Question

${}^{64}Cu$ (half-life-12.8 hr) decays by ${\beta }^{-}$-emission (38%), ${\beta }^{+}$-emission (19%) and electron capture (43%). Calculate partial half lives for electron capture?

• A. $29.78hr$
• B. $28.78hr$
• C. $59.56hr$
• D. $57.56hr$

Answer 1

The correct option is A $29.78hr$

Given, ${\lambda }_{av}=\frac{0.693}{12.8}h{r}^{-1}$
$\therefore {\lambda }_1+{\lambda }_2+{\lambda }_3={\lambda }_{av}=\frac{0.693}{12.8}$
${\lambda }_1+{\lambda }_2+{\lambda }_3=5.41\times {10}^{-2}h{r}^{-1}$ ... (i)
Also for parallel path decay
${\lambda }_1=Fractionalyieldof{}_{30}^{64}\text{Zn}\times {\lambda }_{av}$ ... (ii)
${\lambda }_2=Fractionalyieldof{}_{28}^{64}\text{Ni}\times {\lambda }_{av}$ ... (iii)
${\lambda }_3=Fractionalyieldof{}_{28}^{64}\text{Ni}\times {\lambda }_{av}$ ... (iv)
$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{38}{19}$ ... (v)
and $\frac{{\lambda }_1}{{\lambda }_3}=\frac{38}{43}$ ... (vi)
From eqs. (i), (v) and (vi) : ${\lambda }_1=2.056\times {10}^{-2}$;
${\lambda }_2=1.028\times {10}^{-2};{\lambda }_3=2.327\times {10}^{-2}h{r}^{-1}$
$\therefore t_{1/2}for{\beta }^{-}$-$emission=\frac{0.693}{2.056\times {10}^{-2}}=33.70hr$
$t_{1/2}for{\beta }^{+}$-$emission=\frac{0.693}{1.028\times {10}^{-2}}=67.41hr$
$t_{1/2}forelectroncapture=\frac{0.693}{2.327\times {10}^{-2}}=29.78hr$