Question

If 3/4 quantity of radioactive substance disintegrates in 2 hours. Find out its half-life period.

Answer 1

We have,

$\lambda =\frac{2.303}{t_{1/2}}log\frac{a_0}{a_0-a}$

We have, $t_{3/4}=2hours$

$\therefore$ for $t_{3/4}$ the equation will be:

$\lambda =\frac{2.303}{t_{3/4}}log\frac{a_0}{a_0-\frac{3}{4}}{a}_0$

$\lambda =\frac{2.303}{2}log4$

$\lambda =0.693hou{r}^{-1}$

Now, $t_{1/2}=\frac{0.693}{\lambda }$

$t_{1/2}=\frac{0.693}{0.693}=1hour$