Californium (Cf) is a synthetic radioactive transuranic element in the actinide series with an atomic number Z of 98 and 20 known radioisotopes. Of these only Cf-252, as an intense neutron emitter, is of commercial interest and was found useful in a wide range of specialized areas of science,

Question

Californium (Cf) is a synthetic radioactive transuranic element in the actinide series with an atomic number Z of 98 and 20 known radioisotopes. Of these only Cf-252, as an intense neutron emitter, is of commercial interest and was found useful in a wide range of specialized areas of science, industry, and medicine, such as the study of fission, neutron activation analysis, neutron radiography, well logging, nuclear reactor start up, and brachytherapy of cancer.
Californium-252 decays through two radioactive decay modes: (1) α decay with half life [latex]\left(t_{1/2} ight)_α = 2.73[/latex] years and branching fraction (ratio) of 0.969 and (2) spontaneous fission accompanied by emission of neutrons with half life of [latex] (t_{1/2})_{SF} = 85.5[/latex] years and branching fraction (ratio) of 0.031. Mean neutron fraction [latex]\bar{f}_n[/latex], i.e., the mean number of neutrons emitted per each spontaneous fission decay is [latex]\bar{f}_n[/latex] = 3.8.

(a) Determine the effective half-life [latex]\left(t_{1 / 2} ight)_{ ext {eff }} ext { of a }{ }_{98}^{252} \mathrm{Cf} ext { neutron source. }[/latex]
(b) Determine the monthly decay [latex] ext { (in \%) of a }{ }_{98}^{252} \mathrm{Cf} ext { neutron source. }[/latex]
(c) Determine specific activity [latex]a_{\mathrm{SF}} ext { for spontaneous fission of }{ }_{98}^{252} \mathrm{Cf} ext {. }[/latex]
(d) Determine specific activity [latex]a_\alpha ext { for } \alpha ext { decay of }{ }_{98}^{252} \mathrm{Cf} ext {. }[/latex]
(e) Determine specific activity [latex]a_{ ext {eff }} ext { for decay of }{ }_{98}^{252} \mathrm{Cf} ext {. }[/latex]
(f) Determine the neutron production rate of [latex]{ }_{98}^{252} \mathrm{Cf} ext { in number of neutrons }[/latex] per second.
(g) Calculate Q value for α decay of [latex]{ }_{98}^{252} \mathrm{Cf} ext { into }{ }_{96}^{248} \mathrm{Cm} ext { (curium) }[/latex] and determine the kinetic energy [latex]E_{\mathrm{K}}^\alpha ext { of the emitted } \alpha ext { particle. }[/latex]

Accepted Answer

(a) The effective half-life [latex]\left(t_{1/2} ight)_{eff}[/latex] is calculated via the total decay constant λ of [latex]{ }_{98}^{252} \mathrm{Cf}[/latex] which follows the general rule of radioactivity stipulating that when more than one mode of decay is available to the radioactive nucleus (branching), the total decay constant λ is the sum of the partial decay constants [latex]λ_i[/latex] applicable to each mode. We thus have

[latex]\lambda=\sum_i \lambda_i=\lambda_\alpha+\lambda_{\mathrm{SF}}=\frac{\ln 2}{\left(t_{1 / 2} ight)_\alpha}+\frac{\ln 2}{\left(t_{1 / 2} ight)_{\mathrm{SF}}}=(\ln 2)\left[\frac{1}{\left(t_{1 / 2} ight)_\alpha}+\frac{1}{\left(t_{1 / 2} ight)_{\mathrm{SF}}} ight]=\frac{\ln 2}{\left(t_{1 / 2} ight)_{\mathrm{eff}}}[/latex] (9.135)

The effective half-life [latex]\left(t_{1 / 2} ight)_{ ext {eff }} ext { of a }{ }_{98}^{252} \mathrm{Cf} ext { source is calculated as }[/latex] follows

[latex]\frac{1}{\left(t_{1 / 2} ight)_{\mathrm{eff}}}=\frac{1}{\left(t_{1 / 2} ight)_\alpha}+\frac{1}{\left(t_{1 / 2} ight)_{\mathrm{SF}}}=\frac{1}{2.73 \mathrm{y}}+\frac{1}{85.5 \mathrm{y}}=0.378 \mathrm{y}^{-1}[/latex] (9.154)

resulting in [latex]\left(t_{1 / 2} ight)_{\mathrm{eff}}=2.645[/latex] y.

(b) The monthly decay (in %) of a [latex]{ }_{98}^{252} \mathrm{Cf}[/latex] neutron source is calculated by assuming exponential source decay and an effective half-life [latex]\left(t_{1/2} ight)_{eff}[/latex] = 2.645 y, as determined in (9.154). The monthly rate of source decay is expressed with the ratio [latex]I/I_0[/latex], where [latex]I_0[/latex] is the source intensity at a given distance d from the source on day 0 (time: t = 0) and I is the source intensity at the same distance d from the source on day 30 (time: t = 30 days). [latex]I/I_0[/latex] is given as follows

[latex]\frac{I}{I_0}=e^{-\frac{\ln 2}{\left(t_{1 / 2} ight) ext { eff }} t}=e^{-\frac{(\ln 2) imes 30}{2.645 imes 365}}=e^{-0.0216}=0.979[/latex] (9.155)

indicating that in one month a [latex]{ }_{98}^{252} \mathrm{Cf}[/latex] neutron source will decay by 2.1 %.

(c) Specific activity [latex]a_{\mathrm{SF}} ext { of }{ }_{98}^{252} \mathrm{Cf}[/latex] for spontaneous fission is calculated from activity [latex]\mathcal{A}_{\mathrm{SF}}[/latex] for spontaneous fission defined as [latex]\mathcal{A}_{\mathrm{SF}}=\lambda_{\mathrm{SF}} N[/latex] where N stands for the number of radioactive atoms, [latex]N_{\mathrm{A}} ext { is the Avogadro number }\left(6.022 imes 10^{23} \mathrm{~mol}^{-1} ight) ext {, }[/latex] and A is the atomic mass number in g/mol

[latex]\begin{aligned} a_{\mathrm{SF}} & =\frac{\mathcal{A}_{\mathrm{SF}}}{m}=\frac{\lambda_{\mathrm{SF}} N}{m}=\frac{\lambda_{\mathrm{SF}} N_{\mathrm{A}}}{A}=\frac{\ln 2}{\left(t_{1 / 2} ight)_{SF}} \frac{N_{\mathrm{A}}}{A} \ & =\frac{(\ln 2) imes\left(6.022 imes 10^{23} \mathrm{~mol}^{-1} ight)}{(85.5 \mathrm{y}) imes(365 ext { day } / \mathrm{y}) imes(24 \mathrm{~h} / ext { day }) imes(3600 \mathrm{~s} / \mathrm{h}) imes\left(252 \mathrm{~g} \cdot \mathrm{mol}^{-1} ight)} \ & =6.143 imes 10^{11} \mathrm{~s}^{-1} \cdot \mathrm{g}^{-1}=0.6143 \mathrm{TBq} \cdot \mathrm{g}^{-1} .\quad (9.156) \end{aligned}[/latex]

(d) Specific activity [latex]a_\alpha ext { of }{ }_{98}^{252} \mathrm{Cf} ext { for } \alpha[/latex] decay is calculated from activity [latex]\mathcal{A}_\alpha[/latex] for α decay defined as [latex]\mathcal{A}_α = λ_αN[/latex].

[latex]\begin{aligned} a_\alpha & =\frac{\mathcal{A}_\alpha}{m}=\frac{\lambda_\alpha N}{m}=\frac{\lambda_\alpha N_{\mathrm{A}}}{A}=\frac{\ln 2}{\left(t_{1 / 2} ight)_\alpha} \frac{N_{\mathrm{A}}}{A} \ & =\frac{(\ln 2) imes\left(6.022 imes 10^{23} \mathrm{~mol}^{-1} ight)}{(2.73 \mathrm{y}) imes(365 ext { day } / \mathrm{y}) imes(24 \mathrm{~h} / ext { day }) imes(3600 \mathrm{~s} / \mathrm{h}) imes\left(252 \mathrm{~g} \cdot \mathrm{mol}^{-1} ight)} \ & =19.24 imes 10^{12} \mathrm{~s}^{-1} \cdot \mathrm{g}^{-1}=19.24 \mathrm{TBq} \cdot \mathrm{g}^{-1} .\quad (9.157) \end{aligned}[/latex]

(e) Specific activity [latex]a_{ ext {eff }} ext { of }{ }_{98}^{252} \mathrm{Cf}[/latex] is simply the sum of specific activities for α decay and for spontaneous fission

[latex]a_{\mathrm{eff}}=a_\alpha+a_{\mathrm{SF}}=19.24 \mathrm{TBq} / \mathrm{s}+0.61 \mathrm{TBq} / \mathrm{g}=19.85 \mathrm{TBq} / \mathrm{g}[/latex] (9.158)

We can obtain the same result calculated directly from activity [latex] ext { A of }{ }_{98}^{252} \mathrm{Cf}[/latex] defined as [latex]\mathcal{A} = λN[/latex].

[latex]\begin{aligned} a_{ ext {eff }} & =\frac{\mathcal{A}}{m}=\frac{\lambda N}{m}=\frac{\lambda N_{\mathrm{A}}}{A}=\frac{\ln 2}{\left(t_{1 / 2} ight)_{\mathrm{eff}}} \frac{N_{\mathrm{A}}}{A} \ & =\frac{(\ln 2) imes\left(6.022 imes 10^{23} \mathrm{~mol}^{-1} ight)}{(2.645 \mathrm{y}) imes(365 ext { day } / \mathrm{y}) imes(24 \mathrm{~h} / \mathrm{day}) imes(3600 \mathrm{~s} / \mathrm{h}) imes\left(252 \mathrm{~g} \cdot \mathrm{mol}^{-1} ight)} \ & =19.85 imes 10^{12} \mathrm{~s}^{-1} \cdot \mathrm{g}^{-1}=19.85 \mathrm{TBq} \cdot \mathrm{g}^{-1} \ & =\frac{19.85 imes 10^{12} \mathrm{~Bq} \cdot \mathrm{g}^{-1}}{3.7 imes 10^{10} \mathrm{~Bq} / \mathrm{Ci}}=536.5 \mathrm{Ci} / \mathrm{g} .\quad (9.159) \end{aligned}[/latex]

(f) Neutron production rate in units of [latex]g^{−1}\ ·\ s^{−1}[/latex] is calculated by multiplying the specific activity [latex]a_{\mathrm{SF}} ext { of }{ }_{98}^{252} \mathrm{Cf}[/latex] for spontaneous fission by the neutron factor [latex]\bar{f}_{\mathrm{n}}[/latex] (defined as the mean number of neutrons produced by each spontaneous fission decay) to get

[latex]\bar{f}_{\mathrm{n}} a_{\mathrm{SF}}=3.8 imes\left(0.6143 imes 10^{12} \mathrm{~s}^{-1} \cdot \mathrm{g}^{-1} ight)=2.33 imes 10^{12} \mathrm{~s}^{-1} \cdot \mathrm{g}^{-1}[/latex] (9.160)

(1) Industrial sources contain up to 50 mg of emitting of the order of [latex]∼10^{11}[/latex] neutrons per second [latex]\left[\left(50 imes 10^{-3} \mathrm{~g} ight) imes\left(2.33 imes 10^{12} \mathrm{~s}^{-1} \cdot \mathrm{g}^{-1} ight) \approx 10^{11} \mathrm{~s}^{-1} ight][/latex].
(2) High dose rate brachytherapy (HDR) source requires about 500 µg of [latex]{ }_{98}^{252} \mathrm{Cf} ext { per source emitting of the order of } \sim 10^9[/latex] neutron/s [latex]\left[\left(500 imes 10^{-6} \mathrm{~g} ight) imes \left(2.33 imes 10^{12} \mathrm{~s}^{-1} \cdot \mathrm{g}^{-1} ight) \approx 10^9 \mathrm{~s}^{-1} ight][/latex].

(g) Q value for α decay of [latex]{ }_{98}^{252} \mathrm{Cf} ext { into }{ }_{96}^{248} \mathrm{Cm}[/latex] is calculated in a manner similar to the calculation of Q value for nuclear reactions using either (1) rest energy method or (2) binding energy method.

(1) Rest energy method for α decay: [latex]{ }_{98}^{252} \mathrm{Cf} ightarrow{ }_{96}^{248} \mathrm{Cm}+\alpha[/latex].

[latex]\begin{aligned} Q_\alpha & =\sum_{i, ext { before }} M_i c^2-\sum_{i, ext { after }} M_i c^2=\left[M\left({ }_{98}^{252} \mathrm{Cf} ight) ight]-\left[M\left({ }_{96}^{248} \mathrm{Cm} ight) c^2+m_\alpha c^2 ight] \ & =[234762.4495 \mathrm{MeV}]-[231028.8556 \mathrm{MeV}+3727.3791 \mathrm{MeV}] \ & =[234762.4495 \mathrm{MeV}]-[234756.2347 \mathrm{MeV}]=6.215 \mathrm{MeV} .\quad (9.161) \end{aligned}[/latex]

(2) Binding energy method for α decay: [latex]{ }_{98}^{252} \mathrm{Cf} ightarrow{ }_{96}^{248} \mathrm{Cm}+\alpha[/latex].

[latex]\begin{aligned} Q_\alpha & =\sum_{i, ext { after }} E_{\mathrm{B}}(i)-\sum_{i, ext { before }} E_{\mathrm{B}}(i)=\left[E_{\mathrm{B}}^{\mathrm{Cm}-248}+E_{\mathrm{B}}^{\mathrm{Cm}-248} ight]-\left[E_{\mathrm{B}}^{\mathrm{Cf}-252} ight] \ & =[1859.1902 \mathrm{MeV}+28.2957 \mathrm{MeV}]-[1881.2748 \mathrm{MeV}] \ & =[1887.4859 \mathrm{MeV}]-[1881.2748 \mathrm{MeV}]=6.212 \mathrm{MeV}\quad (9.162) \end{aligned}[/latex]

(3) Q value of the α decay is shared between the α particle and the recoil Cm-248 nucleus in inverse proportions to the rest energies, so that α particle receives about 98 % of the energy available in Q value.

Question 9.6.Q4: Californium (Cf) is a synthetic radioactive transuranic elem......

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