Question

Suresh found that a certain radioactive substance of $10g$ reduced to $5g$ in 3 hours. From this, he calculated the amount of time required by this radioactive substance to reduce from $80g$ to $10g$. The time is _______ hours.

• A. 7
• B. 6
• C. 3
• D. 9

Answer 1

The correct option is D 9

$\ln \frac{A_t}{A_o}=-kt\ln \frac{10}{5}=-k\times 3-k=\frac{1}{3}\ln \frac{10}{5}k=\frac{1}{3}\ln \frac{5}{10}=\frac{1}{3}\ln \frac{1}{2}So,\ln \frac{A_t}{A_o}=-kt\ln \frac{10}{80}=-\frac{1}{3}\ln \frac{1}{2}\times t\ln \frac{1}{8}=-\frac{1}{3}\ln \frac{1}{2}t\ln {\left(\frac{1}{2}\right)}^3=-\frac{1}{3}\ln \frac{1}{2}t3\ln \frac{1}{2}=-\frac{1}{3}\ln \frac{1}{2}\times tSo,t=3\times 3=9hours$