# If the half life of a radioactive substance is 4 days. Calculate the time required to decrease the concentration of 1/8th of original??

Apr 6, 2018

Half life of a radioactive substance is defined as the time taken for it to degrade to 1/2 of its original quantity.

I like looking at the problem this way

$1 \to \frac{1}{2} \to \frac{1}{4} \to \frac{1}{8}$

So it reaches $\frac{1}{8}$th of the original after $3$(counting the number of arrows) half lives.

The time required, therefore, to decrease the concentration of 1/8th of original would be

$= 3 \times 4 \text{ days}$

$= 12 \text{ days}$

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Now, this is the formula approach.

Let ${N}_{0}$ be the original quantity and $N$ be the new quantity then,

$\frac{N}{{N}_{0}} = \frac{1}{2} ^ \left(\frac{t}{T}\right)$

$\frac{N}{{N}_{0}} = \frac{1}{8}$ color(white)(ddddwwwwwwwwwdd $\left[\text{given that } N = {N}_{0} / 8\right]$

$\implies \frac{1}{8} = \frac{1}{2} ^ \left(\frac{t}{T}\right)$

$\implies {2}^{\frac{t}{T}} = 8 = {2}^{3}$

$\implies \frac{t}{T} = 3$

$\implies t = 3 \times 4 \text{ days"= 12 " days}$ color(white)(ddd ["given that " T=4 " days"]