Question

Half-life $\left(t_{1/2}\right)$ for a radioactive decay is $6930sec$. The time required to fall the rate of decay by ${\left(\frac{1}{100}\right)}^{th}$ of it's initial value is:

• A. $46051.7sec$
• B. $20,000sec$
• C. $23030sec$
• D. $noneofthese$

Answer 1

The correct option is A $46051.7sec$

Given,

Half-life, $T^{1/2}=6930\sec$

Half-life, $T^{1/2}=\frac{0.693}{\lambda }$

$\Rightarrow \lambda =\frac{0.693}{6930}{\sec }^{-1}$

Rate of decay, $R=R_0{e}^{-\lambda t}$

$t=\frac{-1}{\lambda }\ln \left(\frac{R}{R_0}\right)=-\frac{6930}{0.693}\ln \left(\frac{1}{100}\right)$

$t=46051.7\sec$

Hence, time required is $46051.7\sec$