$^{6} 4 Cu$ (half-life = $12.8$ hr) decays by $β^{-}$ emission ( $38 %$ ), $β^{+}$ emission ( $19 %$ ) and electron capture ( $43 %$ ). Write the decay product and calculate partial half-lives for each of the decay processes.

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Solution

Let $β^{-}$ emission = $λ_{1}$
$β^{+}$ emission = $λ_{2}$
$k -$ electron emission = $λ_{3}$
$λ_{n e t} = λ_{1} + λ_{2} + λ_{3}$
Fractional yield of path (I) = $\frac{λ_{1}}{λ_{n e t}} = \frac{(t_{1 / 2})_{n e t}}{(t_{1 / 2})_{1}}$

Half life of path I = $\frac{(t_{1 / 2})_{1}}{F r a c t i o n a l y i e l d} = \frac{12.8}{0.38} = 33.684 h r$

Half life of path II = $\frac{12.8}{0.19} = 67.368 h r$

Half life of path III = $\frac{12.8}{0.43} = 29.767 h r$

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(IIT-JEE, 2002)