Question

$C{u}^{64}$ (half life = 12.8 hours) decays by ${\beta }^{⊝}-emission$ (38%), ${\beta }^{\oplus }-emission$(19%), and electron capture (43%). Write the decay products and calculate partial half lives for each of the decay processes.
(IIT-JEE, 2002)

• A. 33.70 hr, 67.41 hr and 29.78 hr
• B. 67.41 hr, 33.70 hr and 29.78 hr
• C. 29.78 hr , 33.70 hr and 67.41 hr
• D. 29.78 hr, 67.41 hr and 33.70 hr

Answer 1

The correct option is A 33.70 hr, 67.41 hr and 29.78 hr

The nuclear reactions are:

${}_{29}C{u}^{64}\to {}_{30}Z{n}^{64}{+}_{-1}{e}^0{(}_{-1}{\beta }^0)$

${}_{29}C{u}^{64}\to {}_{28}N{i}^{64}{+}_{+1}{e}^0{(}_{+1}{\beta }^0)$

${}_{29}C{u}_{-1}^{64}{e}^0\to {+}_{28}N{i}^{64}$

Given, ${\lambda }^{avg}=\frac{0.693}{12.8}h{r}^{-1}$

$\therefore {\lambda }_1+{\lambda }_2+{\lambda }_3={\lambda }_{avg}=\frac{0.693}{12.8}=5.41\times {10}^{-2}h{r}^{-1}$ …(i)

Also for parallel path decay

${\lambda }_1=Fractionalyieldof{}_{30}Z{n}_{64}\times {\lambda }_{avg}$

${\lambda }_2=Fractionalyieldof{}_{28}N{i}_{64}\times {\lambda }_{avg}$

${\lambda }_3=Fractionalyieldof{}_{28}N{i}_{64}\times {\lambda }_{avg}$

$\frac{{\lambda }^1}{{\lambda }^2}=\frac{38}{19}$ …(ii)

and $\frac{{\lambda }^1}{{\lambda }^3}=\frac{38}{43}$

From Eqs. (i), (ii), and (iii) ${\lambda }_1=2.056\times {10}^{2}h{r}^1;$

${\lambda }_2=1.028\times {10}^{-2}\times h{r}^{-1};{\lambda }_3=2.327\times {10}^{-2}h{r}^{-1}$

$t_{1/2}for{\beta }^{⊝}-emission=\frac{0.693}{2.056\times {10}^{-2}}=33.70hr$

$t_{1/2}for{\beta }^{\oplus }-emission=\frac{0.693}{1.028\times {10}^{-2}}=67.41hr$

$t_{1/2}forelectroncapture=\frac{0.693}{2.327\times {10}^{-2}}=29.78hr$