A radio active material is reduced to (1)/(8) of its original...

The initial and final amount of the radioactive substance are related by the formula

$\dfrac{N}{{N}_{0}} = {\left(\dfrac{1}{2}\right)}^{n} . . . \left(1\right)$

where, $n$ is the number of half lives.

Substitute $\dfrac{1}{8}$ for $\dfrac{N}{{N}_{0}}$ into equation (1) and solve to calculate the value of $n$.

$\begin{cases}\dfrac{1}{8} & = & {\left(\dfrac{1}{2}\right)}^{n} \\ & ⇒ & {\left(\dfrac{1}{2}\right)}^{3} = {\left(\dfrac{1}{2}\right)}^{n} \\ & ⇒ & n = 3\end{cases}$

The half-life of the radioactive element can be calculated as follows:

$\dfrac{t}{{t}_{\dfrac{1}{2}}} = n . . . \left(2\right)$

Substitute the values of the known parameters into equation (2) and solve to calculate the half-life.

$\begin{cases}\dfrac{3days}{{t}_{\dfrac{1}{2}}} & = & 3 \\ & ⇒ & t\dfrac{1}{2} = 1day\end{cases}$

The initial and the final mass of the radioactive element are related by

$\dfrac{m}{{m}_{0}} = \left(\dfrac{1}{2}\right)\dfrac{t '}{{t}_{\dfrac{1}{2}}} . . . \left(3\right)$

Substitute the values of the known parameters into equation (3) and solve to calculate the initial mass of the substance.

$\begin{cases}\dfrac{8 \times {10}^{- 3}kg}{{m}_{0}} & = & \left(\dfrac{1}{2}\right)\dfrac{5}{1} \\ & ⇒ & {m}_{0} = {2}^{5} \times 8 \times {10}^{- 3}kg \\ & = & 256g\end{cases}$