Question

The half life period of a radio active element is $30$ days, after $90$ days the following quantity will be left

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

Answer 1

Answer: (A)

$\frac{1}{8}$

Answer 2

All radioactive elements follow first order kinetics, Given, $t_{1 / 2}=30$ days $\therefore$ Rate constant, $k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{30}$ days $^{-1}$ 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 $8=\frac{1}{1-x}$ $\therefore 1-x=\frac{1}{8}$ $N=\frac{N_{0}}{2^{n}}$ (where, $n=$ number of half-life, $N_{0}=$ initial mass of radioactive element, $N=$ final mass) $\therefore N=\frac{N_{0}}{2^{3}}$ $N=\frac{N_{0}}{8}$