A radioactive sample decays 7/8 times its initial quantity in 21 minu…

The correct answer is Option (1) → 7 min

Given:

Fraction decayed = $\frac{7}{8}$ → Remaining fraction = $\frac{1}{8}$

Time elapsed, $t = 21\ \text{min}$

Decay law:

$N = N_0 \left(\frac{1}{2}\right)^{t/T}$

Remaining fraction: $\frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/T}$

Substitute values:

$\frac{1}{8} = \left(\frac{1}{2}\right)^{21/T}$

But $\frac{1}{8} = \left(\frac{1}{2}\right)^3$

So, $\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^{21/T}$ ⇒ $3 = 21/T$

∴ $T = \frac{21}{3} = 7\ \text{min}$

∴ Half-life of the sample = 7 min