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 [take ln(5) = 1.61]

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

Answer 1

The correct option is C

161 days

$T_{\frac{1}{2}}$ = 69.3 days. Therefore $\lambda =\frac{0.693}{T_{\frac{1}{2}}}={10}^{-2}$ per day.
N = 20% of $N_0$ = 0.2 $N_0$.
Now $N=N_0{e}^{\lambda t}$, which gives
$\frac{N_0}{N}=e^{\lambda t}orIn\left(\frac{N_0}{N}\right)=\lambda tort=\frac{In\left(\frac{N_0}{N}\right)}{\lambda }=\frac{In\left(5\right)}{{10}^{-2}}=\frac{1.61}{{10}^{-2}}$
= 161 days
Hence the correct choice is (c)