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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Answer:
Correct Answer: 60. $(67 %)$
Solution:
$$ \begin{aligned} & { } _{29} Cu^{64} \underset{k _1}{38 %}{ } _{-1} \beta^{0}+{ } _{30} Zn^{64} \ & { } _{29} Cu^{64} \underset{k _2}{19 %}+{ } _1 \beta^{0}+{ } _{28} Ni^{64} \ & { } _{29} Cu^{64}+{ } _{-1} e^{0} \stackrel{43 %}{{ } _{k 3}}{ } _{28} Ni^{64} \end{aligned} $$
Above are the parallel reactions occurring from $Cu^{64}$.
$$ \frac{k _1}{k _2}=\frac{38}{19}=2=\frac{T _2}{T _1} \quad \text { and } \quad \frac{k _1}{k _3}=\frac{38}{43}=\frac{T _3}{T _1} $$
$T _1, T _2$ and $T _3$ are the corresponding partial half-lives.
$$ \begin{aligned} \text { Also } & \ \Rightarrow \quad \frac{\ln 2}{T} & =\frac{\ln 2}{T _1}+\frac{\ln 2}{T _2}+\frac{\ln 3}{T _3} \ \Rightarrow \quad \frac{1}{T} & =\frac{1}{T _1}+\frac{1}{T _2}+\frac{1}{T _3}=\frac{1}{T _1}+\frac{1}{2 T _1}+\frac{43}{38 T _1} \ & =\frac{1}{T _1}\left(1+\frac{1}{2}+\frac{43}{38}\right) \ & =\frac{1}{T _1}\left(\frac{38+19+43}{38}\right)=\frac{100}{38 T _1} \ \Rightarrow \quad T _1 & =\frac{100 T}{38}=\frac{100}{38} \times 12.8=33.68 h \ \Rightarrow \quad T _2 & =2 T _1=67.36 h \ T _3 & =\frac{38 T _1}{43}=\frac{38 \times 33.68}{43}=29.76 h \end{aligned} $$
