Q79 / 90 · JEE Main 2021 (16 Mar Shift 2) · Physics

Calculate the time interval between $33 %$ decay and $67 %$ decay if half-life of a substance is $20 \min$ .

Solution

$N_{1} = N_{0} e^{- λ t_{1}}$

$\frac{N_{1}}{N_{0}} = e^{- λ t_{1}}$

$0.67 = e^{- λ t_{1}}$

$\ln (0.67) = - λ t_{1}$

$N_{2} = N_{0} e^{- λ t_{2}}$

$\frac{N_{2}}{N_{0}} = e^{- λ t_{2}}$

$0.33 = e^{- λ t_{2}}$

$\ln (0.33) = - λ t_{2}$

$\ln (0.67) - \ln (0.33) = λ t_{1} - λ t_{2}$

$λ (t_{1} - t_{2}) = \ln (\frac{0.67}{0.33})$

$λ (t_{1} - t_{2}) ≅ \ln 2$

$t_{1} - t_{2} ≃ \frac{\ln 2}{λ} = t_{1 / 2}$

Half life $= t_{1 / 2} = 20$ minutes.

Hence, the correct answer is option: 'B'.