Question

For the following parallel chain reaction

What will be that value of overall half-life of A in minutes?$\left\lbrack Giventhat\frac{\lbrack B\rbrack_{t}}{\lbrack C\rbrack_{t}} = \frac{16}{9} \right\rbrack$

• A. 3.3
• B. 6.3
• C. 3.6
• D. None

Answer 1

Answer: (A)

3.3

Answer 2

We have,

$\frac{\lbrack B\rbrack_{t}}{\lbrack C\rbrack_{t}} = \frac{4k_{1}}{3k_{2}} = \frac{16}{9}$

so, $\frac{k_{l}}{k_{2}} = \frac{4}{3}$

Now,

$\text{k = }\text{k}_{1}\text{ + }\text{k}_{2}\text{ = }\lbrack\text{2 }\text{×}\text{ 1}\text{0}^{\text{-3 }}\text{+ }\frac{3}{4}\text{ }\text{×}\text{ 2 }\text{×}\text{ 1}\text{0}^{\text{-3}}\rbrack\text{ se}\text{c}^{\text{-1}}$

$\text{= }\frac{7}{2}\text{ }\text{×}\text{ 1}\text{0}^{\text{-3}}\text{ se}\text{c}^{\text{-1}}\text{ = }\frac{7 \times \text{1}\text{0}^{\text{-3}} \times 60}{2}\text{ mi}\text{n}^{\text{-1}}\$

so,$T_{\text{1/2}}\text{ =}\frac{\mathcal{l}\text{n2}}{7 \times \text{30} \times \text{1}\text{0}^{\text{-3}}}\text{ min = }\frac{\text{693}}{7 \times \text{30}}\text{ = 3.3 min.}$