Question

An artificial element shows activity of \[12.5{\text{ cps}}\] (counts per second) at its formation as recorded in a G.M. counter. After \[{\text{10 min}}\] activity is \[7.5{\text{ cps}}\]. In the absence of element, G.M.counter records activity Of \[2.5{\text{ cps}}\] . What is its half life period ?
A. half-life \[ = {\text{ 10 min}}\]
B. half-life \[ = {\text{ 13 min}}\]
C. half-life \[ = {\text{ 20 min}}\]
D. None of these

Answer 1

Hint:1. Half life period is the time needed for the decay of one half of the radioactive element.
2. In one half-life period, the activity of G.M. the counter will reduce to one half.

Complete answer:
The initial activity is \[12.5{\text{ cps}}\] .
The activity in the absence of the element is \[2.5{\text{ cps}}\] .
From the initial activity, subtract the activity in the absence of the element to obtain the corrected value of the initial activity.
\[12.5{\text{ cps}} - 2.5{\text{ cps}} = 10{\text{ cps}}\]
The final activity is \[75{\text{ cps}}\] .
The activity in the absence of the element is \[2.5{\text{ cps}}\] .
From the final activity, subtract the activity in the absence of the element to obtain the corrected value of the final activity.
\[7.5{\text{ cps}} - 2.5{\text{ cps}} = 5{\text{ cps}}\]
Thus, in the time period of \[{\text{10 min}}\] the activity reduces to one half from \[10{\text{ cps}}\] to \[5{\text{ cps}}\]
Hence, the half-life period is \[{\text{10 min}}\] .

Hence, the correct option is the option A.

Note:
It is necessary to subtract the activity in the absence of element from the values of initial and final activities. This helps in the background correction. This ensures that only the activities from the element are accounted for and the background noise is eliminated.