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?

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Answer 1 · 2018-02-17T17:36:47

Working formula: $A=A_0 * e^-(kt)$

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

$kt=ln(A_0/A)$

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:

$t_("1/2")=(ln 2)/k$ .

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

$k=ln2/"30 s"$

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

$A=A_0/16$

So

$t=1/(ln 2/"30 s")*ln(A_0/(A_0/16))$

$t=("30 s"/ln2)*ln (2^4)$

$t=("30 s"/ln2)*4 ln2$

$t="30 s" * 4$

$t= "120 s"$

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