Question

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?

Answer 1

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

Explanation:

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.