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

<p>The decay of a radioactive material follows an exponential decay law, which can be expressed as:</p> <p>$N = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T}}$</p> <p>where:</p> <ul> <li>$N$ is the final amount of the material,</li> <li>$N_0$ is the initial amount of the material,</li> <li>$t$ is the elapsed time,</li> <li>$T$ is the half-life of the material.</li> </ul> <p>From the problem, we know that the material is reduced to $1/8$ of its original amount in 3 days. Therefore, the half-life of the material can be calculated as follows:</p> <p>$\frac{1}{8} = \left(\frac{1}{2}\right)^{\frac{3}{T}}$</p> <p>This simplifies to $2^{-3} = 2^{-\frac{3}{T}}$, which gives $T = 1 \, \text{day}$.</p> <p>Knowing the half-life, we can now find the initial amount of the material. We know that after 5 days, $8 \times 10^{-3} \, \text{kg}$ of the material is left. Therefore, we can write:</p> <p>$8 \times 10^{-3} \, \text{kg} = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{5}{1}}$</p> <p>Solving this equation for $N_0$ gives:</p> <p>$$ N_0 = 8 \times 10^{-3} \, \text{kg} \cdot 2^{5} = 256 \times 10^{-3} \, \text{kg} = 0.256 \, \text{kg} = 256 \, \text{g} $$</p>