Q: The half- life period of a radioactive substance is 20 days. What is the time taken for 7/8^{th} of its original

mass to disintegrate ?

Sol: Let the initial mass be m unit.

Mass remaining = m – 7m/8 = m/8

$\large \frac{m}{8} = m (\frac{1}{2})^n $

n = 3

$\large n = \frac{t}{T_{1/2}}$

$\large 3 = \frac{t}{20}$

t = 60 days