A radio active material is reduced to 1/8 of its original amount in 3 days. If 8 × 10⁻³ kg of the material is left after 5 days the initial amount of the material is (a) 64 g (b) 256 g (c) 32 g (d) 40 g

Answer: (b). 1/8 = (1/2)³ in 3 days, so T1/2 = 1 day. In 5 days: N = (N₀)/(2⁵) = N₀/32. N = 8 × 10⁻³ kg = 8 g ⇒ N₀ = 32 × 8 = 256 g. So the answer is (b) 256 g.

A radio active material is reduced to 1/8 of its original amount in 3… | JEE Main 2023 (8 Apr, Shift 2)

Q30JEE Main 2023 · 8 APR · SHIFT 2Easy

Law of Radioactive Decay

A radio active material is reduced to $1/8$ of its original amount in 3 days. If $8 \times 10^{-3}\text{ kg}$ of the material is left after 5 days the initial amount of the material is

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Answer: (b)

$\frac{1}{8} = \left(\frac{1}{2}\right)^3$ in 3 days, so $T_{1/2} = 1$ day.
In 5 days: $N = \frac{N_0}{2^5} = \frac{N_0}{32}$ .
$N = 8 \times 10^{-3}\text{ kg} = 8\text{ g} \Rightarrow N_0 = 32 \times 8 = 256\text{ g}$ .
So the answer is (b) 256 g.

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