Chemical Kinetics - Result Question 60
60. ${ }^{64} Cu$ (half-life $=12.8$ $ h$ ) decays by $\beta$ emission $(38 \%), \beta^{+}$ emission $(19\%)$ and electron capture $(43\%)$. Write the decay products and calculate partial half-lives for each of the decay processes.
(2002)
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Solution:
$\begin{aligned} & { }{29} \mathrm{Cu}^{64} \xrightarrow[k_1]{38 %}{ }{-1} \beta^0+{ }{30} \mathrm{Zn}^{64} \\ & { }{29} \mathrm{Cu}^{64} \xrightarrow[k_2]{19 %}+{ }1 \beta^0+{ }{28} \mathrm{Ni}^{64} \\ & { }{29} \mathrm{Cu}^{64}+{ }{-1} e^0 \xrightarrow[k_3]{43 %}{ }_{28} \mathrm{Ni}^{64}\end{aligned}$
Above are the parallel reactions occurring from $Cu^{64}$.
$ \dfrac{k _1}{k _2}=\dfrac{38}{19}=2=\dfrac{T _2}{T _1} \quad \text { and } \quad \dfrac{k _1}{k _3}=\dfrac{38}{43}=\dfrac{T _3}{T _1} $
$T _1, T _2$ and $T _3$ are the corresponding partial half-lives.
$ \begin{aligned} \text { Also } & \\ \Rightarrow \quad \dfrac{\ln 2}{T} & =\dfrac{\ln 2}{T _1}+\dfrac{\ln 2}{T _2}+\dfrac{\ln 3}{T _3} \\ \Rightarrow \quad \dfrac{1}{T} & =\dfrac{1}{T _1}+\dfrac{1}{T _2}+\dfrac{1}{T _3}=\dfrac{1}{T _1}+\dfrac{1}{2 T _1}+\dfrac{43}{38 T _1} \\ & =\dfrac{1}{T _1}\left(1+\dfrac{1}{2}+\dfrac{43}{38}\right) \\ & =\dfrac{1}{T _1}\left(\dfrac{38+19+43}{38}\right)=\dfrac{100}{38 T _1} \\ \Rightarrow \quad T _1 & =\dfrac{100 T}{38}=\dfrac{100}{38} \times 12.8=33.68 h \\ \Rightarrow \quad T _2 & =2 T _1=67.36 h \\ T _3 & =\dfrac{38 T _1}{43}=\dfrac{38 \times 33.68}{43}=29.76 h \end{aligned} $
BETA
