Question

A radioactive substance is reduced to 1/8 of its original concentration in 24 s. The rate constant of the reaction is :

• A. $\frac{0.69}{16}{s}^{-1}$
• B. $\frac{1}{8}{s}^{-1}$
• C. $\frac{1}{24}{s}^{-1}$
• D. $\frac{ln2}{8}{s}^{-1}$

Answer 1

The correct option is D $\frac{ln2}{8}{s}^{-1}$

Radioactive disintegration or decay is such a kind of chemical reaction that follows first order kinetics. For the first-order reaction we have:-

$K=\frac{1}{t}ln\left(\frac{a}{a-x}\right)$ $-\left(i\right)$

where $K$= rate constant of the reaction.

$t$= time

$a$= amount of concentration of the sample

$x$= amount of sample undergone disintegration in time $t$.

According to question, $t=24$ sec

$(a-x)=\frac{1}{8}a$

Putting values in $e{q}^n\left(i\right)$:-

$K=\frac{1}{24}ln\left(8\right)=\frac{1}{8}ln2s^{-1}$

$=0.0866$ $se{c}^{-1}$