Question

The half-life period of a radioactive element is $30$ days. After $90$ days, which of the following quantity will be left?

• A. $\frac{1}{2}$
• B. $\frac{1}{6}$
• C. $\frac{1}{8}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{1}{8}$

All radioactive elements follow first-order kinetics,
Given, $t_{1/2}=30$ days

$\therefore$ Rate constant, $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{30}$ $day{s}^{-1}$

Now, as we know for first order reaction, $k=\frac{2.303}{t}log\frac{a}{a-x}$
Here, $t=90$ days, $a=1$

$\therefore \frac{0.693}{30}=\frac{2.303}{90}$ $log\frac{1}{1-x}$

or $0.9027=log\frac{1}{1-x}$

or $8=\frac{1}{1-x}$

$\therefore 1-x=\frac{1}{8}$.

Alternate Method:

$N=\frac{N_0}{2^n}$

(where, $n=$ number of half-life, $N_0=$ initial mass of radioactive element, $N=$ final mass)

$\therefore N=\frac{N_0}{2^3}$

$N=\frac{N_0}{8}$.