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AI-তৈরি নমুনা উত্তর

Given:
Half-life, $T_{1/2} = 3.8\text{ days}$

Decay constant, $\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{3.8} \approx 0.1824\text{ day}^{-1}$

When 30% of original atoms decay, the remaining percentage of atoms is:
$N = (1 - 0.30) N_0 = 0.70 N_0$

Using the radioactive decay law:
$N = N_0 e^{-\lambda t}$
$\Rightarrow \frac{N}{N_0} = e^{-\lambda t}$
$\Rightarrow 0.70 = e^{-\lambda t}$
$\Rightarrow -\lambda t = \ln(0.70)$
$\Rightarrow t = -\frac{\ln(0.70)}{\lambda} = -\frac{-0.3567}{0.1824} \approx 1.955\text{ days} \approx 1.96\text{ days}$

Ans: Approximately 1.955 days (or 1.96 days)