Question

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

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

Answer 1

The correct option is C $40$ min

Given that the half-life of a radioactive substance is $20$ min. So, $t_{\frac{1}{2}}=20$ min.

For $20\%$ decay, we have $80\%$ of the substance left, hence

$\frac{80N_0}{100}=N_0{e}^{-\lambda l_{20}}\dots \left(i\right)$

Where $N_0=\text{initial undecayed substance and}{t}_{20}$ is the time taken for $20\%$ decay.

For $80\%$ decay, we have $20\%$ of the substance left, hence

$\frac{20N_0}{100}=N_0{e}^{-\lambda t_{80}}\dots \left(ii\right)$

Dividing eq. (i) and eq. (ii), we get

$4=e^{\lambda (t_{80}-t_{20})}$

$\Rightarrow$ In $4=\lambda (t_{80}-t_{20})$

Taking log on both sides

$\Rightarrow 2ln2=\frac{0.693}{t_{\frac{1}{2}}}(t_{80}-t_{20})$

$\Rightarrow t_{80}-t_{20}=2\times t_{\frac{1}{2}}=40$ min.