Question

A radioactive substance of half life 69.3 days is kept in a container. The time in which 80 % of the substance will disintegrate will be

• A. $1.61$ days
• B. $16.1$days
• C. $161$ days
• D. $1610$ days

Answer 1

The correct option is C $161$ days

$t_1=\frac{0.693}{K_1}$

or, $69.3=\frac{0.693}{K_1}$

or, $K_1=\frac{0.693}{69.3}=0.01{day}^{-1}$

$K_1=$ disintegrating constant

$t=\frac{2.303}{K_1}log\frac{a}{a-x}$

$=\frac{2.303}{0.01}log\left(\frac{100}{20}\right)$

$\approx$$161$ days