Question

The half-life period of a radioacttive substance is $30$ days. What is the time taken for $3/4$th of its original mass to disintegrate?

Answer 1

$\ln \frac{A_t}{A_o}=-ktk=\frac{\ln 2}{t_{1/2}}=\frac{\ln 2}{30}\ln \left(\frac{4A_o}{3A_o}\right)=\frac{\ln 2}{30}\times t\ln \frac{4}{3}=\frac{\ln 2}{30}\times tt=\frac{30}{\ln 2}\ln \frac{4}{3}t=\frac{30(\ln 4-\ln 3)}{\ln 2}$