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

A radioactive material reduces to $$\dfrac{1}{8}$$ of its initial amount in 3 days, and after 5 days, $$8 \times 10^{-3}$$ kg (i.e., 8 g) remains. We need the initial amount.

Using the decay formula $$\dfrac{N}{N_0} = \left(\dfrac{1}{2}\right)^{t/T_{1/2}}$$, we first find the half-life:

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

Since $$\frac{1}{8} = \left(\frac{1}{2}\right)^3$$, we get $$T_{1/2} = 1$$ day.

Now, using 5 days of decay:

$$N = N_0 \left(\frac{1}{2}\right)^{5} = \frac{N_0}{32}$$

$$8 = \frac{N_0}{32}$$

$$N_0 = 256 \text{ g}$$

Hence, the correct answer is Option D.