10^6)(365)(86400)\,\mathrm{s}} = 8.4537\times10^{-15}\,\mathrm{s^{-1}}. \]

The initial activity \(R_0=1.0\times10^9\,\mathrm{Bq}\) corresponds to

\[ N_0=\frac{R_0}{\lambda_R}=3.3655\times10^{14} \]

ruthenium atoms.

The parent and daughter populations obey

\[ \frac{dN_R}{dt}=-\lambda_RN_R, \qquad \frac{dN_T}{dt}=\lambda_RN_R-\lambda_TN_T. \]

With \(N_R(0)=N_0\) and \(N_T(0)=0\), their activities are

\[ R_R(t)=R_0e^{-\lambda_Rt}, \]

\[ R_T(t)= R_0\frac{\lambda_T}{\lambda_R-\lambda_T} \left(e^{-\lambda_Tt}-e^{-\lambda_Rt}\right). \]

Because \(\ce{^{97}Mo}\) is stable,

\[ R_{\mathrm{total}}(t)=R_R(t)+R_T(t). \]

8.4(i)

After \(6.0\) days, the ruthenium activity is

\[ R_R = (1.0\times10^9)\,2^{-6.0/2.7} = 2.1431\times10^8\,\mathrm{Bq}. \]

The daughter activity from the expression above is only

\[ R_T=2.2354\,\mathrm{Bq}. \]

It is negligible at the precision of the supplied data. Thus

\[ \boxed{R_{\mathrm{total}}(6.0\,\mathrm{days}) =2.1\times10^8\,\mathrm{Bq}}. \]

8.4(ii)

After \(6000\) years, essentially all the original ruthenium has decayed. Since its decay took only a few days, the technetium population is, to excellent accuracy,

\[ N_T(t)\simeq N_0\,2^{-t/T_{1/2,T}}. \]

Its surviving fraction is

\[ 2^{-6000/(2.6\times10^6)}=0.99840. \]

Therefore,

\[ \begin{aligned} R_{\mathrm{total}} &\simeq \lambda_TN_0\,2^{-6000/(2.6\times10^6)}\\ &=(8.4537\times10^{-15}\,\mathrm{s^{-1}}) (3.3655\times10^{14})(0.99840)\\ &=2.8406\,\mathrm{Bq}. \end{aligned} \]

Hence

\[ \boxed{R_{\mathrm{total}}(6000\,\mathrm{years})=2.8\,\mathrm{Bq}}. \]

The solution was checked against the official solution.