# The half life of a radioactive element is 30 seconds. In what period of time would the activity of the sample be reduced to one - sixteenth of the original activity?

Feb 17, 2018

Working formula: $A = {A}_{0} \cdot {e}^{-} \left(k t\right)$

#### Explanation:

You can simplify the above working formula to the following form:

$k t = \ln \left({A}_{0} / A\right)$

where ${A}_{0}$ is the activity of the element at time $t = 0$ and $A$ is the activity at a given time.

Now, we know that the formula for half-life (which can be obtained from the above equation itself) is given by:

${t}_{\text{1/2}} = \frac{\ln 2}{k}$.

So, here the half-life is $30$ seconds. Using this information, we find the decay constant $k$.

$k = \ln \frac{2}{\text{30 s}}$

Now, using this value of $k$, we get the time at which the activity of the element becomes $\frac{1}{16}$ of original activity, i.e

$A = {A}_{0} / 16$

So

$t = \frac{1}{\ln \frac{2}{\text{30 s}}} \cdot \ln \left({A}_{0} / \left({A}_{0} / 16\right)\right)$

$t = \left(\frac{\text{30 s}}{\ln} 2\right) \cdot \ln \left({2}^{4}\right)$

$t = \left(\frac{\text{30 s}}{\ln} 2\right) \cdot 4 \ln 2$

$t = \text{30 s} \cdot 4$

$t = \text{120 s}$

Therefore, the activity of the radioactive element becomes one-sixteenth of its original activity at $120$ seconds.