Question

Half-life of a radioactive substance is $20$ minute. The time between $20\%$ and $80\%$ decay will be :

• A. 20 min
• B. 30 min
• C. 40 min
• D. 25 min

Answer 1

Answer: (C)

Given that the half-life of a radioactive substance is $20min.$ So, $t_{1/2}=20min$.
For $20\%$ decay, we have $80\%$ of the substance left, hence
$\frac{80N_0}{100}=N_0{e}^{-\lambda t_{20}}...(i)$
where $N_0=$ initial undecayed substance and $t_{20}$ is the time taken for $20\%$ decay.
For $80\%$ decay, we have $20\%$ of the substance left, hence
$\frac{20N_0}{100}=N_0{e}^{-\lambda t_{80}}...(ii)$
Dividing Eq. (i) and Eq. (ii), we get
$4=e^{\lambda \left(t_{80}-t_{20}\right)}$
$\Rightarrow \ln 4=\lambda \left(t_{80}-t_{20}\right)$
(taking log on both sides)
$\Rightarrow 2\ln 2=\frac{0.693}{t_{1/2}}\left(t_{80}-t_{20}\right)$
$\Rightarrow t_{80}-t_{20}=2\times t_{1/2}=40min$