Question

The ratio activity of an element becomes \[\dfrac{1}{{64}}\,th\] of its original value in 60 sec. Then the half life period is
A. 5 Sec
B. 10 Sec
C. 20 Sec
D. 30 Sec

Answer 1

Hint: Activity of a radioactive sample is defined as the number of disintegration occurring per second in the sample and it is measured in Becquerel (Bq) or Curie (Ci). For a sample containing a number of elements, activity is equal to the sum of individual values of each element. Relation between initial \[({A_0})\]and final activity (A) is given by \[A = {A_0}{\left( {\dfrac{1}{2}} \right)^n}\].




Formula used:
\[A = {A_0}{\left( {\dfrac{1}{2}} \right)^n}\]and \[n = \dfrac{t}{T}\]
Here, A = Final activity of substance after time t, \[{A_0} = \]Initial activity of substance
          T = Half life and n = Number of half lives



Complete answer:
Given,
Ratio of activity \[\dfrac{A}{{{A_0}}} = \dfrac{1}{{64}}\] and t =60 sec
To calculate half life T
As we know that ratio of final and initial activity of a radioactive substance is given by,
\[A = {A_0}{\left( {\dfrac{1}{2}} \right)^n} \Rightarrow \dfrac{A}{{{A_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}\]
Substituting given values in above equation we get,

\[\dfrac{1}{{64}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{60}}{T}}}\]
Above equation can be rewritten as,
\[{\left( {\dfrac{1}{2}} \right)^6} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{60}}{T}}}\]
As the base in both side of equation is equal, therefore exponent on both sides will also be equal,
i.e. \[6 = \dfrac{{60}}{T} \Rightarrow T = 10\,\sec \]
Hence, the half life period of the given element will be 10 sec.
Therefore, option B is the correct option.




Note: Alternatively this problem can be solved by using the equation \[A = {A_0}{e^{ - \lambda t}}\], where\[\lambda \] is the decay constant. Initial activity of the element is directly proportional to final activity of the element which means that if initial activity increases final activity will also increase or decrease.